Variable value is not changing

Question

This is my code for calculating primes. Basically I take a number divide by all the previous primes that i calculated to check if it is a prime or not. I was not getting correct answer. I kept changing the code to figure out the mistake. I found that i value is not initializing to 0 whenever it enters the loop. Instead i is 1,2,3. I cannot understand why this is happening. I'm using visual studio c. Please explain why is this happening ???

#include<stdio.h>
#include<conio.h>
int main()
{
int primes[10000];
int i,j,k;
int primecheck;
 primes[0]=2;

k=1;
for(j=3;;j+=2)
{
  primecheck = 1;
  i=0;
    for(i=0;i<k;i++);
   {

       printf("%d\n%d\n%d\n",i,j,j%primes[i]);// this helped me figure 
                                              // out the problem
       if(j%primes[i] == 0)
       { 
           primecheck = 0;
           printf (" %d, %d\n",j,primes[i]);
       }
       printf("%d\n",k);

   }
   if (primecheck == 1)
   {
       primes[k]=j;
       printf("%d prime is %d\n",k+1,primes[k]);
       k++;
       if(k == 5)
       {
           break;

       }
   }
}
printf("%d",primes[k-1]);
getch();

}

score 0 · Accepted Answer · answered Oct 21 '16 at 05:43

0

Don't use the dummy loop.

for(i=0;i<k;i++);

Remove the semicolon at the end and try. For current code its not entering the below the section under above for loop for any iteration, it is treated as a dummy loop instead. So just try

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

}

answered Oct 21 '16 at 05:43

Stubborn

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Similar answer : http://stackoverflow.com/questions/37482904/why-didnt-the-compiler-warn-me-about-an-empty-if-statement – AlphaGoku Oct 21 '16 at 06:23
@Vector9 For a different question. – Stubborn Oct 21 '16 at 06:26
Yes. But the same problem that the user overlooked – AlphaGoku Oct 21 '16 at 06:26
@Vector9 that's true man. – Stubborn Oct 21 '16 at 06:31

Variable value is not changing

1 Answers1