convert string to integer and getback same string from integer value in c

-2

I want to convert string "hello world" to some integer value and get back "hello world" string from converted integer value. I tried but getting issue while converting int to string.

I converted the string to an integer using this code:

char str[50]; 

int i, len; int result=0; 

printf("Enter string to be converted: "); 

gets(str); 

len = strlen(str); 

for(i=0; i<len; i++){ 

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

} 

printf("%dn", result);

and then I tried converting it back like this:

printf("%dn", result); 

int rem; int j = 0; 

char result_str[50]; 

while (result !=0) { 

    result = result - '0' ; 

    rem = result%10 ; 

    result = result/10; 

    result_str[j] = 'a' + (rem -1); 

    printf("%c",result_str[j]); j++; 

} 

printf("result string = %s",result_str);

Sample output:

$ ./a.out 

Enter string to be converted: hello

619663

619663

eccfhresult string = eccfh

When I try the string "hello world" I get a segmentation fault.

edited Nov 13 '18 at 7:08

Broman

6,246112241

asked Nov 13 '18 at 6:17

Vijay Dubey

3

How us what you tried and how it failed. Also show us your debugging effort.
– Sourav Ghosh
Nov 13 '18 at 6:22

What have you tried so far? Please show and somebody could help!
– Mike
Nov 13 '18 at 6:23

1

@VijayDubey Please stop posting code in comments and edit your question instead.
– Swordfish
Nov 13 '18 at 6:33

add a comment |

-2

I want to convert string "hello world" to some integer value and get back "hello world" string from converted integer value. I tried but getting issue while converting int to string.

I converted the string to an integer using this code:

char str[50]; 

int i, len; int result=0; 

printf("Enter string to be converted: "); 

gets(str); 

len = strlen(str); 

for(i=0; i<len; i++){ 

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

} 

printf("%dn", result);

and then I tried converting it back like this:

printf("%dn", result); 

int rem; int j = 0; 

char result_str[50]; 

while (result !=0) { 

    result = result - '0' ; 

    rem = result%10 ; 

    result = result/10; 

    result_str[j] = 'a' + (rem -1); 

    printf("%c",result_str[j]); j++; 

} 

printf("result string = %s",result_str);

Sample output:

$ ./a.out 

Enter string to be converted: hello

619663

619663

eccfhresult string = eccfh

When I try the string "hello world" I get a segmentation fault.

edited Nov 13 '18 at 7:08

Broman

6,246112241

asked Nov 13 '18 at 6:17

Vijay Dubey

3

How us what you tried and how it failed. Also show us your debugging effort.
– Sourav Ghosh
Nov 13 '18 at 6:22

What have you tried so far? Please show and somebody could help!
– Mike
Nov 13 '18 at 6:23

1

@VijayDubey Please stop posting code in comments and edit your question instead.
– Swordfish
Nov 13 '18 at 6:33

add a comment |

-2

I want to convert string "hello world" to some integer value and get back "hello world" string from converted integer value. I tried but getting issue while converting int to string.

I converted the string to an integer using this code:

char str[50]; 

int i, len; int result=0; 

printf("Enter string to be converted: "); 

gets(str); 

len = strlen(str); 

for(i=0; i<len; i++){ 

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

} 

printf("%dn", result);

and then I tried converting it back like this:

printf("%dn", result); 

int rem; int j = 0; 

char result_str[50]; 

while (result !=0) { 

    result = result - '0' ; 

    rem = result%10 ; 

    result = result/10; 

    result_str[j] = 'a' + (rem -1); 

    printf("%c",result_str[j]); j++; 

} 

printf("result string = %s",result_str);

Sample output:

$ ./a.out 

Enter string to be converted: hello

619663

619663

eccfhresult string = eccfh

When I try the string "hello world" I get a segmentation fault.

edited Nov 13 '18 at 7:08

Broman

6,246112241

asked Nov 13 '18 at 6:17

Vijay Dubey

I want to convert string "hello world" to some integer value and get back "hello world" string from converted integer value. I tried but getting issue while converting int to string.

