Trying to find prime numbers












2














I'm a first year student in a programming university and my first assignment is to find the sum of prime numbers between 3990000000 and 4010000000. The problem is everything I do, when I run the program it says the sum is 0 with a return value of 25. I've been trying to debug this code but with no luck, could someone help me?



My code is:



#include <stdio.h>

#define STARTNUMBER 3990000000 

#define ENDNUMBER 4010000000



int main() {

  unsigned int num;

  int j, c, flag, sum = 0;

  flag = 1;

  c = 5;

  j = 7;

  for (num = STARTNUMBER; num <= ENDNUMBER; num++) {

    if (num % 2 == 0) { /*if number mod 2 equals zero go to next number*/

      flag = 0;

      break;

    }

    if (num % 3 == 0) { /*if number mod 3 equals zero go to next number*/

      flag = 0;

      break;

    } else

      /*check if number is prime with the sequences 5+6+6...<=sqrt(number) and 7+6+6..<=sqrt(number)*/

      while (c * c <= num && j * j <= num && flag == 1) { 

        if (num % c == 0 || num % j == 0) {

          flag = 0;

          break;

        }

        c += 6;

        j += 6;

      }

    if (flag == 1)

      sum++;

  }

  printf("There are %d prime numbers", sum);

}





c







share|improve this question




















  • 8




    You break out of the loop in your first iteration.
    – tkausl
    Nov 12 at 14:22






  • 1




    Are you looking for the sum of the prime numbers (as you've stated), or just for the number of prime numbers (as you've done in the code)?
    – Giovanni Cerretani
    Nov 12 at 14:32










  • Like @tkausl already said, you need to revisit the topic of for loops and how the break instruction works exactly
    – aPhilRa
    Nov 12 at 14:35






  • 1




    As for the "return value of 25", it is the number of characters printed by printf (which is missing a newline at the end, BTW), since your main is missing a return statement of its own.
    – Arkku
    Nov 12 at 14:38










  • Once you have fixed your breaking logic, you will see that you must reset the values of flag, c and j for each new num. At the moment, you just set these values once at the beginning.
    – M Oehm
    Nov 12 at 14:41
















2














I'm a first year student in a programming university and my first assignment is to find the sum of prime numbers between 3990000000 and 4010000000. The problem is everything I do, when I run the program it says the sum is 0 with a return value of 25. I've been trying to debug this code but with no luck, could someone help me?



My code is:



#include <stdio.h>

#define STARTNUMBER 3990000000 

#define ENDNUMBER 4010000000



int main() {

  unsigned int num;

  int j, c, flag, sum = 0;

  flag = 1;

  c = 5;

  j = 7;

  for (num = STARTNUMBER; num <= ENDNUMBER; num++) {

    if (num % 2 == 0) { /*if number mod 2 equals zero go to next number*/

      flag = 0;

      break;

    }

    if (num % 3 == 0) { /*if number mod 3 equals zero go to next number*/

      flag = 0;

      break;

    } else

      /*check if number is prime with the sequences 5+6+6...<=sqrt(number) and 7+6+6..<=sqrt(number)*/

      while (c * c <= num && j * j <= num && flag == 1) { 

        if (num % c == 0 || num % j == 0) {

          flag = 0;

          break;

        }

        c += 6;

        j += 6;

      }

    if (flag == 1)

      sum++;

  }

  printf("There are %d prime numbers", sum);

}





c







share|improve this question




















  • 8




    You break out of the loop in your first iteration.
    – tkausl
    Nov 12 at 14:22






  • 1




    Are you looking for the sum of the prime numbers (as you've stated), or just for the number of prime numbers (as you've done in the code)?
    – Giovanni Cerretani
    Nov 12 at 14:32










  • Like @tkausl already said, you need to revisit the topic of for loops and how the break instruction works exactly
    – aPhilRa
    Nov 12 at 14:35






  • 1




    As for the "return value of 25", it is the number of characters printed by printf (which is missing a newline at the end, BTW), since your main is missing a return statement of its own.
    – Arkku
    Nov 12 at 14:38










  • Once you have fixed your breaking logic, you will see that you must reset the values of flag, c and j for each new num. At the moment, you just set these values once at the beginning.
    – M Oehm
    Nov 12 at 14:41














2












2








2







I'm a first year student in a programming university and my first assignment is to find the sum of prime numbers between 3990000000 and 4010000000. The problem is everything I do, when I run the program it says the sum is 0 with a return value of 25. I've been trying to debug this code but with no luck, could someone help me?



