How to convert a string to integer in C?

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218

I am trying to find out if there is an alternative way of converting string to integer in C.

I regularly pattern the following in my code.

char s = "45";



int num = atoi(s);

So, is there a better way or another way?

edited May 1 '18 at 13:10

Rodrigo de Azevedo

780517

asked Aug 11 '11 at 6:30

user618677

1,76551721

20

Your tags and title say you want a solution in C, but your question says C or C++. Which one do you want?

– In silico
Aug 11 '11 at 6:32

1

@Yann, Sorry for that confusion. I will prefer C .

– user618677
Aug 11 '11 at 16:16

1

It works, but it's not the recommended way, because there is no way to handle errors. Never use this in production code unless you can trust the input 100%.

– Uwe Geuder
Nov 5 '15 at 17:47

1

Define 'better', and state clearly why you need another way.

– user207421
Jun 18 '18 at 10:35

1

@EJP Just to improve myself.

– user618677
Aug 3 '18 at 1:32

|
show 1 more comment

218

I am trying to find out if there is an alternative way of converting string to integer in C.

I regularly pattern the following in my code.

char s = "45";



int num = atoi(s);

So, is there a better way or another way?

edited May 1 '18 at 13:10

Rodrigo de Azevedo

780517

asked Aug 11 '11 at 6:30

user618677

1,76551721

20

Your tags and title say you want a solution in C, but your question says C or C++. Which one do you want?

– In silico
Aug 11 '11 at 6:32

1

@Yann, Sorry for that confusion. I will prefer C .

– user618677
Aug 11 '11 at 16:16

1

It works, but it's not the recommended way, because there is no way to handle errors. Never use this in production code unless you can trust the input 100%.

– Uwe Geuder
Nov 5 '15 at 17:47

1

Define 'better', and state clearly why you need another way.

– user207421
Jun 18 '18 at 10:35

1

@EJP Just to improve myself.

– user618677
Aug 3 '18 at 1:32

|
show 1 more comment

218

I am trying to find out if there is an alternative way of converting string to integer in C.

I regularly pattern the following in my code.

char s = "45";



int num = atoi(s);

So, is there a better way or another way?

edited May 1 '18 at 13:10

Rodrigo de Azevedo

780517

asked Aug 11 '11 at 6:30

user618677

1,76551721

I am trying to find out if there is an alternative way of converting string to integer in C.

I regularly pattern the following in my code.

char s = "45";



int num = atoi(s);

So, is there a better way or another way?

c string atoi

edited May 1 '18 at 13:10

Rodrigo de Azevedo

780517

asked Aug 11 '11 at 6:30

user618677

1,76551721

edited May 1 '18 at 13:10

Rodrigo de Azevedo

780517

asked Aug 11 '11 at 6:30

user618677

1,76551721

edited May 1 '18 at 13:10

Rodrigo de Azevedo

780517

edited May 1 '18 at 13:10

Rodrigo de Azevedo

780517

edited May 1 '18 at 13:10

Rodrigo de Azevedo

780517

asked Aug 11 '11 at 6:30

user618677

1,76551721

asked Aug 11 '11 at 6:30

user618677

1,76551721

asked Aug 11 '11 at 6:30

user618677

1,76551721

20

Your tags and title say you want a solution in C, but your question says C or C++. Which one do you want?

– In silico
Aug 11 '11 at 6:32

1

@Yann, Sorry for that confusion. I will prefer C .

– user618677
Aug 11 '11 at 16:16

1

It works, but it's not the recommended way, because there is no way to handle errors. Never use this in production code unless you can trust the input 100%.

– Uwe Geuder
Nov 5 '15 at 17:47

1

Define 'better', and state clearly why you need another way.

– user207421
Jun 18 '18 at 10:35

1

@EJP Just to improve myself.

– user618677
Aug 3 '18 at 1:32

|
show 1 more comment

20

Your tags and title say you want a solution in C, but your question says C or C++. Which one do you want?

– In silico
Aug 11 '11 at 6:32

1

@Yann, Sorry for that confusion. I will prefer C .

– user618677
Aug 11 '11 at 16:16

1

It works, but it's not the recommended way, because there is no way to handle errors. Never use this in production code unless you can trust the input 100%.

– Uwe Geuder
Nov 5 '15 at 17:47

1

Define 'better', and state clearly why you need another way.

– user207421
Jun 18 '18 at 10:35

1

@EJP Just to improve myself.

– user618677
Aug 3 '18 at 1:32

Your tags and title say you want a solution in C, but your question says C or C++. Which one do you want?

– In silico
Aug 11 '11 at 6:32

@Yann, Sorry for that confusion. I will prefer C .

– user618677
Aug 11 '11 at 16:16

It works, but it's not the recommended way, because there is no way to handle errors. Never use this in production code unless you can trust the input 100%.

– Uwe Geuder
Nov 5 '15 at 17:47

Define 'better', and state clearly why you need another way.

– user207421
Jun 18 '18 at 10:35

@EJP Just to improve myself.

– user618677
Aug 3 '18 at 1:32

|
show 1 more comment

11 Answers
11

active

oldest

votes

158

There is strtol which is better IMO. Also I have taken a liking in strtonum, so use it if you have it (but remember it's not portable):

long long

     strtonum(const char *nptr, long long minval, long long maxval,

     const char **errstr);

EDIT

You might also be interested in strtoumax and strtoimax which are standard functions in C99. For example you could say:

uintmax_t num = strtoumax(s, NULL, 10);

if (num == UINTMAX_MAX && errno == ERANGE)

    /* Could not convert. */

Anyway, stay away from atoi:

The call atoi(str) shall be equivalent to:
(int) strtol(str, (char **)NULL, 10)
except that the handling of errors may differ. If the value cannot be
represented, the behavior is undefined.

edited Oct 11 '15 at 8:11

Spikatrix

17.7k62762

answered Aug 11 '11 at 6:32

cnicutar

147k19276329

what do I need to include for strtonum? I keep getting an implicit declaration warning

– jsj
Mar 30 '13 at 1:57

@trideceth12 On systems where it's available it should be declared in #<stdlib.h>. However, you could use the standard strtoumax alternative.

– cnicutar
Mar 30 '13 at 10:21

3

This answer doesn't seem shorter than the questioner's first code.

– Azurespot
Sep 3 '14 at 4:46

10

@NoniA. Conciseness is always good, but not at the expense of correctness.

– cnicutar
Sep 3 '14 at 9:33

5

Not so much wrong as unsafe. atoi() works if input is valid. But what if you do atoi("cat")? strtol() has defined behavior if the value cannot be represented as a long, atoi() does not.

– Daniel B.
Jul 22 '15 at 19:13

|
show 2 more comments

Robust C89 strtol-based solution

With:

no undefined behavior (as could be had with the atoi family)

a stricter definition of integer than strtol (e.g. no leading whitespace nor trailing trash chars)

classification of the error case (e.g. to give useful error messages to users)

a "testsuite"

#include <assert.h>

#include <ctype.h>

#include <errno.h>

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>



typedef enum {

    STR2INT_SUCCESS,

    STR2INT_OVERFLOW,

    STR2INT_UNDERFLOW,

    STR2INT_INCONVERTIBLE

} str2int_errno;



/* Convert string s to int out.

 *

 * @param[out] out The converted int. Cannot be NULL.

 *

 * @param[in] s Input string to be converted.

 *

 *     The format is the same as strtol,

 *     except that the following are inconvertible:

 *

 *     - empty string

 *     - leading whitespace

 *     - any trailing characters that are not part of the number

 *

 *     Cannot be NULL.

 *

 * @param[in] base Base to interpret string in. Same range as strtol (2 to 36).

 *

 * @return Indicates if the operation succeeded, or why it failed.

 */

str2int_errno str2int(int *out, char *s, int base) {

    char *end;

    if (s[0] == '' || isspace(s[0]))

        return STR2INT_INCONVERTIBLE;

    errno = 0;

    long l = strtol(s, &end, base);

    /* Both checks are needed because INT_MAX == LONG_MAX is possible. */

    if (l > INT_MAX || (errno == ERANGE && l == LONG_MAX))

        return STR2INT_OVERFLOW;

    if (l < INT_MIN || (errno == ERANGE && l == LONG_MIN))

        return STR2INT_UNDERFLOW;

    if (*end != '')

        return STR2INT_INCONVERTIBLE;

    *out = l;

    return STR2INT_SUCCESS;

}



int main(void) {

    int i;

    /* Lazy to calculate this size properly. */

    char s[256];



    /* Simple case. */

    assert(str2int(&i, "11", 10) == STR2INT_SUCCESS);

    assert(i == 11);



    /* Negative number . */

    assert(str2int(&i, "-11", 10) == STR2INT_SUCCESS);

    assert(i == -11);



    /* Different base. */

    assert(str2int(&i, "11", 16) == STR2INT_SUCCESS);

    assert(i == 17);



    /* 0 */

    assert(str2int(&i, "0", 10) == STR2INT_SUCCESS);

    assert(i == 0);



    /* INT_MAX. */

    sprintf(s, "%d", INT_MAX);

    assert(str2int(&i, s, 10) == STR2INT_SUCCESS);

    assert(i == INT_MAX);



    /* INT_MIN. */

    sprintf(s, "%d", INT_MIN);

    assert(str2int(&i, s, 10) == STR2INT_SUCCESS);

    assert(i == INT_MIN);



    /* Leading and trailing space. */

    assert(str2int(&i, " 1", 10) == STR2INT_INCONVERTIBLE);

    assert(str2int(&i, "1 ", 10) == STR2INT_INCONVERTIBLE);



    /* Trash characters. */

    assert(str2int(&i, "a10", 10) == STR2INT_INCONVERTIBLE);

    assert(str2int(&i, "10a", 10) == STR2INT_INCONVERTIBLE);



    /* int overflow.

     *

     * `if` needed to avoid undefined behaviour

     * on `INT_MAX + 1` if INT_MAX == LONG_MAX.

     */

    if (INT_MAX < LONG_MAX) {

        sprintf(s, "%ld", (long int)INT_MAX + 1L);

        assert(str2int(&i, s, 10) == STR2INT_OVERFLOW);

    }



    /* int underflow */

    if (LONG_MIN < INT_MIN) {

        sprintf(s, "%ld", (long int)INT_MIN - 1L);

        assert(str2int(&i, s, 10) == STR2INT_UNDERFLOW);

    }



    /* long overflow */

    sprintf(s, "%ld0", LONG_MAX);

    assert(str2int(&i, s, 10) == STR2INT_OVERFLOW);



    /* long underflow */

    sprintf(s, "%ld0", LONG_MIN);

    assert(str2int(&i, s, 10) == STR2INT_UNDERFLOW);



    return EXIT_SUCCESS;

}

GitHub upstream.

Based on: https://stackoverflow.com/a/6154614/895245

edited Sep 1 '18 at 9:53

answered Oct 16 '12 at 21:46

Ciro Santilli 新疆改造中心996ICU六四事件

149k34560478

3

Nice robust str2int(). Pedantic: use isspace((unsigned char) s[0]).

– chux
May 30 '16 at 12:09

@chux thanks! Can you explain a bit more why the (unsigned char) cast could make a difference?

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:00

1

@chux thanks! OMG, C is so hard :-) greenend.org.uk/rjk/tech/cfu.html

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:51

IAR C compiler warns that l > INT_MAX and l < INT_MIN are pointless integer comparison since either result is always false. What happens if I change them to l >= INT_MAX and l <= INT_MIN to clear the warnings? On ARM C, long and int are 32-bit signed Basic data types in ARM C and C++

– ecle
Jan 26 '17 at 2:23

@ecle changing code to l >= INT_MAX incurs incorrect functionality: Example returning STR2INT_OVERFLOW with input "32767" and 16-bit int. Use a conditional compile. Example.

