CODEWITHSUNDEEP
cfor-loop
Uclydde
Uclydde

Reputation: 1414

For-loop not working as intended

When I run this code, the printf() function seems to give a random large number, as if it is calling an array that is out of bounds. What is going on here?

#include <stdio.h>
#include <math.h>

int main(void) 
{
    int test_num = 1000;
    int factors[16];

    for(int i = 1, j = 0; i < test_num; i++, j++) {
        if(test_num % i == 0)
            factors[j] = i;
    }

    printf("%d", factors[2]);

    return 0;
}

Upvotes: 1

Views: 139

Answers (3)

static_cast
static_cast

Reputation: 1178

When j == 2, i == 3 and 1000 % 3 does not equal 0. It equals 1. Not passing your if statement condition. Therefore factors[2] will be undefined (since you didn't initialize your array). Hence the large number.

Upvotes: 0

Mad Physicist
Mad Physicist

Reputation: 114508

Most likely, the problem is that you are incrementing j even when you don't assign i.

The sequence of factors you get is 1, 2, 4, 5, 8, 10, ... You probably want to assign those to the indices 0-5 (inclusive), not 0, 1, 3, 4, 7, 9, etc.

Change your loop as follows:

for(int i = 1, j = 0; i < test_num && j < 16; i++) {
    if(test_num % i == 0) {
        factors[j] = i;
        j++;
    }
}

The main point is only to increment j when i fits your criterion. You also want to make sure that you don't go out of bounds (&& j < 16).

Upvotes: 3

jh314
jh314

Reputation: 27812

If you print out i and j in the loop, notice that j is never 2.

Thus, factors[2] is never initialized, so you will print out junk.

Upvotes: 0

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