My code is:



#include <stdio.h>

#define STARTNUMBER 3990000000 

#define ENDNUMBER 4010000000



int main() {

  unsigned int num;

  int j, c, flag, sum = 0;

  flag = 1;

  c = 5;

  j = 7;

  for (num = STARTNUMBER; num <= ENDNUMBER; num++) {

    if (num % 2 == 0) { /*if number mod 2 equals zero go to next number*/

      flag = 0;

      break;

    }

    if (num % 3 == 0) { /*if number mod 3 equals zero go to next number*/

      flag = 0;

      break;

    } else

      /*check if number is prime with the sequences 5+6+6...<=sqrt(number) and 7+6+6..<=sqrt(number)*/

      while (c * c <= num && j * j <= num && flag == 1) { 

        if (num % c == 0 || num % j == 0) {

          flag = 0;

          break;

        }

        c += 6;

        j += 6;

      }

    if (flag == 1)

      sum++;

  }

  printf("There are %d prime numbers", sum);

}





c







share|improve this question















I'm a first year student in a programming university and my first assignment is to find the sum of prime numbers between 3990000000 and 4010000000. The problem is everything I do, when I run the program it says the sum is 0 with a return value of 25. I've been trying to debug this code but with no luck, could someone help me?



My code is:



#include <stdio.h>

#define STARTNUMBER 3990000000 

#define ENDNUMBER 4010000000



int main() {

  unsigned int num;

  int j, c, flag, sum = 0;

  flag = 1;

  c = 5;

  j = 7;

  for (num = STARTNUMBER; num <= ENDNUMBER; num++) {

    if (num % 2 == 0) { /*if number mod 2 equals zero go to next number*/

      flag = 0;

      break;

    }

    if (num % 3 == 0) { /*if number mod 3 equals zero go to next number*/

      flag = 0;

      break;

    } else

      /*check if number is prime with the sequences 5+6+6...<=sqrt(number) and 7+6+6..<=sqrt(number)*/

      while (c * c <= num && j * j <= num && flag == 1) { 

        if (num % c == 0 || num % j == 0) {

          flag = 0;

          break;

        }

        c += 6;

        j += 6;

      }

    if (flag == 1)

      sum++;

  }

  printf("There are %d prime numbers", sum);

}





c




c






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 14:59









chux

80.2k870147




80.2k870147










asked Nov 12 at 14:21









stefnto

111




111








  • 8




    You break out of the loop in your first iteration.
    – tkausl
    Nov 12 at 14:22






  • 1




    Are you looking for the sum of the prime numbers (as you've stated), or just for the number of prime numbers (as you've done in the code)?
    – Giovanni Cerretani
    Nov 12 at 14:32










  • Like @tkausl already said, you need to revisit the topic of for loops and how the break instruction works exactly
    – aPhilRa
    Nov 12 at 14:35






  • 1




    As for the "return value of 25", it is the number of characters printed by printf (which is missing a newline at the end, BTW), since your main is missing a return statement of its own.
    – Arkku
    Nov 12 at 14:38










  • Once you have fixed your breaking logic, you will see that you must reset the values of flag, c and j for each new num. At the moment, you just set these values once at the beginning.
    – M Oehm
    Nov 12 at 14:41














  • 8




    You break out of the loop in your first iteration.
    – tkausl
    Nov 12 at 14:22






  • 1




    Are you looking for the sum of the prime numbers (as you've stated), or just for the number of prime numbers (as you've done in the code)?
    – Giovanni Cerretani
    Nov 12 at 14:32










  • Like @tkausl already said, you need to revisit the topic of for loops and how the break instruction works exactly
    – aPhilRa
    Nov 12 at 14:35






  • 1




    As for the "return value of 25", it is the number of characters printed by printf (which is missing a newline at the end, BTW), since your main is missing a return statement of its own.
    – Arkku
    Nov 12 at 14:38










  • Once you have fixed your breaking logic, you will see that you must reset the values of flag, c and j for each new num. At the moment, you just set these values once at the beginning.
    – M Oehm
    Nov 12 at 14:41








8




8




You break out of the loop in your first iteration.
– tkausl
Nov 12 at 14:22




You break out of the loop in your first iteration.
– tkausl
Nov 12 at 14:22




1




1




Are you looking for the sum of the prime numbers (as you've stated), or just for the number of prime numbers (as you've done in the code)?
– Giovanni Cerretani
Nov 12 at 14:32




Are you looking for the sum of the prime numbers (as you've stated), or just for the number of prime numbers (as you've done in the code)?
– Giovanni Cerretani
Nov 12 at 14:32