– chux
Dec 20 '17 at 12:45

|
show 3 more comments

Don't use functions from ato... group. These are broken and virtually useless. A moderately better solution would be to use sscanf, although it is not perfect either.

To convert string to integer, functions from strto... group should be used. In your specific case it would be strtol function.

edited Aug 11 '11 at 22:09

answered Aug 11 '11 at 6:34

AnT

262k34424666

7

sscanf actually has undefined behavior if it tries to convert a number outside the range of its type (for example, sscanf("999999999999999999999", "%d", &n)).

– Keith Thompson
Aug 11 '11 at 6:50

1

@Keith Thompson: That's exactly what I mean. atoi provides no meaningful success/failure feedback and has undefined behavior on overflow. sscanf provides success/failure feedback of sorts (the return value, which is what makes it "moderately better"), but still has undefined behavior on overflow. Only strtol is a viable solution.

– AnT
Aug 11 '11 at 6:55

1

Agreed; I just wanted to emphasize the potentially fatal problem with sscanf. (Though I confess I sometimes use atoi, usually for programs that I don't expect to survive more than 10 minute before I delete the source.)

– Keith Thompson
Aug 11 '11 at 6:58

add a comment |

You can code a little atoi() for fun:

int my_getnbr(char *str)

{

  int result;

  int puiss;



  result = 0;

  puiss = 1;

  while (('-' == (*str)) || ((*str) == '+'))

  {

      if (*str == '-')

        puiss = puiss * -1;

      str++;

  }

  while ((*str >= '0') && (*str <= '9'))

  {

      result = (result * 10) + ((*str) - '0');

      str++;

  }

  return (result * puiss);

}

You can also make it recursive wich can old in 3 lines =)

edited Sep 1 '18 at 18:11

vishal dharankar

4,68443767

answered Aug 11 '11 at 8:20

jDourlens

1,57511529

Thanks so much .. But could you tell me how the below code works? code ((*str) - '0') code

– user618677
Aug 11 '11 at 16:15

a character has an ascii value. If you are uner linux type: man ascii in the shell or if not go to:table-ascii.com. You will see that the character '0' = 68 (i think) for a int. So to get the number of '9' (it's '0' + 9) so you get 9 = '9' - '0'. You get it?

– jDourlens
Aug 11 '11 at 16:31

1

1) The code allows "----1" 2) Has undefined behavior with int overflow when the result should be INT_MIN. Consider my_getnbr("-2147483648")

– chux
May 30 '16 at 12:12

Thanks for precision, it was just for showing a little example. As it's said it for fun and learning. You should definetly use standart lib for this kind of tasks. Faster and safer!

– jDourlens
Jun 1 '16 at 0:12

add a comment |

Just wanted to share a solution for unsigned long aswell.

unsigned long ToUInt(char* str)

{

    unsigned long mult = 1;

    unsigned long re = 0;

    int len = strlen(str);

    for(int i = len -1 ; i >= 0 ; i--)

    {

        re = re + ((int)str[i] -48)*mult;

        mult = mult*10;

    }

    return re;

}

answered Oct 18 '16 at 19:26

Jacob

1488

1

Doesn't handle overflow. Also, the parameter should be const char *.

– Roland Illig
Jan 21 '17 at 17:35

1

Plus, what's that 48 mean? Are you assuming that's the value of '0' where the code will run? Please don't inflict such broad assumptions on the world!

– Toby Speight
Jun 18 '18 at 16:15

@TobySpeight Yes I assume 48 represent '0' in the ascii table.

– Jacob
Jun 20 '18 at 9:17

1

Not all the world is ASCII - just use '0' like you should.

– Toby Speight
Jun 20 '18 at 10:42

add a comment |

This function will help you

int strtoint_n(char* str, int n)

{

    int sign = 1;

    int place = 1;

    int ret = 0;



    int i;

    for (i = n-1; i >= 0; i--, place *= 10)

    {

        int c = str[i];

        switch (c)

        {

            case '-':

                if (i == 0) sign = -1;

                else return -1;

                break;

            default:

                if (c >= '0' && c <= '9')   ret += (c - '0') * place;

                else return -1;

        }

    }



    return sign * ret;

}



int strtoint(char* str)

{

    char* temp = str;

    int n = 0;

    while (*temp != '')

    {

        n++;

        temp++;

    }

    return strtoint_n(str, n);

}

Ref: http://amscata.blogspot.com/2013/09/strnumstr-version-2.html

edited Aug 7 '15 at 16:35

MC93

762714

answered Sep 12 '13 at 11:31

Amith Chinthaka

5431122

1

Why do this though? One of the biggest problems with atoi and friends is that if there's overflow, it's undefined behavior. Your function does not check for this. strtol and friends do.

– chad
Sep 2 '15 at 19:45

1

Yup. Since C is not Python I hope that the people who use C language are aware of these kind of overflow errors. Everything has it's own limits.

– Amith Chinthaka
Sep 3 '15 at 9:04

add a comment |

Ok, I had the same problem.I came up with this solution.It worked for me the best.I did try atoi() but didn't work well for me.So here is my solution:

void splitInput(int arr, int sizeArr, char num)

{

    for(int i = 0; i < sizeArr; i++)

        // We are subtracting 48 because the numbers in ASCII starts at 48.

        arr[i] = (int)num[i] - 48;

}

answered Sep 5 '15 at 15:36

Khaled Mohammad

1347

add a comment |

//I think this way we could go :

int my_atoi(const char* snum)

{

 int nInt(0);

 int index(0);

 while(snum[index])

 {

    if(!nInt)

        nInt= ( (int) snum[index]) - 48;

    else

    {

        nInt = (nInt *= 10) + ((int) snum[index] - 48);

    }

    index++;

 }

 return(nInt);

}



int main()

{

    printf("Returned number is: %dn", my_atoi("676987"));

    return 0;

}

answered Nov 6 '15 at 13:05

Aditya Kumar

673

Code does not compile in C. Why nInt = (nInt *= 10) + ((int) snum[index] - 48); vs. nInt = nInt*10 + snum[index] - '0'; if(!nInt) not needed.

– chux
May 30 '16 at 12:19

add a comment |

-1

You can always roll your own!

#include <stdio.h>

#include <string.h>

#include <math.h>



int my_atoi(const char* snum)

{

    int idx, strIdx = 0, accum = 0, numIsNeg = 0;

    const unsigned int NUMLEN = (int)strlen(snum);



    /* Check if negative number and flag it. */

    if(snum[0] == 0x2d)

        numIsNeg = 1;



    for(idx = NUMLEN - 1; idx >= 0; idx--)

    {

        /* Only process numbers from 0 through 9. */

        if(snum[strIdx] >= 0x30 && snum[strIdx] <= 0x39)

            accum += (snum[strIdx] - 0x30) * pow(10, idx);



        strIdx++;

    }



    /* Check flag to see if originally passed -ve number and convert result if so. */

    if(!numIsNeg)

        return accum;

    else

        return accum * -1;

}



int main()

{

    /* Tests... */

    printf("Returned number is: %dn", my_atoi("34574"));

    printf("Returned number is: %dn", my_atoi("-23"));



    return 0;

}

This will do what you want without clutter.

answered Jan 3 '15 at 0:42

ButchDean

746

2

But... why? This doesn't check for overflow and simply ignores garbage values. There's no reason not to use the strto... family of functions. They are portable and significantly better.

– chad
Sep 2 '15 at 19:41

1

Strange to use 0x2d, 0x30 instead of '-', '0'. Does not allow '+' sign. Why (int) cast in (int)strlen(snum)? UB if input is "". UB when result is INT_MIN due to int overflow with accum += (snum[strIdx] - 0x30) * pow(10, idx);

– chux
May 30 '16 at 12:24

@chux - This code is demonstration code. There are easy fixes to what you described as potential issues.

– ButchDean
Jun 7 '16 at 4:12

1

@ButchDean What you describe as "demonstration code" will be used by others who have no clue about all the details. Only the negative score and the comments on this answer protect them now. In my opinion, "demonstration code" must have much higher quality.

– Roland Illig
Jan 21 '17 at 17:39

@RolandIllig Rather than being all critical, wouldn't it be more helpful to others to actually put up your own solution?

– ButchDean
Jan 24 '17 at 17:35

|
show 6 more comments

-3

In C++, you can use a such function:

template <typename T>

T to(const std::string & s)

{

    std::istringstream stm(s);

    T result;

    stm >> result;



    if(stm.tellg() != s.size())

        throw error;



    return result;

}

This can help you to convert any string to any type such as float, int, double...

answered Aug 11 '11 at 6:41

neodelphi

2,06411017

1

There's already a similar question covering C++, where the problems with this approach are explained.

– Ben Voigt
Feb 20 '15 at 17:28

add a comment |

-4

Yes, you can store the integer directly:

int num = 45;

If you must parse a string, atoi or strol is going to win the "shortest amount of code" contest.

answered Aug 11 '11 at 6:32

Yann Ramin

30.6k14776

If you want to do it safely, strtol() actually requires a fair amount of code. It can return LONG_MIN or LONG_MAX either if that's the actual converted value or if there's an underflow or overflow, and it can return 0 either if that's the actual value or if there was no number to convert. You need to set errno = 0 before the call, and check the endptr.

– Keith Thompson
Aug 11 '11 at 6:54

The solutions given to parse, are no viable solutions.

– BananaAcid
Feb 1 '18 at 21:15

add a comment |

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11 Answers
11

active

oldest

votes

11 Answers
11

active

oldest

votes

158

There is strtol which is better IMO. Also I have taken a liking in strtonum, so use it if you have it (but remember it's not portable):

long long

     strtonum(const char *nptr, long long minval, long long maxval,

     const char **errstr);

EDIT

You might also be interested in strtoumax and strtoimax which are standard functions in C99. For example you could say:

uintmax_t num = strtoumax(s, NULL, 10);

if (num == UINTMAX_MAX && errno == ERANGE)

    /* Could not convert. */

Anyway, stay away from atoi:

The call atoi(str) shall be equivalent to:
(int) strtol(str, (char **)NULL, 10)
except that the handling of errors may differ. If the value cannot be
represented, the behavior is undefined.

edited Oct 11 '15 at 8:11

Spikatrix

17.7k62762

answered Aug 11 '11 at 6:32

cnicutar

147k19276329

what do I need to include for strtonum? I keep getting an implicit declaration warning

– jsj
Mar 30 '13 at 1:57

@trideceth12 On systems where it's available it should be declared in #<stdlib.h>. However, you could use the standard strtoumax alternative.

– cnicutar
Mar 30 '13 at 10:21

3

This answer doesn't seem shorter than the questioner's first code.

– Azurespot
Sep 3 '14 at 4:46

10

@NoniA. Conciseness is always good, but not at the expense of correctness.

– cnicutar
Sep 3 '14 at 9:33

5

Not so much wrong as unsafe. atoi() works if input is valid. But what if you do atoi("cat")? strtol() has defined behavior if the value cannot be represented as a long, atoi() does not.

– Daniel B.
Jul 22 '15 at 19:13

|
show 2 more comments

158

There is strtol which is better IMO. Also I have taken a liking in strtonum, so use it if you have it (but remember it's not portable):

long long

     strtonum(const char *nptr, long long minval, long long maxval,

     const char **errstr);

EDIT

You might also be interested in strtoumax and strtoimax which are standard functions in C99. For example you could say:

uintmax_t num = strtoumax(s, NULL, 10);

if (num == UINTMAX_MAX && errno == ERANGE)

    /* Could not convert. */

Anyway, stay away from atoi:

The call atoi(str) shall be equivalent to:
(int) strtol(str, (char **)NULL, 10)
except that the handling of errors may differ. If the value cannot be
represented, the behavior is undefined.

edited Oct 11 '15 at 8:11

Spikatrix

17.7k62762

answered Aug 11 '11 at 6:32

cnicutar

147k19276329

what do I need to include for strtonum? I keep getting an implicit declaration warning

– jsj
Mar 30 '13 at 1:57

@trideceth12 On systems where it's available it should be declared in #<stdlib.h>. However, you could use the standard strtoumax alternative.

