why is the output contains 4 time 0's?

Question

why is the output contains 4 time 0's.The main call again and again until if condition become false and then it should be exit from if block.

#include <stdio.h>
    int main()
    {
        static int i=5;
        if(--i)
        {           
            main();
            printf("%d ",i);
        }
    }

What output were you expecting? Did you try working it out by hand? — Shawn, Sep 17 '19 at 07:45
`printf()` is what comes after the recursive call to `main()`, and thus it gets called when the stack starts unwinding. — Shawn, Sep 17 '19 at 07:52

score 0 · Answer 1 · answered Sep 17 '19 at 07:50

Note the following.

the int i is static.
you are calling main reursively.
In the if condition you have a pre decrement of i

Every time you call main, the value of i will be the same as the previous call. So i will decrement each time. Since this is --i it will be 4 the first time and go to 0.

After the innermost main function returns(i==0), the printf of the main before that will be executed.

But i is static and has a value of 0. So you get 4 zeros printed for each of the main functions.

why is the output contains 4 time 0's?

1 Answers1