Like @tkausl already said, you need to revisit the topic of for loops and how the break instruction works exactly
– aPhilRa
Nov 12 at 14:35




Like @tkausl already said, you need to revisit the topic of for loops and how the break instruction works exactly
– aPhilRa
Nov 12 at 14:35




1




1




As for the "return value of 25", it is the number of characters printed by printf (which is missing a newline at the end, BTW), since your main is missing a return statement of its own.
– Arkku
Nov 12 at 14:38




As for the "return value of 25", it is the number of characters printed by printf (which is missing a newline at the end, BTW), since your main is missing a return statement of its own.
– Arkku
Nov 12 at 14:38












Once you have fixed your breaking logic, you will see that you must reset the values of flag, c and j for each new num. At the moment, you just set these values once at the beginning.
– M Oehm
Nov 12 at 14:41




Once you have fixed your breaking logic, you will see that you must reset the values of flag, c and j for each new num. At the moment, you just set these values once at the beginning.
– M Oehm
Nov 12 at 14:41












1 Answer
1






active

oldest

votes


















0














You are asking for the sum of the prime number, even if your code is just printing how many they are. Assuming you've misunderstood the exercise, I try to show a possible problem of your original question, glimpsing a possible trick in your exercise since the interval is very close to 232.



Assuming also you are on a 64-bit environment, if there are at least two prime numbers in that inteval, the sum is going to be greater than INT_MAX (231 - 1). An int is not sufficient to store a value, and also unsigned int is not sufficient since UINT_MAX is 232 - 1.



Finally, assuming you've solved your problems with the break statements already described in the comments, try to store your sum variable into a unsigned long int, and replace the last part of the loop with 



if (flag==1)

    sum += num;





share|improve this answer























  • Yeah my mistake, by saying the sum of numbers, I meant how many numbers are between the startnumber and endnumber, thus storing the sum variable into unsigned int is enough. As for the use of the break command, doesn't the loop stop and continues to the next number if the command where break is in (in this case those if commands) is false? (flag=0). I've tried removing the break commands and leaving it with the flag command, but it didn't help.
    – stefnto
    Nov 12 at 15:46






  • 2




    You should call that a "count" not a "sum". Variable names are important--you and other humans may have to read the code years later, and might reasonably (and mistakenly) assume that "sum" is a sum.
    – Lee Daniel Crocker
    Nov 12 at 19:21











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1 Answer
1






active

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0














You are asking for the sum of the prime number, even if your code is just printing how many they are. Assuming you've misunderstood the exercise, I try to show a possible problem of your original question, glimpsing a possible trick in your exercise since the interval is very close to 232.



Assuming also you are on a 64-bit environment, if there are at least two prime numbers in that inteval, the sum is going to be greater than INT_MAX (231 - 1). An int is not sufficient to store a value, and also unsigned int is not sufficient since UINT_MAX is 232 - 1.



Finally, assuming you've solved your problems with the break statements already described in the comments, try to store your sum variable into a unsigned long int, and replace the last part of the loop with 



if (flag==1)

    sum += num;





share|improve this answer























  • Yeah my mistake, by saying the sum of numbers, I meant how many numbers are between the startnumber and endnumber, thus storing the sum variable into unsigned int is enough. As for the use of the break command, doesn't the loop stop and continues to the next number if the command where break is in (in this case those if commands) is false? (flag=0). I've tried removing the break commands and leaving it with the flag command, but it didn't help.
    – stefnto
    Nov 12 at 15:46






  • 2




    You should call that a "count" not a "sum". Variable names are important--you and other humans may have to read the code years later, and might reasonably (and mistakenly) assume that "sum" is a sum.
    – Lee Daniel Crocker
    Nov 12 at 19:21
















0














You are asking for the sum of the prime number, even if your code is just printing how many they are. Assuming you've misunderstood the exercise, I try to show a possible problem of your original question, glimpsing a possible trick in your exercise since the interval is very close to 232.



Assuming also you are on a 64-bit environment, if there are at least two prime numbers in that inteval, the sum is going to be greater than INT_MAX (231 - 1). An int is not sufficient to store a value, and also unsigned int is not sufficient since UINT_MAX is 232 - 1.