– cnicutar
Mar 30 '13 at 10:21

3

This answer doesn't seem shorter than the questioner's first code.

– Azurespot
Sep 3 '14 at 4:46

10

@NoniA. Conciseness is always good, but not at the expense of correctness.

– cnicutar
Sep 3 '14 at 9:33

5

Not so much wrong as unsafe. atoi() works if input is valid. But what if you do atoi("cat")? strtol() has defined behavior if the value cannot be represented as a long, atoi() does not.

– Daniel B.
Jul 22 '15 at 19:13

|
show 2 more comments

158

There is strtol which is better IMO. Also I have taken a liking in strtonum, so use it if you have it (but remember it's not portable):

long long

     strtonum(const char *nptr, long long minval, long long maxval,

     const char **errstr);

EDIT

You might also be interested in strtoumax and strtoimax which are standard functions in C99. For example you could say:

uintmax_t num = strtoumax(s, NULL, 10);

if (num == UINTMAX_MAX && errno == ERANGE)

    /* Could not convert. */

Anyway, stay away from atoi:

The call atoi(str) shall be equivalent to:
(int) strtol(str, (char **)NULL, 10)
except that the handling of errors may differ. If the value cannot be
represented, the behavior is undefined.

edited Oct 11 '15 at 8:11

Spikatrix

17.7k62762

answered Aug 11 '11 at 6:32

cnicutar

147k19276329

There is strtol which is better IMO. Also I have taken a liking in strtonum, so use it if you have it (but remember it's not portable):

long long

     strtonum(const char *nptr, long long minval, long long maxval,

     const char **errstr);

EDIT

You might also be interested in strtoumax and strtoimax which are standard functions in C99. For example you could say:

uintmax_t num = strtoumax(s, NULL, 10);

if (num == UINTMAX_MAX && errno == ERANGE)

    /* Could not convert. */

Anyway, stay away from atoi:

The call atoi(str) shall be equivalent to:
(int) strtol(str, (char **)NULL, 10)
except that the handling of errors may differ. If the value cannot be
represented, the behavior is undefined.

edited Oct 11 '15 at 8:11

Spikatrix

17.7k62762

answered Aug 11 '11 at 6:32

cnicutar

147k19276329

edited Oct 11 '15 at 8:11

Spikatrix

17.7k62762

edited Oct 11 '15 at 8:11

Spikatrix

17.7k62762

edited Oct 11 '15 at 8:11

Spikatrix

17.7k62762

answered Aug 11 '11 at 6:32

cnicutar

147k19276329

answered Aug 11 '11 at 6:32

cnicutar

147k19276329

answered Aug 11 '11 at 6:32

cnicutar

147k19276329

what do I need to include for strtonum? I keep getting an implicit declaration warning

– jsj
Mar 30 '13 at 1:57

@trideceth12 On systems where it's available it should be declared in #<stdlib.h>. However, you could use the standard strtoumax alternative.

– cnicutar
Mar 30 '13 at 10:21

3

This answer doesn't seem shorter than the questioner's first code.

– Azurespot
Sep 3 '14 at 4:46

10

@NoniA. Conciseness is always good, but not at the expense of correctness.

– cnicutar
Sep 3 '14 at 9:33

5

Not so much wrong as unsafe. atoi() works if input is valid. But what if you do atoi("cat")? strtol() has defined behavior if the value cannot be represented as a long, atoi() does not.

– Daniel B.
Jul 22 '15 at 19:13

|
show 2 more comments

what do I need to include for strtonum? I keep getting an implicit declaration warning

– jsj
Mar 30 '13 at 1:57

@trideceth12 On systems where it's available it should be declared in #<stdlib.h>. However, you could use the standard strtoumax alternative.

– cnicutar
Mar 30 '13 at 10:21

3

This answer doesn't seem shorter than the questioner's first code.

– Azurespot
Sep 3 '14 at 4:46

10

@NoniA. Conciseness is always good, but not at the expense of correctness.

– cnicutar
Sep 3 '14 at 9:33

5

Not so much wrong as unsafe. atoi() works if input is valid. But what if you do atoi("cat")? strtol() has defined behavior if the value cannot be represented as a long, atoi() does not.

– Daniel B.
Jul 22 '15 at 19:13

what do I need to include for strtonum? I keep getting an implicit declaration warning

– jsj
Mar 30 '13 at 1:57

@trideceth12 On systems where it's available it should be declared in #<stdlib.h>. However, you could use the standard strtoumax alternative.

– cnicutar
Mar 30 '13 at 10:21

This answer doesn't seem shorter than the questioner's first code.

– Azurespot
Sep 3 '14 at 4:46

@NoniA. Conciseness is always good, but not at the expense of correctness.

– cnicutar
Sep 3 '14 at 9:33

Not so much wrong as unsafe. atoi() works if input is valid. But what if you do atoi("cat")? strtol() has defined behavior if the value cannot be represented as a long, atoi() does not.

– Daniel B.
Jul 22 '15 at 19:13

|
show 2 more comments

Robust C89 strtol-based solution

With:

no undefined behavior (as could be had with the atoi family)

a stricter definition of integer than strtol (e.g. no leading whitespace nor trailing trash chars)

classification of the error case (e.g. to give useful error messages to users)

a "testsuite"

#include <assert.h>

#include <ctype.h>

#include <errno.h>

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>



typedef enum {

    STR2INT_SUCCESS,

    STR2INT_OVERFLOW,

    STR2INT_UNDERFLOW,

    STR2INT_INCONVERTIBLE

} str2int_errno;



/* Convert string s to int out.

 *

 * @param[out] out The converted int. Cannot be NULL.

 *

 * @param[in] s Input string to be converted.

 *

 *     The format is the same as strtol,

 *     except that the following are inconvertible:

 *

 *     - empty string

 *     - leading whitespace

 *     - any trailing characters that are not part of the number

 *

 *     Cannot be NULL.

 *

 * @param[in] base Base to interpret string in. Same range as strtol (2 to 36).

 *

 * @return Indicates if the operation succeeded, or why it failed.

 */

str2int_errno str2int(int *out, char *s, int base) {

    char *end;

    if (s[0] == '' || isspace(s[0]))

        return STR2INT_INCONVERTIBLE;

    errno = 0;

    long l = strtol(s, &end, base);

    /* Both checks are needed because INT_MAX == LONG_MAX is possible. */

    if (l > INT_MAX || (errno == ERANGE && l == LONG_MAX))

        return STR2INT_OVERFLOW;

    if (l < INT_MIN || (errno == ERANGE && l == LONG_MIN))

        return STR2INT_UNDERFLOW;

    if (*end != '')

        return STR2INT_INCONVERTIBLE;

    *out = l;

    return STR2INT_SUCCESS;

}



int main(void) {

    int i;

    /* Lazy to calculate this size properly. */

    char s[256];



    /* Simple case. */

    assert(str2int(&i, "11", 10) == STR2INT_SUCCESS);

    assert(i == 11);



    /* Negative number . */

    assert(str2int(&i, "-11", 10) == STR2INT_SUCCESS);

    assert(i == -11);



    /* Different base. */

    assert(str2int(&i, "11", 16) == STR2INT_SUCCESS);

    assert(i == 17);



    /* 0 */

    assert(str2int(&i, "0", 10) == STR2INT_SUCCESS);

    assert(i == 0);



    /* INT_MAX. */

    sprintf(s, "%d", INT_MAX);

    assert(str2int(&i, s, 10) == STR2INT_SUCCESS);

    assert(i == INT_MAX);



    /* INT_MIN. */

    sprintf(s, "%d", INT_MIN);

    assert(str2int(&i, s, 10) == STR2INT_SUCCESS);

    assert(i == INT_MIN);



    /* Leading and trailing space. */

    assert(str2int(&i, " 1", 10) == STR2INT_INCONVERTIBLE);

    assert(str2int(&i, "1 ", 10) == STR2INT_INCONVERTIBLE);



    /* Trash characters. */

    assert(str2int(&i, "a10", 10) == STR2INT_INCONVERTIBLE);

    assert(str2int(&i, "10a", 10) == STR2INT_INCONVERTIBLE);



    /* int overflow.

     *

     * `if` needed to avoid undefined behaviour

     * on `INT_MAX + 1` if INT_MAX == LONG_MAX.

     */

    if (INT_MAX < LONG_MAX) {

        sprintf(s, "%ld", (long int)INT_MAX + 1L);

        assert(str2int(&i, s, 10) == STR2INT_OVERFLOW);

    }



    /* int underflow */

    if (LONG_MIN < INT_MIN) {

        sprintf(s, "%ld", (long int)INT_MIN - 1L);

        assert(str2int(&i, s, 10) == STR2INT_UNDERFLOW);

    }



    /* long overflow */

    sprintf(s, "%ld0", LONG_MAX);

    assert(str2int(&i, s, 10) == STR2INT_OVERFLOW);



    /* long underflow */

    sprintf(s, "%ld0", LONG_MIN);

    assert(str2int(&i, s, 10) == STR2INT_UNDERFLOW);



    return EXIT_SUCCESS;

}

GitHub upstream.

Based on: https://stackoverflow.com/a/6154614/895245

edited Sep 1 '18 at 9:53

answered Oct 16 '12 at 21:46

Ciro Santilli 新疆改造中心996ICU六四事件

149k34560478

3

Nice robust str2int(). Pedantic: use isspace((unsigned char) s[0]).

– chux
May 30 '16 at 12:09

@chux thanks! Can you explain a bit more why the (unsigned char) cast could make a difference?

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:00

1

@chux thanks! OMG, C is so hard :-) greenend.org.uk/rjk/tech/cfu.html

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:51

IAR C compiler warns that l > INT_MAX and l < INT_MIN are pointless integer comparison since either result is always false. What happens if I change them to l >= INT_MAX and l <= INT_MIN to clear the warnings? On ARM C, long and int are 32-bit signed Basic data types in ARM C and C++

– ecle
Jan 26 '17 at 2:23

@ecle changing code to l >= INT_MAX incurs incorrect functionality: Example returning STR2INT_OVERFLOW with input "32767" and 16-bit int. Use a conditional compile. Example.

– chux
Dec 20 '17 at 12:45

|
show 3 more comments

Robust C89 strtol-based solution

With:

no undefined behavior (as could be had with the atoi family)

a stricter definition of integer than strtol (e.g. no leading whitespace nor trailing trash chars)

classification of the error case (e.g. to give useful error messages to users)

a "testsuite"

#include <assert.h>

#include <ctype.h>

#include <errno.h>

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>



typedef enum {

    STR2INT_SUCCESS,

    STR2INT_OVERFLOW,

    STR2INT_UNDERFLOW,

    STR2INT_INCONVERTIBLE

} str2int_errno;



/* Convert string s to int out.

 *

 * @param[out] out The converted int. Cannot be NULL.

 *

 * @param[in] s Input string to be converted.

 *

 *     The format is the same as strtol,

 *     except that the following are inconvertible:

 *

 *     - empty string

 *     - leading whitespace

 *     - any trailing characters that are not part of the number

 *

 *     Cannot be NULL.

 *

 * @param[in] base Base to interpret string in. Same range as strtol (2 to 36).

 *

 * @return Indicates if the operation succeeded, or why it failed.

