从int返回函数不返回任何内容时，c中出现意外输出

Question

    #include <stdio.h>

    int foo(int a)
    {
        int i;
        for(i=2;i<=a;i++)
        {
            if(i%5==0)
            {
                return;
            }
        }
    }

    int main()
    {
        int c = foo(10);
        printf("%d",c);
    }

why is 5 getting printed when it is not even mentioned what to return?

   #include <stdio.h>

    int foo(int a)
    {
        int i;
        for(i=2;i<=a;i++)
        {
            return; //no variable is attached with return
        }
    }
    int main()
    {
        int c = foo(10);
        printf("%d",c);
    }

and this one is returning 2. which is the first value of i when the loop breaks due to return statement. but where is it mentioned that the function has to return i ??

Code executed in Linux.

Answer 1

Not returning a value from a function designed to return an int invokes Undefined Behavior . Anything can happen. Quote from the C11 standard:

6.9.1 Function definitions

[...]

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

Answer 2

You're not returning an Integer in a method designed to return an Integer, there must be a catch that is returning the last integer created.

If you don't want to return anything, simply return void.