Finally, assuming you've solved your problems with the break statements already described in the comments, try to store your sum variable into a unsigned long int, and replace the last part of the loop with 



if (flag==1)

    sum += num;





share|improve this answer























  • Yeah my mistake, by saying the sum of numbers, I meant how many numbers are between the startnumber and endnumber, thus storing the sum variable into unsigned int is enough. As for the use of the break command, doesn't the loop stop and continues to the next number if the command where break is in (in this case those if commands) is false? (flag=0). I've tried removing the break commands and leaving it with the flag command, but it didn't help.
    – stefnto
    Nov 12 at 15:46






  • 2




    You should call that a "count" not a "sum". Variable names are important--you and other humans may have to read the code years later, and might reasonably (and mistakenly) assume that "sum" is a sum.
    – Lee Daniel Crocker
    Nov 12 at 19:21














0












0








0






You are asking for the sum of the prime number, even if your code is just printing how many they are. Assuming you've misunderstood the exercise, I try to show a possible problem of your original question, glimpsing a possible trick in your exercise since the interval is very close to 232.



Assuming also you are on a 64-bit environment, if there are at least two prime numbers in that inteval, the sum is going to be greater than INT_MAX (231 - 1). An int is not sufficient to store a value, and also unsigned int is not sufficient since UINT_MAX is 232 - 1.



Finally, assuming you've solved your problems with the break statements already described in the comments, try to store your sum variable into a unsigned long int, and replace the last part of the loop with 



if (flag==1)

    sum += num;





share|improve this answer














You are asking for the sum of the prime number, even if your code is just printing how many they are. Assuming you've misunderstood the exercise, I try to show a possible problem of your original question, glimpsing a possible trick in your exercise since the interval is very close to 232.



Assuming also you are on a 64-bit environment, if there are at least two prime numbers in that inteval, the sum is going to be greater than INT_MAX (231 - 1). An int is not sufficient to store a value, and also unsigned int is not sufficient since UINT_MAX is 232 - 1.



Finally, assuming you've solved your problems with the break statements already described in the comments, try to store your sum variable into a unsigned long int, and replace the last part of the loop with 



if (flag==1)

    sum += num;






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 12 at 15:01

























answered Nov 12 at 14:28









Giovanni Cerretani

581316




581316












  • Yeah my mistake, by saying the sum of numbers, I meant how many numbers are between the startnumber and endnumber, thus storing the sum variable into unsigned int is enough. As for the use of the break command, doesn't the loop stop and continues to the next number if the command where break is in (in this case those if commands) is false? (flag=0). I've tried removing the break commands and leaving it with the flag command, but it didn't help.
    – stefnto
    Nov 12 at 15:46






  • 2




    You should call that a "count" not a "sum". Variable names are important--you and other humans may have to read the code years later, and might reasonably (and mistakenly) assume that "sum" is a sum.
    – Lee Daniel Crocker
    Nov 12 at 19:21


















  • Yeah my mistake, by saying the sum of numbers, I meant how many numbers are between the startnumber and endnumber, thus storing the sum variable into unsigned int is enough. As for the use of the break command, doesn't the loop stop and continues to the next number if the command where break is in (in this case those if commands) is false? (flag=0). I've tried removing the break commands and leaving it with the flag command, but it didn't help.
    – stefnto
    Nov 12 at 15:46






  • 2




    You should call that a "count" not a "sum". Variable names are important--you and other humans may have to read the code years later, and might reasonably (and mistakenly) assume that "sum" is a sum.
    – Lee Daniel Crocker
    Nov 12 at 19:21
















Yeah my mistake, by saying the sum of numbers, I meant how many numbers are between the startnumber and endnumber, thus storing the sum variable into unsigned int is enough. As for the use of the break command, doesn't the loop stop and continues to the next number if the command where break is in (in this case those if commands) is false? (flag=0). I've tried removing the break commands and leaving it with the flag command, but it didn't help.
– stefnto
Nov 12 at 15:46




Yeah my mistake, by saying the sum of numbers, I meant how many numbers are between the startnumber and endnumber, thus storing the sum variable into unsigned int is enough. As for the use of the break command, doesn't the loop stop and continues to the next number if the command where break is in (in this case those if commands) is false? (flag=0). I've tried removing the break commands and leaving it with the flag command, but it didn't help.
– stefnto
Nov 12 at 15:46




2




2




You should call that a "count" not a "sum". Variable names are important--you and other humans may have to read the code years later, and might reasonably (and mistakenly) assume that "sum" is a sum.
– Lee Daniel Crocker
Nov 12 at 19:21




You should call that a "count" not a "sum". Variable names are important--you and other humans may have to read the code years later, and might reasonably (and mistakenly) assume that "sum" is a sum.
– Lee Daniel Crocker
Nov 12 at 19:21


















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