 */

str2int_errno str2int(int *out, char *s, int base) {

    char *end;

    if (s[0] == '' || isspace(s[0]))

        return STR2INT_INCONVERTIBLE;

    errno = 0;

    long l = strtol(s, &end, base);

    /* Both checks are needed because INT_MAX == LONG_MAX is possible. */

    if (l > INT_MAX || (errno == ERANGE && l == LONG_MAX))

        return STR2INT_OVERFLOW;

    if (l < INT_MIN || (errno == ERANGE && l == LONG_MIN))

        return STR2INT_UNDERFLOW;

    if (*end != '')

        return STR2INT_INCONVERTIBLE;

    *out = l;

    return STR2INT_SUCCESS;

}



int main(void) {

    int i;

    /* Lazy to calculate this size properly. */

    char s[256];



    /* Simple case. */

    assert(str2int(&i, "11", 10) == STR2INT_SUCCESS);

    assert(i == 11);



    /* Negative number . */

    assert(str2int(&i, "-11", 10) == STR2INT_SUCCESS);

    assert(i == -11);



    /* Different base. */

    assert(str2int(&i, "11", 16) == STR2INT_SUCCESS);

    assert(i == 17);



    /* 0 */

    assert(str2int(&i, "0", 10) == STR2INT_SUCCESS);

    assert(i == 0);



    /* INT_MAX. */

    sprintf(s, "%d", INT_MAX);

    assert(str2int(&i, s, 10) == STR2INT_SUCCESS);

    assert(i == INT_MAX);



    /* INT_MIN. */

    sprintf(s, "%d", INT_MIN);

    assert(str2int(&i, s, 10) == STR2INT_SUCCESS);

    assert(i == INT_MIN);



    /* Leading and trailing space. */

    assert(str2int(&i, " 1", 10) == STR2INT_INCONVERTIBLE);

    assert(str2int(&i, "1 ", 10) == STR2INT_INCONVERTIBLE);



    /* Trash characters. */

    assert(str2int(&i, "a10", 10) == STR2INT_INCONVERTIBLE);

    assert(str2int(&i, "10a", 10) == STR2INT_INCONVERTIBLE);



    /* int overflow.

     *

     * `if` needed to avoid undefined behaviour

     * on `INT_MAX + 1` if INT_MAX == LONG_MAX.

     */

    if (INT_MAX < LONG_MAX) {

        sprintf(s, "%ld", (long int)INT_MAX + 1L);

        assert(str2int(&i, s, 10) == STR2INT_OVERFLOW);

    }



    /* int underflow */

    if (LONG_MIN < INT_MIN) {

        sprintf(s, "%ld", (long int)INT_MIN - 1L);

        assert(str2int(&i, s, 10) == STR2INT_UNDERFLOW);

    }



    /* long overflow */

    sprintf(s, "%ld0", LONG_MAX);

    assert(str2int(&i, s, 10) == STR2INT_OVERFLOW);



    /* long underflow */

    sprintf(s, "%ld0", LONG_MIN);

    assert(str2int(&i, s, 10) == STR2INT_UNDERFLOW);



    return EXIT_SUCCESS;

}

GitHub upstream.

Based on: https://stackoverflow.com/a/6154614/895245

edited Sep 1 '18 at 9:53

answered Oct 16 '12 at 21:46

Ciro Santilli 新疆改造中心996ICU六四事件

149k34560478

3

Nice robust str2int(). Pedantic: use isspace((unsigned char) s[0]).

– chux
May 30 '16 at 12:09

@chux thanks! Can you explain a bit more why the (unsigned char) cast could make a difference?

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:00

1

@chux thanks! OMG, C is so hard :-) greenend.org.uk/rjk/tech/cfu.html

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:51

IAR C compiler warns that l > INT_MAX and l < INT_MIN are pointless integer comparison since either result is always false. What happens if I change them to l >= INT_MAX and l <= INT_MIN to clear the warnings? On ARM C, long and int are 32-bit signed Basic data types in ARM C and C++

– ecle
Jan 26 '17 at 2:23

@ecle changing code to l >= INT_MAX incurs incorrect functionality: Example returning STR2INT_OVERFLOW with input "32767" and 16-bit int. Use a conditional compile. Example.

– chux
Dec 20 '17 at 12:45

|
show 3 more comments

Robust C89 strtol-based solution

With:

no undefined behavior (as could be had with the atoi family)

a stricter definition of integer than strtol (e.g. no leading whitespace nor trailing trash chars)

classification of the error case (e.g. to give useful error messages to users)

a "testsuite"

#include <assert.h>

#include <ctype.h>

#include <errno.h>

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>



typedef enum {

    STR2INT_SUCCESS,

    STR2INT_OVERFLOW,

    STR2INT_UNDERFLOW,

    STR2INT_INCONVERTIBLE

} str2int_errno;



/* Convert string s to int out.

 *

 * @param[out] out The converted int. Cannot be NULL.

 *

 * @param[in] s Input string to be converted.

 *

 *     The format is the same as strtol,

 *     except that the following are inconvertible:

 *

 *     - empty string

 *     - leading whitespace

 *     - any trailing characters that are not part of the number

 *

 *     Cannot be NULL.

 *

 * @param[in] base Base to interpret string in. Same range as strtol (2 to 36).

 *

 * @return Indicates if the operation succeeded, or why it failed.

 */

str2int_errno str2int(int *out, char *s, int base) {

    char *end;

    if (s[0] == '' || isspace(s[0]))

        return STR2INT_INCONVERTIBLE;

    errno = 0;

    long l = strtol(s, &end, base);

    /* Both checks are needed because INT_MAX == LONG_MAX is possible. */

    if (l > INT_MAX || (errno == ERANGE && l == LONG_MAX))

        return STR2INT_OVERFLOW;

    if (l < INT_MIN || (errno == ERANGE && l == LONG_MIN))

        return STR2INT_UNDERFLOW;

    if (*end != '')

        return STR2INT_INCONVERTIBLE;

    *out = l;

    return STR2INT_SUCCESS;

}



int main(void) {

    int i;

    /* Lazy to calculate this size properly. */

    char s[256];



    /* Simple case. */

    assert(str2int(&i, "11", 10) == STR2INT_SUCCESS);

    assert(i == 11);



    /* Negative number . */

    assert(str2int(&i, "-11", 10) == STR2INT_SUCCESS);

    assert(i == -11);



    /* Different base. */

    assert(str2int(&i, "11", 16) == STR2INT_SUCCESS);

    assert(i == 17);



    /* 0 */

    assert(str2int(&i, "0", 10) == STR2INT_SUCCESS);

    assert(i == 0);



    /* INT_MAX. */

    sprintf(s, "%d", INT_MAX);

    assert(str2int(&i, s, 10) == STR2INT_SUCCESS);

    assert(i == INT_MAX);



    /* INT_MIN. */

    sprintf(s, "%d", INT_MIN);

    assert(str2int(&i, s, 10) == STR2INT_SUCCESS);

    assert(i == INT_MIN);



    /* Leading and trailing space. */

    assert(str2int(&i, " 1", 10) == STR2INT_INCONVERTIBLE);

    assert(str2int(&i, "1 ", 10) == STR2INT_INCONVERTIBLE);



    /* Trash characters. */

    assert(str2int(&i, "a10", 10) == STR2INT_INCONVERTIBLE);

    assert(str2int(&i, "10a", 10) == STR2INT_INCONVERTIBLE);



    /* int overflow.

     *

     * `if` needed to avoid undefined behaviour

     * on `INT_MAX + 1` if INT_MAX == LONG_MAX.

     */

    if (INT_MAX < LONG_MAX) {

        sprintf(s, "%ld", (long int)INT_MAX + 1L);

        assert(str2int(&i, s, 10) == STR2INT_OVERFLOW);

    }



    /* int underflow */

    if (LONG_MIN < INT_MIN) {

        sprintf(s, "%ld", (long int)INT_MIN - 1L);

        assert(str2int(&i, s, 10) == STR2INT_UNDERFLOW);

    }



    /* long overflow */

    sprintf(s, "%ld0", LONG_MAX);

    assert(str2int(&i, s, 10) == STR2INT_OVERFLOW);



    /* long underflow */

    sprintf(s, "%ld0", LONG_MIN);

    assert(str2int(&i, s, 10) == STR2INT_UNDERFLOW);



    return EXIT_SUCCESS;

}

GitHub upstream.

Based on: https://stackoverflow.com/a/6154614/895245

edited Sep 1 '18 at 9:53

answered Oct 16 '12 at 21:46

Ciro Santilli 新疆改造中心996ICU六四事件

149k34560478

Robust C89 strtol-based solution

With:

no undefined behavior (as could be had with the atoi family)

a stricter definition of integer than strtol (e.g. no leading whitespace nor trailing trash chars)

classification of the error case (e.g. to give useful error messages to users)

a "testsuite"

#include <assert.h>

#include <ctype.h>

#include <errno.h>

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>



typedef enum {

    STR2INT_SUCCESS,

    STR2INT_OVERFLOW,

    STR2INT_UNDERFLOW,

    STR2INT_INCONVERTIBLE

} str2int_errno;



/* Convert string s to int out.

 *

 * @param[out] out The converted int. Cannot be NULL.

 *

 * @param[in] s Input string to be converted.

 *

 *     The format is the same as strtol,

 *     except that the following are inconvertible:

 *

 *     - empty string

 *     - leading whitespace

 *     - any trailing characters that are not part of the number

 *

 *     Cannot be NULL.

 *

 * @param[in] base Base to interpret string in. Same range as strtol (2 to 36).

 *

 * @return Indicates if the operation succeeded, or why it failed.

 */

str2int_errno str2int(int *out, char *s, int base) {

    char *end;

    if (s[0] == '' || isspace(s[0]))

        return STR2INT_INCONVERTIBLE;

    errno = 0;

    long l = strtol(s, &end, base);

    /* Both checks are needed because INT_MAX == LONG_MAX is possible. */

    if (l > INT_MAX || (errno == ERANGE && l == LONG_MAX))

        return STR2INT_OVERFLOW;

    if (l < INT_MIN || (errno == ERANGE && l == LONG_MIN))

        return STR2INT_UNDERFLOW;

    if (*end != '')

        return STR2INT_INCONVERTIBLE;

    *out = l;

    return STR2INT_SUCCESS;

}



int main(void) {

    int i;

    /* Lazy to calculate this size properly. */

    char s[256];



    /* Simple case. */

    assert(str2int(&i, "11", 10) == STR2INT_SUCCESS);

    assert(i == 11);



    /* Negative number . */

    assert(str2int(&i, "-11", 10) == STR2INT_SUCCESS);

    assert(i == -11);



    /* Different base. */

    assert(str2int(&i, "11", 16) == STR2INT_SUCCESS);

    assert(i == 17);



    /* 0 */

    assert(str2int(&i, "0", 10) == STR2INT_SUCCESS);

    assert(i == 0);



    /* INT_MAX. */

    sprintf(s, "%d", INT_MAX);

    assert(str2int(&i, s, 10) == STR2INT_SUCCESS);

    assert(i == INT_MAX);



    /* INT_MIN. */

    sprintf(s, "%d", INT_MIN);

    assert(str2int(&i, s, 10) == STR2INT_SUCCESS);

    assert(i == INT_MIN);



    /* Leading and trailing space. */

    assert(str2int(&i, " 1", 10) == STR2INT_INCONVERTIBLE);

    assert(str2int(&i, "1 ", 10) == STR2INT_INCONVERTIBLE);



    /* Trash characters. */

    assert(str2int(&i, "a10", 10) == STR2INT_INCONVERTIBLE);

    assert(str2int(&i, "10a", 10) == STR2INT_INCONVERTIBLE);



    /* int overflow.

     *

     * `if` needed to avoid undefined behaviour

     * on `INT_MAX + 1` if INT_MAX == LONG_MAX.

     */

    if (INT_MAX < LONG_MAX) {

        sprintf(s, "%ld", (long int)INT_MAX + 1L);

        assert(str2int(&i, s, 10) == STR2INT_OVERFLOW);

    }



    /* int underflow */

    if (LONG_MIN < INT_MIN) {

        sprintf(s, "%ld", (long int)INT_MIN - 1L);

        assert(str2int(&i, s, 10) == STR2INT_UNDERFLOW);

    }



    /* long overflow */

    sprintf(s, "%ld0", LONG_MAX);

    assert(str2int(&i, s, 10) == STR2INT_OVERFLOW);



    /* long underflow */

    sprintf(s, "%ld0", LONG_MIN);

    assert(str2int(&i, s, 10) == STR2INT_UNDERFLOW);



    return EXIT_SUCCESS;

}

GitHub upstream.

