Project 7 Euler C

Question

I'm stuck at Question 7 of Project Euler. I have this piece of code.

#include <stdio.h>

int main (void)
{
    int contador = 0, i, n, variavel = 0;
    for (i = 0; contador == 1000; i++)
    {
        for(n = 0; n == i; n++)
        {
            if (i % n == 0)
            {
                variavel = i;
                contador++;
            }
        }
    }
    printf("%d\n", variavel);
}

It always prints 0. Why is that?

---

PS: I wrote 1000 but the answer must be the 10001st prime number.

Answer 1

A for loop consist of 3 parts

for( init ; cond ; step )

When the execution reaches the loop,

1. init is executed.

2. cond is evaluated.

   • If false, break the loop
   • If true, proceed to the next step
3. Execute the body of the loop.
4. Do step(in many cases, this is increment/decrement)
5. Goto step 2

So, in your code, when the execution reaches

for (i = 0; contador == 1000; i++)

i is set to 0. Then the condition contador == 1000 is checked. It is false as contador is initialized to 0. So, the loop breaks and the execution reaches the printf which prints the value of variavel which is 0 and then

return 0;

executes. This ends the execution of your program.

Your inner for loop has a somewhat similar issue. If the condition of the outer for loop is corrected, then the inner for loop executes. n is set to zero and then, the condition n==i is checked. It will be true only when i=0,i.e, it will be true only in the first iteration of the outer for loop.

You need to correct these mistakes.

Answer 2

The reason it is always zero is that the inner loop condition is satisfied in the first iteration:

for(n=0; n==i; n++)

Additionally, your outer loop will never run. It is written to say that it should only loop when contador == 1000, which can never happen as written.