In for loop, in C language, variable int i has different value then expected

Question

Can you please help me why int i becomes -4912 (see the picture) when it should start from 1? Thank you very much.

int main(void)
{
    long long n = 4003600000000014;
    int i;
    int num;
    for (i = 1; i < 15; i += 2)
    {
        int digit = (n / pow(10, i));
        printf("%i\n", digit);
    }
}

enter image description here

When do you check the value of i? Before it's assigned to by the for loop? What happens if you step once in the debugger? — Some programmer dude
– Some programmer dude, Commented Apr 14, 2021 at 7:28
Is an int on your system large enough to hold the values returned by that equation? — Shawn
– Shawn, Commented Apr 14, 2021 at 7:31
Seems that i isn't initialized yet. Make sure the break point is at line 19 — 4386427
– 4386427, Commented Apr 14, 2021 at 7:31
The first chapters of any C programming book typically addresses numerical limits of integers. Do you think that 4003600000000014 / 10^1 will fit inside an int? — Lundin
– Lundin, Commented Apr 14, 2021 at 7:32
line 3 doesnt initilize i so it gets a random stack value until it is set to 1 in the for-loop. — Eben
– Eben, Commented Apr 14, 2021 at 7:33

Paul Ogilvie · Accepted Answer · 2021-04-14 07:41:00Z

3

In your picture, the debugger is at the for loop. At that point the for loop initialization has not happened yet and i will have an arbitrary value.

Tell the debugger to "step". The initialization of the for loop will then be executed and you will see thati is now 1.

answered Apr 14, 2021 at 7:41

Paul Ogilvie

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Tanguy Bron · Accepted Answer · 2021-04-14 07:37:12Z

-1

you should use %d in your printf and not %i I think.

Just try to initialise i : int i = 0; for example

or declare i in your loop : for(int i = 1; i < n; i++)

edited Apr 14, 2021 at 7:37

answered Apr 14, 2021 at 7:34

Tanguy Bron

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2 Comments

Shawn Over a year ago

They're equivalent. %d is more common, but %i works fine.

Tanguy Bron Over a year ago

yeah, with some researches i saw it. thanks

Eben · Accepted Answer · 2021-04-14 07:41:10Z

-1

Here you can read more about uninitialized variables. But as a short summary. A variable gets whatever random value is in the memory location until it is set. C/C++ does not initialize variables to 0 by default, it is the responsibility of the programmer. Uninitialized variables are sources to many bugs and are usually pointed out by compilers if the warnings are configured to be on.

answered Apr 14, 2021 at 7:41

Eben

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