Based on: https://stackoverflow.com/a/6154614/895245

edited Sep 1 '18 at 9:53

answered Oct 16 '12 at 21:46

Ciro Santilli 新疆改造中心996ICU六四事件

149k34560478

edited Sep 1 '18 at 9:53

answered Oct 16 '12 at 21:46

Ciro Santilli 新疆改造中心996ICU六四事件

149k34560478

answered Oct 16 '12 at 21:46

Ciro Santilli 新疆改造中心996ICU六四事件

149k34560478

answered Oct 16 '12 at 21:46

Ciro Santilli 新疆改造中心996ICU六四事件

149k34560478

3

Nice robust str2int(). Pedantic: use isspace((unsigned char) s[0]).

– chux
May 30 '16 at 12:09

@chux thanks! Can you explain a bit more why the (unsigned char) cast could make a difference?

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:00

1

@chux thanks! OMG, C is so hard :-) greenend.org.uk/rjk/tech/cfu.html

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:51

IAR C compiler warns that l > INT_MAX and l < INT_MIN are pointless integer comparison since either result is always false. What happens if I change them to l >= INT_MAX and l <= INT_MIN to clear the warnings? On ARM C, long and int are 32-bit signed Basic data types in ARM C and C++

– ecle
Jan 26 '17 at 2:23

@ecle changing code to l >= INT_MAX incurs incorrect functionality: Example returning STR2INT_OVERFLOW with input "32767" and 16-bit int. Use a conditional compile. Example.

– chux
Dec 20 '17 at 12:45

|
show 3 more comments

3

Nice robust str2int(). Pedantic: use isspace((unsigned char) s[0]).

– chux
May 30 '16 at 12:09

@chux thanks! Can you explain a bit more why the (unsigned char) cast could make a difference?

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:00

1

@chux thanks! OMG, C is so hard :-) greenend.org.uk/rjk/tech/cfu.html

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:51

IAR C compiler warns that l > INT_MAX and l < INT_MIN are pointless integer comparison since either result is always false. What happens if I change them to l >= INT_MAX and l <= INT_MIN to clear the warnings? On ARM C, long and int are 32-bit signed Basic data types in ARM C and C++

– ecle
Jan 26 '17 at 2:23

@ecle changing code to l >= INT_MAX incurs incorrect functionality: Example returning STR2INT_OVERFLOW with input "32767" and 16-bit int. Use a conditional compile. Example.

– chux
Dec 20 '17 at 12:45

Nice robust str2int(). Pedantic: use isspace((unsigned char) s[0]).

– chux
May 30 '16 at 12:09

@chux thanks! Can you explain a bit more why the (unsigned char) cast could make a difference?

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:00

@chux thanks! OMG, C is so hard :-) greenend.org.uk/rjk/tech/cfu.html

– Ciro Santilli 新疆改造中心996ICU六四事件
May 30 '16 at 13:51

IAR C compiler warns that l > INT_MAX and l < INT_MIN are pointless integer comparison since either result is always false. What happens if I change them to l >= INT_MAX and l <= INT_MIN to clear the warnings? On ARM C, long and int are 32-bit signed Basic data types in ARM C and C++

– ecle
Jan 26 '17 at 2:23

@ecle changing code to l >= INT_MAX incurs incorrect functionality: Example returning STR2INT_OVERFLOW with input "32767" and 16-bit int. Use a conditional compile. Example.

– chux
Dec 20 '17 at 12:45

|
show 3 more comments

Don't use functions from ato... group. These are broken and virtually useless. A moderately better solution would be to use sscanf, although it is not perfect either.

To convert string to integer, functions from strto... group should be used. In your specific case it would be strtol function.

edited Aug 11 '11 at 22:09

answered Aug 11 '11 at 6:34

AnT

262k34424666

7

sscanf actually has undefined behavior if it tries to convert a number outside the range of its type (for example, sscanf("999999999999999999999", "%d", &n)).

– Keith Thompson
Aug 11 '11 at 6:50

1

@Keith Thompson: That's exactly what I mean. atoi provides no meaningful success/failure feedback and has undefined behavior on overflow. sscanf provides success/failure feedback of sorts (the return value, which is what makes it "moderately better"), but still has undefined behavior on overflow. Only strtol is a viable solution.

– AnT
Aug 11 '11 at 6:55

1

Agreed; I just wanted to emphasize the potentially fatal problem with sscanf. (Though I confess I sometimes use atoi, usually for programs that I don't expect to survive more than 10 minute before I delete the source.)

– Keith Thompson
Aug 11 '11 at 6:58

add a comment |

Don't use functions from ato... group. These are broken and virtually useless. A moderately better solution would be to use sscanf, although it is not perfect either.

To convert string to integer, functions from strto... group should be used. In your specific case it would be strtol function.

edited Aug 11 '11 at 22:09

answered Aug 11 '11 at 6:34

AnT

262k34424666

7

sscanf actually has undefined behavior if it tries to convert a number outside the range of its type (for example, sscanf("999999999999999999999", "%d", &n)).

– Keith Thompson
Aug 11 '11 at 6:50

1

@Keith Thompson: That's exactly what I mean. atoi provides no meaningful success/failure feedback and has undefined behavior on overflow. sscanf provides success/failure feedback of sorts (the return value, which is what makes it "moderately better"), but still has undefined behavior on overflow. Only strtol is a viable solution.

– AnT
Aug 11 '11 at 6:55

1

Agreed; I just wanted to emphasize the potentially fatal problem with sscanf. (Though I confess I sometimes use atoi, usually for programs that I don't expect to survive more than 10 minute before I delete the source.)

– Keith Thompson
Aug 11 '11 at 6:58

add a comment |

Don't use functions from ato... group. These are broken and virtually useless. A moderately better solution would be to use sscanf, although it is not perfect either.

To convert string to integer, functions from strto... group should be used. In your specific case it would be strtol function.

edited Aug 11 '11 at 22:09

answered Aug 11 '11 at 6:34

AnT

262k34424666

Don't use functions from ato... group. These are broken and virtually useless. A moderately better solution would be to use sscanf, although it is not perfect either.

To convert string to integer, functions from strto... group should be used. In your specific case it would be strtol function.

edited Aug 11 '11 at 22:09

answered Aug 11 '11 at 6:34

AnT

262k34424666

edited Aug 11 '11 at 22:09

answered Aug 11 '11 at 6:34

AnT

262k34424666

answered Aug 11 '11 at 6:34

AnT

262k34424666

answered Aug 11 '11 at 6:34

AnT

262k34424666

7

sscanf actually has undefined behavior if it tries to convert a number outside the range of its type (for example, sscanf("999999999999999999999", "%d", &n)).

– Keith Thompson
Aug 11 '11 at 6:50

1

@Keith Thompson: That's exactly what I mean. atoi provides no meaningful success/failure feedback and has undefined behavior on overflow. sscanf provides success/failure feedback of sorts (the return value, which is what makes it "moderately better"), but still has undefined behavior on overflow. Only strtol is a viable solution.

– AnT
Aug 11 '11 at 6:55

1

Agreed; I just wanted to emphasize the potentially fatal problem with sscanf. (Though I confess I sometimes use atoi, usually for programs that I don't expect to survive more than 10 minute before I delete the source.)

– Keith Thompson
Aug 11 '11 at 6:58

add a comment |

7

sscanf actually has undefined behavior if it tries to convert a number outside the range of its type (for example, sscanf("999999999999999999999", "%d", &n)).

– Keith Thompson
Aug 11 '11 at 6:50

1

@Keith Thompson: That's exactly what I mean. atoi provides no meaningful success/failure feedback and has undefined behavior on overflow. sscanf provides success/failure feedback of sorts (the return value, which is what makes it "moderately better"), but still has undefined behavior on overflow. Only strtol is a viable solution.

– AnT
Aug 11 '11 at 6:55

1

Agreed; I just wanted to emphasize the potentially fatal problem with sscanf. (Though I confess I sometimes use atoi, usually for programs that I don't expect to survive more than 10 minute before I delete the source.)

– Keith Thompson
Aug 11 '11 at 6:58

sscanf actually has undefined behavior if it tries to convert a number outside the range of its type (for example, sscanf("999999999999999999999", "%d", &n)).

– Keith Thompson
Aug 11 '11 at 6:50

@Keith Thompson: That's exactly what I mean. atoi provides no meaningful success/failure feedback and has undefined behavior on overflow. sscanf provides success/failure feedback of sorts (the return value, which is what makes it "moderately better"), but still has undefined behavior on overflow. Only strtol is a viable solution.

– AnT
Aug 11 '11 at 6:55

Agreed; I just wanted to emphasize the potentially fatal problem with sscanf. (Though I confess I sometimes use atoi, usually for programs that I don't expect to survive more than 10 minute before I delete the source.)

– Keith Thompson
Aug 11 '11 at 6:58

add a comment |

You can code a little atoi() for fun:

int my_getnbr(char *str)

{

  int result;

  int puiss;



  result = 0;

  puiss = 1;

  while (('-' == (*str)) || ((*str) == '+'))

  {

      if (*str == '-')

        puiss = puiss * -1;

      str++;

  }

  while ((*str >= '0') && (*str <= '9'))

  {

      result = (result * 10) + ((*str) - '0');

      str++;

  }

  return (result * puiss);

}

You can also make it recursive wich can old in 3 lines =)

edited Sep 1 '18 at 18:11

vishal dharankar

4,68443767

answered Aug 11 '11 at 8:20

jDourlens

1,57511529

Thanks so much .. But could you tell me how the below code works? code ((*str) - '0') code

– user618677
Aug 11 '11 at 16:15

a character has an ascii value. If you are uner linux type: man ascii in the shell or if not go to:table-ascii.com. You will see that the character '0' = 68 (i think) for a int. So to get the number of '9' (it's '0' + 9) so you get 9 = '9' - '0'. You get it?

– jDourlens
Aug 11 '11 at 16:31

1

1) The code allows "----1" 2) Has undefined behavior with int overflow when the result should be INT_MIN. Consider my_getnbr("-2147483648")

– chux
May 30 '16 at 12:12

Thanks for precision, it was just for showing a little example. As it's said it for fun and learning. You should definetly use standart lib for this kind of tasks. Faster and safer!

– jDourlens
Jun 1 '16 at 0:12

add a comment |

You can code a little atoi() for fun:

int my_getnbr(char *str)

{

  int result;

  int puiss;



  result = 0;

  puiss = 1;

  while (('-' == (*str)) || ((*str) == '+'))

  {

      if (*str == '-')

        puiss = puiss * -1;

      str++;

  }

  while ((*str >= '0') && (*str <= '9'))

  {

      result = (result * 10) + ((*str) - '0');

      str++;

  }

  return (result * puiss);

}

You can also make it recursive wich can old in 3 lines =)

edited Sep 1 '18 at 18:11

vishal dharankar

4,68443767

answered Aug 11 '11 at 8:20

jDourlens

1,57511529

Thanks so much .. But could you tell me how the below code works? code ((*str) - '0') code

– user618677
Aug 11 '11 at 16:15

a character has an ascii value. If you are uner linux type: man ascii in the shell or if not go to:table-ascii.com. You will see that the character '0' = 68 (i think) for a int. So to get the number of '9' (it's '0' + 9) so you get 9 = '9' - '0'. You get it?

– jDourlens
Aug 11 '11 at 16:31

1

1) The code allows "----1" 2) Has undefined behavior with int overflow when the result should be INT_MIN. Consider my_getnbr("-2147483648")

– chux
May 30 '16 at 12:12

Thanks for precision, it was just for showing a little example. As it's said it for fun and learning. You should definetly use standart lib for this kind of tasks. Faster and safer!