I converted the string to an integer using this code:

char str[50]; 

int i, len; int result=0; 

printf("Enter string to be converted: "); 

gets(str); 

len = strlen(str); 

for(i=0; i<len; i++){ 

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

} 

printf("%dn", result);

and then I tried converting it back like this:

printf("%dn", result); 

int rem; int j = 0; 

char result_str[50]; 

while (result !=0) { 

    result = result - '0' ; 

    rem = result%10 ; 

    result = result/10; 

    result_str[j] = 'a' + (rem -1); 

    printf("%c",result_str[j]); j++; 

} 

printf("result string = %s",result_str);

Sample output:

$ ./a.out 

Enter string to be converted: hello

619663

619663

eccfhresult string = eccfh

When I try the string "hello world" I get a segmentation fault.

edited Nov 13 '18 at 7:08

Broman

6,246112241

asked Nov 13 '18 at 6:17

Vijay Dubey

edited Nov 13 '18 at 7:08

Broman

6,246112241

asked Nov 13 '18 at 6:17

Vijay Dubey

edited Nov 13 '18 at 7:08

Broman

6,246112241

edited Nov 13 '18 at 7:08

Broman

6,246112241

edited Nov 13 '18 at 7:08

Broman

6,246112241

asked Nov 13 '18 at 6:17

Vijay Dubey

asked Nov 13 '18 at 6:17

Vijay Dubey

asked Nov 13 '18 at 6:17

Vijay Dubey

3

How us what you tried and how it failed. Also show us your debugging effort.
– Sourav Ghosh
Nov 13 '18 at 6:22

What have you tried so far? Please show and somebody could help!
– Mike
Nov 13 '18 at 6:23

1

@VijayDubey Please stop posting code in comments and edit your question instead.
– Swordfish
Nov 13 '18 at 6:33

add a comment |

3

How us what you tried and how it failed. Also show us your debugging effort.
– Sourav Ghosh
Nov 13 '18 at 6:22

What have you tried so far? Please show and somebody could help!
– Mike
Nov 13 '18 at 6:23

1

@VijayDubey Please stop posting code in comments and edit your question instead.
– Swordfish
Nov 13 '18 at 6:33

How us what you tried and how it failed. Also show us your debugging effort.
– Sourav Ghosh
Nov 13 '18 at 6:22

What have you tried so far? Please show and somebody could help!
– Mike
Nov 13 '18 at 6:23

@VijayDubey Please stop posting code in comments and edit your question instead.
– Swordfish
Nov 13 '18 at 6:33

add a comment |

1 Answer
1

active

oldest

votes

There is absolutely no way to convert a string to an int and then back again the way you want.

An int is typically 4 bytes, and you're using a char[50] which is 50 bytes. This means that an int can have 2³² different values, while a char[50] can have 2⁴⁰⁰ different values. So it is simply impossible to map them one to one.

Let's take an example outside the realm of code and computers and just focus on numbers. Can you imagine a method to convert a two-digit number to a one-digit number and back? If this was possible, then we would not need two digits in the first place. If such a method existed, you would be able to store an infinite amount of data in a single bit.

You can convert a char[4] to an int and back. It's actually really easy. This code will print abcd.

char str[4] = { 'a', 'b', 'c', 'd'};

// Convert to int

int result = *(int*) str;



char newstr[4];



// Convert back

char * ptr = (char*) &result;

for(int i=0; i<4; i++)

    newstr[i] = ptr[i];



printf("%.4sn", newstr);

Note that I completely ignored termination of the string, and just instructed printf to stop after four characters.

Oh, and never use the gets function. It's dangerous.

edited Nov 13 '18 at 7:04

answered Nov 13 '18 at 6:50

Broman

6,246112241

1

@VijayDubey To do what?
– Broman
Nov 13 '18 at 7:04

@Broman is completely right, its nonsense, what are you trying to do, there is no technical way. Theoretically, you can implement two options - use ascii code of every and each character in the string , or use pointers and print addresses, but addresses of pointers are not exactly the int numbers
– xxxvodnikxxx
Nov 13 '18 at 7:09

Thanks @Broman.
– Vijay Dubey
Nov 13 '18 at 10:58

add a comment |

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There is absolutely no way to convert a string to an int and then back again the way you want.

You can convert a char[4] to an int and back. It's actually really easy. This code will print abcd.

char str[4] = { 'a', 'b', 'c', 'd'};

// Convert to int

int result = *(int*) str;



char newstr[4];



// Convert back

char * ptr = (char*) &result;

for(int i=0; i<4; i++)

    newstr[i] = ptr[i];



printf("%.4sn", newstr);

Note that I completely ignored termination of the string, and just instructed printf to stop after four characters.

