why the function returns value to the previous function when it is not returned in the code?

Question

why the following code is returning the output even if I am not returning the output of variable a to the previous function.

int fact(int n)
{
    int a;
    if (n <= 1) 
        return 1;
    else
        a = n*fact(n-1);
}
int main()
{
    int c=fact(5);
    printf("%d",c);
}

Welcome to the wonderful world of [*undefined behavior*](https://en.wikipedia.org/wiki/Undefined_behavior). In short, if you declare a function to return a value, you *must* return a value, anything else is wrong. — Some programmer dude, Jul 24 '18 at 05:32

score 2 · Answer 1 · answered Jul 24 '18 at 05:35

2

Quoting C11, chapter §6.9.1

If the } that terminates a function is reached, and the value of the function call is used by the caller, the behavior is undefined.

So, your program exhibits undefined behavior.

answered Jul 24 '18 at 05:35

Sourav Ghosh

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why the function returns value to the previous function when it is not returned in the code?

1 Answers1