– jDourlens
Jun 1 '16 at 0:12

add a comment |

You can code a little atoi() for fun:

int my_getnbr(char *str)

{

  int result;

  int puiss;



  result = 0;

  puiss = 1;

  while (('-' == (*str)) || ((*str) == '+'))

  {

      if (*str == '-')

        puiss = puiss * -1;

      str++;

  }

  while ((*str >= '0') && (*str <= '9'))

  {

      result = (result * 10) + ((*str) - '0');

      str++;

  }

  return (result * puiss);

}

You can also make it recursive wich can old in 3 lines =)

edited Sep 1 '18 at 18:11

vishal dharankar

4,68443767

answered Aug 11 '11 at 8:20

jDourlens

1,57511529

You can code a little atoi() for fun:

int my_getnbr(char *str)

{

  int result;

  int puiss;



  result = 0;

  puiss = 1;

  while (('-' == (*str)) || ((*str) == '+'))

  {

      if (*str == '-')

        puiss = puiss * -1;

      str++;

  }

  while ((*str >= '0') && (*str <= '9'))

  {

      result = (result * 10) + ((*str) - '0');

      str++;

  }

  return (result * puiss);

}

You can also make it recursive wich can old in 3 lines =)

edited Sep 1 '18 at 18:11

vishal dharankar

4,68443767

answered Aug 11 '11 at 8:20

jDourlens

1,57511529

edited Sep 1 '18 at 18:11

vishal dharankar

4,68443767

edited Sep 1 '18 at 18:11

vishal dharankar

4,68443767

edited Sep 1 '18 at 18:11

vishal dharankar

4,68443767

answered Aug 11 '11 at 8:20

jDourlens

1,57511529

answered Aug 11 '11 at 8:20

jDourlens

1,57511529

answered Aug 11 '11 at 8:20

jDourlens

1,57511529

Thanks so much .. But could you tell me how the below code works? code ((*str) - '0') code

– user618677
Aug 11 '11 at 16:15

a character has an ascii value. If you are uner linux type: man ascii in the shell or if not go to:table-ascii.com. You will see that the character '0' = 68 (i think) for a int. So to get the number of '9' (it's '0' + 9) so you get 9 = '9' - '0'. You get it?

– jDourlens
Aug 11 '11 at 16:31

1

1) The code allows "----1" 2) Has undefined behavior with int overflow when the result should be INT_MIN. Consider my_getnbr("-2147483648")

– chux
May 30 '16 at 12:12

Thanks for precision, it was just for showing a little example. As it's said it for fun and learning. You should definetly use standart lib for this kind of tasks. Faster and safer!

– jDourlens
Jun 1 '16 at 0:12

add a comment |

Thanks so much .. But could you tell me how the below code works? code ((*str) - '0') code

– user618677
Aug 11 '11 at 16:15

a character has an ascii value. If you are uner linux type: man ascii in the shell or if not go to:table-ascii.com. You will see that the character '0' = 68 (i think) for a int. So to get the number of '9' (it's '0' + 9) so you get 9 = '9' - '0'. You get it?

– jDourlens
Aug 11 '11 at 16:31

1

1) The code allows "----1" 2) Has undefined behavior with int overflow when the result should be INT_MIN. Consider my_getnbr("-2147483648")

– chux
May 30 '16 at 12:12

Thanks for precision, it was just for showing a little example. As it's said it for fun and learning. You should definetly use standart lib for this kind of tasks. Faster and safer!

– jDourlens
Jun 1 '16 at 0:12

Thanks so much .. But could you tell me how the below code works? code ((*str) - '0') code

– user618677
Aug 11 '11 at 16:15

a character has an ascii value. If you are uner linux type: man ascii in the shell or if not go to:table-ascii.com. You will see that the character '0' = 68 (i think) for a int. So to get the number of '9' (it's '0' + 9) so you get 9 = '9' - '0'. You get it?

– jDourlens
Aug 11 '11 at 16:31

1) The code allows "----1" 2) Has undefined behavior with int overflow when the result should be INT_MIN. Consider my_getnbr("-2147483648")

– chux
May 30 '16 at 12:12

Thanks for precision, it was just for showing a little example. As it's said it for fun and learning. You should definetly use standart lib for this kind of tasks. Faster and safer!

– jDourlens
Jun 1 '16 at 0:12

add a comment |

Just wanted to share a solution for unsigned long aswell.

unsigned long ToUInt(char* str)

{

    unsigned long mult = 1;

    unsigned long re = 0;

    int len = strlen(str);

    for(int i = len -1 ; i >= 0 ; i--)

    {

        re = re + ((int)str[i] -48)*mult;

        mult = mult*10;

    }

    return re;

}

answered Oct 18 '16 at 19:26

Jacob

1488

1

Doesn't handle overflow. Also, the parameter should be const char *.

– Roland Illig
Jan 21 '17 at 17:35

1

Plus, what's that 48 mean? Are you assuming that's the value of '0' where the code will run? Please don't inflict such broad assumptions on the world!

– Toby Speight
Jun 18 '18 at 16:15

@TobySpeight Yes I assume 48 represent '0' in the ascii table.

– Jacob
Jun 20 '18 at 9:17

1

Not all the world is ASCII - just use '0' like you should.

– Toby Speight
Jun 20 '18 at 10:42

add a comment |

Just wanted to share a solution for unsigned long aswell.

unsigned long ToUInt(char* str)

{

    unsigned long mult = 1;

    unsigned long re = 0;

    int len = strlen(str);

    for(int i = len -1 ; i >= 0 ; i--)

    {

        re = re + ((int)str[i] -48)*mult;

        mult = mult*10;

    }

    return re;

}

answered Oct 18 '16 at 19:26

Jacob

1488

1

Doesn't handle overflow. Also, the parameter should be const char *.

– Roland Illig
Jan 21 '17 at 17:35

1

Plus, what's that 48 mean? Are you assuming that's the value of '0' where the code will run? Please don't inflict such broad assumptions on the world!

– Toby Speight
Jun 18 '18 at 16:15

@TobySpeight Yes I assume 48 represent '0' in the ascii table.

– Jacob
Jun 20 '18 at 9:17

1

Not all the world is ASCII - just use '0' like you should.

– Toby Speight
Jun 20 '18 at 10:42

add a comment |

Just wanted to share a solution for unsigned long aswell.

unsigned long ToUInt(char* str)

{

    unsigned long mult = 1;

    unsigned long re = 0;

    int len = strlen(str);

    for(int i = len -1 ; i >= 0 ; i--)

    {

        re = re + ((int)str[i] -48)*mult;

        mult = mult*10;

    }

    return re;

}

answered Oct 18 '16 at 19:26

Jacob

1488

Just wanted to share a solution for unsigned long aswell.

unsigned long ToUInt(char* str)

{

    unsigned long mult = 1;

    unsigned long re = 0;

    int len = strlen(str);

    for(int i = len -1 ; i >= 0 ; i--)

    {

        re = re + ((int)str[i] -48)*mult;

        mult = mult*10;

    }

    return re;

}

answered Oct 18 '16 at 19:26

Jacob

1488

answered Oct 18 '16 at 19:26

Jacob

1488

answered Oct 18 '16 at 19:26

Jacob

1488

answered Oct 18 '16 at 19:26

Jacob

1488

1

Doesn't handle overflow. Also, the parameter should be const char *.

– Roland Illig
Jan 21 '17 at 17:35

1

Plus, what's that 48 mean? Are you assuming that's the value of '0' where the code will run? Please don't inflict such broad assumptions on the world!

– Toby Speight
Jun 18 '18 at 16:15

@TobySpeight Yes I assume 48 represent '0' in the ascii table.

– Jacob
Jun 20 '18 at 9:17

1

Not all the world is ASCII - just use '0' like you should.

– Toby Speight
Jun 20 '18 at 10:42

add a comment |

1

Doesn't handle overflow. Also, the parameter should be const char *.

– Roland Illig
Jan 21 '17 at 17:35

1

Plus, what's that 48 mean? Are you assuming that's the value of '0' where the code will run? Please don't inflict such broad assumptions on the world!

– Toby Speight
Jun 18 '18 at 16:15

@TobySpeight Yes I assume 48 represent '0' in the ascii table.

– Jacob
Jun 20 '18 at 9:17

1

Not all the world is ASCII - just use '0' like you should.

– Toby Speight
Jun 20 '18 at 10:42

Doesn't handle overflow. Also, the parameter should be const char *.

– Roland Illig
Jan 21 '17 at 17:35

Plus, what's that 48 mean? Are you assuming that's the value of '0' where the code will run? Please don't inflict such broad assumptions on the world!

– Toby Speight
Jun 18 '18 at 16:15

@TobySpeight Yes I assume 48 represent '0' in the ascii table.

– Jacob
Jun 20 '18 at 9:17

Not all the world is ASCII - just use '0' like you should.

– Toby Speight
Jun 20 '18 at 10:42

add a comment |

This function will help you

int strtoint_n(char* str, int n)

{

    int sign = 1;

    int place = 1;

    int ret = 0;



    int i;

    for (i = n-1; i >= 0; i--, place *= 10)

    {

        int c = str[i];

        switch (c)

        {

            case '-':

                if (i == 0) sign = -1;

                else return -1;

                break;

            default:

                if (c >= '0' && c <= '9')   ret += (c - '0') * place;

                else return -1;

        }

    }



    return sign * ret;

}



int strtoint(char* str)

{

    char* temp = str;

    int n = 0;

    while (*temp != '')

    {

        n++;

        temp++;

    }

    return strtoint_n(str, n);

}

Ref: http://amscata.blogspot.com/2013/09/strnumstr-version-2.html

edited Aug 7 '15 at 16:35

MC93

762714

answered Sep 12 '13 at 11:31

Amith Chinthaka

5431122

1

Why do this though? One of the biggest problems with atoi and friends is that if there's overflow, it's undefined behavior. Your function does not check for this. strtol and friends do.

– chad
Sep 2 '15 at 19:45

1

Yup. Since C is not Python I hope that the people who use C language are aware of these kind of overflow errors. Everything has it's own limits.

– Amith Chinthaka
Sep 3 '15 at 9:04

add a comment |

This function will help you

int strtoint_n(char* str, int n)

{

    int sign = 1;

    int place = 1;

    int ret = 0;



    int i;

    for (i = n-1; i >= 0; i--, place *= 10)

    {

        int c = str[i];

        switch (c)

        {

            case '-':

                if (i == 0) sign = -1;

                else return -1;

                break;

            default:

                if (c >= '0' && c <= '9')   ret += (c - '0') * place;

                else return -1;

        }

    }



    return sign * ret;

}



int strtoint(char* str)

{

    char* temp = str;

    int n = 0;

    while (*temp != '')

    {

        n++;

        temp++;

    }

    return strtoint_n(str, n);

}

Ref: http://amscata.blogspot.com/2013/09/strnumstr-version-2.html

edited Aug 7 '15 at 16:35

MC93

762714

answered Sep 12 '13 at 11:31

Amith Chinthaka

5431122

1

Why do this though? One of the biggest problems with atoi and friends is that if there's overflow, it's undefined behavior. Your function does not check for this. strtol and friends do.

– chad
Sep 2 '15 at 19:45

1

Yup. Since C is not Python I hope that the people who use C language are aware of these kind of overflow errors. Everything has it's own limits.