Oh, and never use the gets function. It's dangerous.

edited Nov 13 '18 at 7:04

answered Nov 13 '18 at 6:50

Broman

6,246112241

1

@VijayDubey To do what?
– Broman
Nov 13 '18 at 7:04

@Broman is completely right, its nonsense, what are you trying to do, there is no technical way. Theoretically, you can implement two options - use ascii code of every and each character in the string , or use pointers and print addresses, but addresses of pointers are not exactly the int numbers
– xxxvodnikxxx
Nov 13 '18 at 7:09

Thanks @Broman.
– Vijay Dubey
Nov 13 '18 at 10:58

add a comment |

There is absolutely no way to convert a string to an int and then back again the way you want.

You can convert a char[4] to an int and back. It's actually really easy. This code will print abcd.

char str[4] = { 'a', 'b', 'c', 'd'};

// Convert to int

int result = *(int*) str;



char newstr[4];



// Convert back

char * ptr = (char*) &result;

for(int i=0; i<4; i++)

    newstr[i] = ptr[i];



printf("%.4sn", newstr);

Note that I completely ignored termination of the string, and just instructed printf to stop after four characters.

Oh, and never use the gets function. It's dangerous.

edited Nov 13 '18 at 7:04

answered Nov 13 '18 at 6:50

Broman

6,246112241

1

@VijayDubey To do what?
– Broman
Nov 13 '18 at 7:04

@Broman is completely right, its nonsense, what are you trying to do, there is no technical way. Theoretically, you can implement two options - use ascii code of every and each character in the string , or use pointers and print addresses, but addresses of pointers are not exactly the int numbers
– xxxvodnikxxx
Nov 13 '18 at 7:09

Thanks @Broman.
– Vijay Dubey
Nov 13 '18 at 10:58

add a comment |

There is absolutely no way to convert a string to an int and then back again the way you want.

You can convert a char[4] to an int and back. It's actually really easy. This code will print abcd.

char str[4] = { 'a', 'b', 'c', 'd'};

// Convert to int

int result = *(int*) str;



char newstr[4];



// Convert back

char * ptr = (char*) &result;

for(int i=0; i<4; i++)

    newstr[i] = ptr[i];



printf("%.4sn", newstr);

Note that I completely ignored termination of the string, and just instructed printf to stop after four characters.

Oh, and never use the gets function. It's dangerous.

edited Nov 13 '18 at 7:04

answered Nov 13 '18 at 6:50

Broman

6,246112241

There is absolutely no way to convert a string to an int and then back again the way you want.

You can convert a char[4] to an int and back. It's actually really easy. This code will print abcd.

char str[4] = { 'a', 'b', 'c', 'd'};

// Convert to int

int result = *(int*) str;



char newstr[4];



// Convert back

char * ptr = (char*) &result;

for(int i=0; i<4; i++)

    newstr[i] = ptr[i];



printf("%.4sn", newstr);

Note that I completely ignored termination of the string, and just instructed printf to stop after four characters.

Oh, and never use the gets function. It's dangerous.

edited Nov 13 '18 at 7:04

answered Nov 13 '18 at 6:50

Broman

6,246112241

edited Nov 13 '18 at 7:04

answered Nov 13 '18 at 6:50

Broman

6,246112241

answered Nov 13 '18 at 6:50

Broman

6,246112241

answered Nov 13 '18 at 6:50

Broman

6,246112241

1

@VijayDubey To do what?
– Broman
Nov 13 '18 at 7:04

@Broman is completely right, its nonsense, what are you trying to do, there is no technical way. Theoretically, you can implement two options - use ascii code of every and each character in the string , or use pointers and print addresses, but addresses of pointers are not exactly the int numbers
– xxxvodnikxxx
Nov 13 '18 at 7:09

Thanks @Broman.
– Vijay Dubey
Nov 13 '18 at 10:58

add a comment |

1

@VijayDubey To do what?
– Broman
Nov 13 '18 at 7:04

@Broman is completely right, its nonsense, what are you trying to do, there is no technical way. Theoretically, you can implement two options - use ascii code of every and each character in the string , or use pointers and print addresses, but addresses of pointers are not exactly the int numbers
– xxxvodnikxxx
Nov 13 '18 at 7:09

Thanks @Broman.
– Vijay Dubey
Nov 13 '18 at 10:58

@VijayDubey To do what?
– Broman
Nov 13 '18 at 7:04

@Broman is completely right, its nonsense, what are you trying to do, there is no technical way. Theoretically, you can implement two options - use ascii code of every and each character in the string , or use pointers and print addresses, but addresses of pointers are not exactly the int numbers
– xxxvodnikxxx
Nov 13 '18 at 7:09

Thanks @Broman.
– Vijay Dubey
Nov 13 '18 at 10:58

add a comment |

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