Why am I getting 'Floating point exception: 8'

Question

I'm trying to calculate all the prime numbers from 0 - 100 and I'm getting a floating point exception, could anyone tell me why? (If it helps I'm using gcc)

#include <stdio.h>

int main(void)
{
  int nums[100], i;

  for(i=0;i<100;i++)
    nums[i] = i;

  int j,k,l,z;

  for(i=1;i<100;i++)
    for(j=2;j<100;j++)
      if((nums[i] % nums[j]) == 0)
       {
        nums[j] = 0;
       }

  for(i=0;i<100;i++)
    if(nums[i] != 0)
      break;

  for(z=0;z<100;z++)
    {  
      for(k=i;k<100;k++)
       for(l = (k+2);l < 100;l++)
         if((nums[k] % nums[l]) == 0)
           nums[k] = 0;
    }

  for(i=0;i<100;i++)
    if(nums[i] != 0)
      printf("%d,",nums[i]);

  printf("\n");

  return 0;
}

Answer 1

Well, it's really hard to understand what your code is doing. But still

for(i=1;i<100;i++)
    for(j=2;j<100;j++)
        if((nums[i] % nums[j]) == 0)
        {
            nums[j] = 0;
        }

After this, many values of nums will be 0.(You can print and check)

So, Later when you are doing

for(z=0;z<100;z++)
{  
  for(k=i;k<100;k++)
   for(l = (k+2);l < 100;l++)
     if((nums[k] % nums[l]) == 0) //Part where division by 0 occurs
       nums[k] = 0;
}

There will be a division by 0, which is giving the floating point exception

Edited

Infact, there will be a floating point exception in the first two for loops only.. When i=2 and j=2, nums[2] will get updated to value 0. Then later when for i=4 and j=2. There will be a division by 0, because num[2] is already 0, thus causing the floating point exception