– Amith Chinthaka
Sep 3 '15 at 9:04

add a comment |

This function will help you

int strtoint_n(char* str, int n)

{

    int sign = 1;

    int place = 1;

    int ret = 0;



    int i;

    for (i = n-1; i >= 0; i--, place *= 10)

    {

        int c = str[i];

        switch (c)

        {

            case '-':

                if (i == 0) sign = -1;

                else return -1;

                break;

            default:

                if (c >= '0' && c <= '9')   ret += (c - '0') * place;

                else return -1;

        }

    }



    return sign * ret;

}



int strtoint(char* str)

{

    char* temp = str;

    int n = 0;

    while (*temp != '')

    {

        n++;

        temp++;

    }

    return strtoint_n(str, n);

}

Ref: http://amscata.blogspot.com/2013/09/strnumstr-version-2.html

edited Aug 7 '15 at 16:35

MC93

762714

answered Sep 12 '13 at 11:31

Amith Chinthaka

5431122

This function will help you

int strtoint_n(char* str, int n)

{

    int sign = 1;

    int place = 1;

    int ret = 0;



    int i;

    for (i = n-1; i >= 0; i--, place *= 10)

    {

        int c = str[i];

        switch (c)

        {

            case '-':

                if (i == 0) sign = -1;

                else return -1;

                break;

            default:

                if (c >= '0' && c <= '9')   ret += (c - '0') * place;

                else return -1;

        }

    }



    return sign * ret;

}



int strtoint(char* str)

{

    char* temp = str;

    int n = 0;

    while (*temp != '')

    {

        n++;

        temp++;

    }

    return strtoint_n(str, n);

}

Ref: http://amscata.blogspot.com/2013/09/strnumstr-version-2.html

edited Aug 7 '15 at 16:35

MC93

762714

answered Sep 12 '13 at 11:31

Amith Chinthaka

5431122

edited Aug 7 '15 at 16:35

MC93

762714

edited Aug 7 '15 at 16:35

MC93

762714

edited Aug 7 '15 at 16:35

MC93

762714

answered Sep 12 '13 at 11:31

Amith Chinthaka

5431122

answered Sep 12 '13 at 11:31

Amith Chinthaka

5431122

answered Sep 12 '13 at 11:31

Amith Chinthaka

5431122

1

Why do this though? One of the biggest problems with atoi and friends is that if there's overflow, it's undefined behavior. Your function does not check for this. strtol and friends do.

– chad
Sep 2 '15 at 19:45

1

Yup. Since C is not Python I hope that the people who use C language are aware of these kind of overflow errors. Everything has it's own limits.

– Amith Chinthaka
Sep 3 '15 at 9:04

add a comment |

1

Why do this though? One of the biggest problems with atoi and friends is that if there's overflow, it's undefined behavior. Your function does not check for this. strtol and friends do.

– chad
Sep 2 '15 at 19:45

1

Yup. Since C is not Python I hope that the people who use C language are aware of these kind of overflow errors. Everything has it's own limits.

– Amith Chinthaka
Sep 3 '15 at 9:04

Why do this though? One of the biggest problems with atoi and friends is that if there's overflow, it's undefined behavior. Your function does not check for this. strtol and friends do.

– chad
Sep 2 '15 at 19:45

Yup. Since C is not Python I hope that the people who use C language are aware of these kind of overflow errors. Everything has it's own limits.

– Amith Chinthaka
Sep 3 '15 at 9:04

add a comment |

Ok, I had the same problem.I came up with this solution.It worked for me the best.I did try atoi() but didn't work well for me.So here is my solution:

void splitInput(int arr, int sizeArr, char num)

{

    for(int i = 0; i < sizeArr; i++)

        // We are subtracting 48 because the numbers in ASCII starts at 48.

        arr[i] = (int)num[i] - 48;

}

answered Sep 5 '15 at 15:36

Khaled Mohammad

1347

add a comment |

Ok, I had the same problem.I came up with this solution.It worked for me the best.I did try atoi() but didn't work well for me.So here is my solution:

void splitInput(int arr, int sizeArr, char num)

{

    for(int i = 0; i < sizeArr; i++)

        // We are subtracting 48 because the numbers in ASCII starts at 48.

        arr[i] = (int)num[i] - 48;

}

answered Sep 5 '15 at 15:36

Khaled Mohammad

1347

add a comment |

Ok, I had the same problem.I came up with this solution.It worked for me the best.I did try atoi() but didn't work well for me.So here is my solution:

void splitInput(int arr, int sizeArr, char num)

{

    for(int i = 0; i < sizeArr; i++)

        // We are subtracting 48 because the numbers in ASCII starts at 48.

        arr[i] = (int)num[i] - 48;

}

answered Sep 5 '15 at 15:36

Khaled Mohammad

1347

Ok, I had the same problem.I came up with this solution.It worked for me the best.I did try atoi() but didn't work well for me.So here is my solution:

void splitInput(int arr, int sizeArr, char num)

{

    for(int i = 0; i < sizeArr; i++)

        // We are subtracting 48 because the numbers in ASCII starts at 48.

        arr[i] = (int)num[i] - 48;

}

answered Sep 5 '15 at 15:36

Khaled Mohammad

1347

answered Sep 5 '15 at 15:36

Khaled Mohammad

1347

answered Sep 5 '15 at 15:36

Khaled Mohammad

1347

answered Sep 5 '15 at 15:36

Khaled Mohammad

1347

add a comment |

//I think this way we could go :

int my_atoi(const char* snum)

{

 int nInt(0);

 int index(0);

 while(snum[index])

 {

    if(!nInt)

        nInt= ( (int) snum[index]) - 48;

    else

    {

        nInt = (nInt *= 10) + ((int) snum[index] - 48);

    }

    index++;

 }

 return(nInt);

}



int main()

{

    printf("Returned number is: %dn", my_atoi("676987"));

    return 0;

}

answered Nov 6 '15 at 13:05

Aditya Kumar

673

Code does not compile in C. Why nInt = (nInt *= 10) + ((int) snum[index] - 48); vs. nInt = nInt*10 + snum[index] - '0'; if(!nInt) not needed.

– chux
May 30 '16 at 12:19

add a comment |

//I think this way we could go :

int my_atoi(const char* snum)

{

 int nInt(0);

 int index(0);

 while(snum[index])

 {

    if(!nInt)

        nInt= ( (int) snum[index]) - 48;

    else

    {

        nInt = (nInt *= 10) + ((int) snum[index] - 48);

    }

    index++;

 }

 return(nInt);

}



int main()

{

    printf("Returned number is: %dn", my_atoi("676987"));

    return 0;

}

answered Nov 6 '15 at 13:05

Aditya Kumar

673

Code does not compile in C. Why nInt = (nInt *= 10) + ((int) snum[index] - 48); vs. nInt = nInt*10 + snum[index] - '0'; if(!nInt) not needed.

– chux
May 30 '16 at 12:19

add a comment |

//I think this way we could go :

int my_atoi(const char* snum)

{

 int nInt(0);

 int index(0);

 while(snum[index])

 {

    if(!nInt)

        nInt= ( (int) snum[index]) - 48;

    else

    {

        nInt = (nInt *= 10) + ((int) snum[index] - 48);

    }

    index++;

 }

 return(nInt);

}



int main()

{

    printf("Returned number is: %dn", my_atoi("676987"));

    return 0;

}

answered Nov 6 '15 at 13:05

Aditya Kumar

673

//I think this way we could go :

int my_atoi(const char* snum)

{

 int nInt(0);

 int index(0);

 while(snum[index])

 {

    if(!nInt)

        nInt= ( (int) snum[index]) - 48;

    else

    {

        nInt = (nInt *= 10) + ((int) snum[index] - 48);

    }

    index++;

 }

 return(nInt);

}



int main()

{

    printf("Returned number is: %dn", my_atoi("676987"));

    return 0;

}

answered Nov 6 '15 at 13:05

Aditya Kumar

673

answered Nov 6 '15 at 13:05

Aditya Kumar

673

answered Nov 6 '15 at 13:05

Aditya Kumar

673

answered Nov 6 '15 at 13:05

Aditya Kumar

673

Code does not compile in C. Why nInt = (nInt *= 10) + ((int) snum[index] - 48); vs. nInt = nInt*10 + snum[index] - '0'; if(!nInt) not needed.

– chux
May 30 '16 at 12:19

add a comment |

Code does not compile in C. Why nInt = (nInt *= 10) + ((int) snum[index] - 48); vs. nInt = nInt*10 + snum[index] - '0'; if(!nInt) not needed.

– chux
May 30 '16 at 12:19

Code does not compile in C. Why nInt = (nInt *= 10) + ((int) snum[index] - 48); vs. nInt = nInt*10 + snum[index] - '0'; if(!nInt) not needed.

– chux
May 30 '16 at 12:19

add a comment |

-1

You can always roll your own!

#include <stdio.h>

#include <string.h>

#include <math.h>



int my_atoi(const char* snum)

{

    int idx, strIdx = 0, accum = 0, numIsNeg = 0;

    const unsigned int NUMLEN = (int)strlen(snum);



    /* Check if negative number and flag it. */

    if(snum[0] == 0x2d)

        numIsNeg = 1;



    for(idx = NUMLEN - 1; idx >= 0; idx--)

    {

        /* Only process numbers from 0 through 9. */

        if(snum[strIdx] >= 0x30 && snum[strIdx] <= 0x39)

            accum += (snum[strIdx] - 0x30) * pow(10, idx);



        strIdx++;

    }



    /* Check flag to see if originally passed -ve number and convert result if so. */

    if(!numIsNeg)

        return accum;

    else

        return accum * -1;

}



int main()

{

    /* Tests... */

    printf("Returned number is: %dn", my_atoi("34574"));

    printf("Returned number is: %dn", my_atoi("-23"));



    return 0;

}

This will do what you want without clutter.

answered Jan 3 '15 at 0:42

ButchDean

746

2

But... why? This doesn't check for overflow and simply ignores garbage values. There's no reason not to use the strto... family of functions. They are portable and significantly better.

– chad
Sep 2 '15 at 19:41

1

Strange to use 0x2d, 0x30 instead of '-', '0'. Does not allow '+' sign. Why (int) cast in (int)strlen(snum)? UB if input is "". UB when result is INT_MIN due to int overflow with accum += (snum[strIdx] - 0x30) * pow(10, idx);

– chux
May 30 '16 at 12:24

@chux - This code is demonstration code. There are easy fixes to what you described as potential issues.

– ButchDean
Jun 7 '16 at 4:12

1

@ButchDean What you describe as "demonstration code" will be used by others who have no clue about all the details. Only the negative score and the comments on this answer protect them now. In my opinion, "demonstration code" must have much higher quality.

– Roland Illig
Jan 21 '17 at 17:39

@RolandIllig Rather than being all critical, wouldn't it be more helpful to others to actually put up your own solution?

– ButchDean
Jan 24 '17 at 17:35

|
show 6 more comments

-1

You can always roll your own!

#include <stdio.h>

#include <string.h>

#include <math.h>



int my_atoi(const char* snum)

{

    int idx, strIdx = 0, accum = 0, numIsNeg = 0;

    const unsigned int NUMLEN = (int)strlen(snum);



    /* Check if negative number and flag it. */

    if(snum[0] == 0x2d)

        numIsNeg = 1;



    for(idx = NUMLEN - 1; idx >= 0; idx--)

    {

        /* Only process numbers from 0 through 9. */

        if(snum[strIdx] >= 0x30 && snum[strIdx] <= 0x39)

            accum += (snum[strIdx] - 0x30) * pow(10, idx);



        strIdx++;

    }



    /* Check flag to see if originally passed -ve number and convert result if so. */

    if(!numIsNeg)

        return accum;

    else

        return accum * -1;

}



int main()

{

    /* Tests... */

    printf("Returned number is: %dn", my_atoi("34574"));

    printf("Returned number is: %dn", my_atoi("-23"));



    return 0;

}

This will do what you want without clutter.

answered Jan 3 '15 at 0:42

ButchDean

746

2

But... why? This doesn't check for overflow and simply ignores garbage values. There's no reason not to use the strto... family of functions. They are portable and significantly better.

– chad
Sep 2 '15 at 19:41

1

Strange to use 0x2d, 0x30 instead of '-', '0'. Does not allow '+' sign. Why (int) cast in (int)strlen(snum)? UB if input is "". UB when result is INT_MIN due to int overflow with accum += (snum[strIdx] - 0x30) * pow(10, idx);

– chux
May 30 '16 at 12:24

@chux - This code is demonstration code. There are easy fixes to what you described as potential issues.

– ButchDean
Jun 7 '16 at 4:12

1

@ButchDean What you describe as "demonstration code" will be used by others who have no clue about all the details. Only the negative score and the comments on this answer protect them now. In my opinion, "demonstration code" must have much higher quality.

– Roland Illig
Jan 21 '17 at 17:39

@RolandIllig Rather than being all critical, wouldn't it be more helpful to others to actually put up your own solution?

– ButchDean
Jan 24 '17 at 17:35

|
show 6 more comments

-1

You can always roll your own!

#include <stdio.h>

#include <string.h>

#include <math.h>



int my_atoi(const char* snum)

{

    int idx, strIdx = 0, accum = 0, numIsNeg = 0;

    const unsigned int NUMLEN = (int)strlen(snum);



    /* Check if negative number and flag it. */

    if(snum[0] == 0x2d)

        numIsNeg = 1;



    for(idx = NUMLEN - 1; idx >= 0; idx--)

    {

        /* Only process numbers from 0 through 9. */

        if(snum[strIdx] >= 0x30 && snum[strIdx] <= 0x39)

            accum += (snum[strIdx] - 0x30) * pow(10, idx);



        strIdx++;

    }



    /* Check flag to see if originally passed -ve number and convert result if so. */

    if(!numIsNeg)

        return accum;

    else

        return accum * -1;

}



int main()

{

    /* Tests... */

    printf("Returned number is: %dn", my_atoi("34574"));

    printf("Returned number is: %dn", my_atoi("-23"));



    return 0;

}

This will do what you want without clutter.

answered Jan 3 '15 at 0:42

ButchDean

746

You can always roll your own!

#include <stdio.h>

#include <string.h>

#include <math.h>



int my_atoi(const char* snum)

{

    int idx, strIdx = 0, accum = 0, numIsNeg = 0;

    const unsigned int NUMLEN = (int)strlen(snum);



    /* Check if negative number and flag it. */

    if(snum[0] == 0x2d)

        numIsNeg = 1;



    for(idx = NUMLEN - 1; idx >= 0; idx--)

    {

        /* Only process numbers from 0 through 9. */

        if(snum[strIdx] >= 0x30 && snum[strIdx] <= 0x39)

            accum += (snum[strIdx] - 0x30) * pow(10, idx);



        strIdx++;

    }



    /* Check flag to see if originally passed -ve number and convert result if so. */

    if(!numIsNeg)

        return accum;

    else

        return accum * -1;

}



int main()

{

    /* Tests... */

    printf("Returned number is: %dn", my_atoi("34574"));

    printf("Returned number is: %dn", my_atoi("-23"));



    return 0;

}

This will do what you want without clutter.

answered Jan 3 '15 at 0:42

ButchDean

746

answered Jan 3 '15 at 0:42

ButchDean

746

answered Jan 3 '15 at 0:42

ButchDean

746

answered Jan 3 '15 at 0:42

ButchDean

746

2

But... why? This doesn't check for overflow and simply ignores garbage values. There's no reason not to use the strto... family of functions. They are portable and significantly better.

– chad
Sep 2 '15 at 19:41

1

Strange to use 0x2d, 0x30 instead of '-', '0'. Does not allow '+' sign. Why (int) cast in (int)strlen(snum)? UB if input is "". UB when result is INT_MIN due to int overflow with accum += (snum[strIdx] - 0x30) * pow(10, idx);

– chux
May 30 '16 at 12:24

@chux - This code is demonstration code. There are easy fixes to what you described as potential issues.

– ButchDean
Jun 7 '16 at 4:12

1

@ButchDean What you describe as "demonstration code" will be used by others who have no clue about all the details. Only the negative score and the comments on this answer protect them now. In my opinion, "demonstration code" must have much higher quality.

– Roland Illig
Jan 21 '17 at 17:39

@RolandIllig Rather than being all critical, wouldn't it be more helpful to others to actually put up your own solution?

– ButchDean
Jan 24 '17 at 17:35

|
show 6 more comments

2

But... why? This doesn't check for overflow and simply ignores garbage values. There's no reason not to use the strto... family of functions. They are portable and significantly better.

– chad
Sep 2 '15 at 19:41

1

Strange to use 0x2d, 0x30 instead of '-', '0'. Does not allow '+' sign. Why (int) cast in (int)strlen(snum)? UB if input is "". UB when result is INT_MIN due to int overflow with accum += (snum[strIdx] - 0x30) * pow(10, idx);

– chux
May 30 '16 at 12:24

@chux - This code is demonstration code. There are easy fixes to what you described as potential issues.

– ButchDean
Jun 7 '16 at 4:12

1

@ButchDean What you describe as "demonstration code" will be used by others who have no clue about all the details. Only the negative score and the comments on this answer protect them now. In my opinion, "demonstration code" must have much higher quality.

– Roland Illig
Jan 21 '17 at 17:39

@RolandIllig Rather than being all critical, wouldn't it be more helpful to others to actually put up your own solution?

– ButchDean
Jan 24 '17 at 17:35

But... why? This doesn't check for overflow and simply ignores garbage values. There's no reason not to use the strto... family of functions. They are portable and significantly better.

– chad
Sep 2 '15 at 19:41

Strange to use 0x2d, 0x30 instead of '-', '0'. Does not allow '+' sign. Why (int) cast in (int)strlen(snum)? UB if input is "". UB when result is INT_MIN due to int overflow with accum += (snum[strIdx] - 0x30) * pow(10, idx);

– chux
May 30 '16 at 12:24

@chux - This code is demonstration code. There are easy fixes to what you described as potential issues.

– ButchDean
Jun 7 '16 at 4:12

@ButchDean What you describe as "demonstration code" will be used by others who have no clue about all the details. Only the negative score and the comments on this answer protect them now. In my opinion, "demonstration code" must have much higher quality.

– Roland Illig
Jan 21 '17 at 17:39

@RolandIllig Rather than being all critical, wouldn't it be more helpful to others to actually put up your own solution?

– ButchDean
Jan 24 '17 at 17:35

|
show 6 more comments

-3

In C++, you can use a such function:

template <typename T>

T to(const std::string & s)

{

    std::istringstream stm(s);

    T result;

    stm >> result;



    if(stm.tellg() != s.size())

        throw error;



    return result;

}

This can help you to convert any string to any type such as float, int, double...

answered Aug 11 '11 at 6:41

neodelphi

2,06411017

1

There's already a similar question covering C++, where the problems with this approach are explained.

– Ben Voigt
Feb 20 '15 at 17:28

add a comment |

-3

In C++, you can use a such function:

template <typename T>

T to(const std::string & s)

{

    std::istringstream stm(s);

    T result;

    stm >> result;



    if(stm.tellg() != s.size())

        throw error;



    return result;

}

This can help you to convert any string to any type such as float, int, double...

answered Aug 11 '11 at 6:41

neodelphi

2,06411017

1

There's already a similar question covering C++, where the problems with this approach are explained.

– Ben Voigt
Feb 20 '15 at 17:28

add a comment |

-3

In C++, you can use a such function:

template <typename T>

T to(const std::string & s)

{

    std::istringstream stm(s);

    T result;

    stm >> result;



    if(stm.tellg() != s.size())

        throw error;



    return result;

}

This can help you to convert any string to any type such as float, int, double...

answered Aug 11 '11 at 6:41

neodelphi

2,06411017

In C++, you can use a such function:

template <typename T>

T to(const std::string & s)

{

    std::istringstream stm(s);

    T result;

    stm >> result;



    if(stm.tellg() != s.size())

        throw error;



    return result;

}

This can help you to convert any string to any type such as float, int, double...

answered Aug 11 '11 at 6:41

neodelphi

2,06411017

answered Aug 11 '11 at 6:41

neodelphi

2,06411017

answered Aug 11 '11 at 6:41

neodelphi

2,06411017

answered Aug 11 '11 at 6:41

neodelphi

2,06411017

1

There's already a similar question covering C++, where the problems with this approach are explained.

– Ben Voigt
Feb 20 '15 at 17:28

add a comment |

1

There's already a similar question covering C++, where the problems with this approach are explained.

– Ben Voigt
Feb 20 '15 at 17:28

There's already a similar question covering C++, where the problems with this approach are explained.

– Ben Voigt
Feb 20 '15 at 17:28

add a comment |

-4

Yes, you can store the integer directly:

int num = 45;

If you must parse a string, atoi or strol is going to win the "shortest amount of code" contest.

answered Aug 11 '11 at 6:32

Yann Ramin

30.6k14776

If you want to do it safely, strtol() actually requires a fair amount of code. It can return LONG_MIN or LONG_MAX either if that's the actual converted value or if there's an underflow or overflow, and it can return 0 either if that's the actual value or if there was no number to convert. You need to set errno = 0 before the call, and check the endptr.

– Keith Thompson
Aug 11 '11 at 6:54

The solutions given to parse, are no viable solutions.

– BananaAcid
Feb 1 '18 at 21:15

add a comment |

-4

Yes, you can store the integer directly:

int num = 45;

If you must parse a string, atoi or strol is going to win the "shortest amount of code" contest.

answered Aug 11 '11 at 6:32

Yann Ramin

30.6k14776

If you want to do it safely, strtol() actually requires a fair amount of code. It can return LONG_MIN or LONG_MAX either if that's the actual converted value or if there's an underflow or overflow, and it can return 0 either if that's the actual value or if there was no number to convert. You need to set errno = 0 before the call, and check the endptr.

– Keith Thompson
Aug 11 '11 at 6:54

The solutions given to parse, are no viable solutions.

– BananaAcid
Feb 1 '18 at 21:15

add a comment |

-4

Yes, you can store the integer directly:

int num = 45;

If you must parse a string, atoi or strol is going to win the "shortest amount of code" contest.

answered Aug 11 '11 at 6:32

Yann Ramin

30.6k14776

Yes, you can store the integer directly:

int num = 45;

If you must parse a string, atoi or strol is going to win the "shortest amount of code" contest.

answered Aug 11 '11 at 6:32

Yann Ramin

30.6k14776

answered Aug 11 '11 at 6:32

Yann Ramin

30.6k14776

answered Aug 11 '11 at 6:32

Yann Ramin

30.6k14776

answered Aug 11 '11 at 6:32

Yann Ramin

30.6k14776

If you want to do it safely, strtol() actually requires a fair amount of code. It can return LONG_MIN or LONG_MAX either if that's the actual converted value or if there's an underflow or overflow, and it can return 0 either if that's the actual value or if there was no number to convert. You need to set errno = 0 before the call, and check the endptr.

– Keith Thompson
Aug 11 '11 at 6:54

The solutions given to parse, are no viable solutions.

– BananaAcid
Feb 1 '18 at 21:15

add a comment |

If you want to do it safely, strtol() actually requires a fair amount of code. It can return LONG_MIN or LONG_MAX either if that's the actual converted value or if there's an underflow or overflow, and it can return 0 either if that's the actual value or if there was no number to convert. You need to set errno = 0 before the call, and check the endptr.

– Keith Thompson
Aug 11 '11 at 6:54

The solutions given to parse, are no viable solutions.

– BananaAcid
Feb 1 '18 at 21:15

If you want to do it safely, strtol() actually requires a fair amount of code. It can return LONG_MIN or LONG_MAX either if that's the actual converted value or if there's an underflow or overflow, and it can return 0 either if that's the actual value or if there was no number to convert. You need to set errno = 0 before the call, and check the endptr.

– Keith Thompson
Aug 11 '11 at 6:54

The solutions given to parse, are no viable solutions.

– BananaAcid
Feb 1 '18 at 21:15

add a comment |

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