Why does recursion return the first call in the stack and not the last?

Question

I cant for the life of me figure out why this returns 0 rather than 5. "i" keeps getting incremented before it hits the last return statement, however it always returns 0, from the first call in the stack. I would think that since the most recent call on the stack hits the return in the block "i == 5" first it would return and print 5.

Returns 0

public static void main(String[] args) {
        System.out.println(incrementI(0));
}       


public static int incrementI(int i) {
    if (i == 5){
        return i;           
    } else {
         incrementI(i + 1);
    }

    return i;       
}

Returns 5

public static int incrementI(int i) {
    if (i == 5){
        return i;           
    } else {
        return incrementI(i + 1);
    }               
}

What's something() ?? Is that supposed to be incrementI()? — Pointy
– Pointy, Commented Oct 10, 2016 at 0:06
Yes, something is suppose to be inrementI(), quick typo, Thanks! — NightSkyCode
– NightSkyCode, Commented Oct 10, 2016 at 0:44

Pavel · Accepted Answer · 2016-10-10 00:23:00Z

3

Imagine you wrote such function:

public static int square(int x) {
    pow(x, 2);
    return x;
}

It will compute the square of x. But then computed square will not affect anything and the function will return just x.

Now look carefully at your code

if (i == 5){
    return i;           
} else {
    something(i + 1);
}
return i;

If i is not 5, something(i + 1) will be called. Then it will return some value but this value will not affect to anything. And then incrementI function will return just i, in your case it's 0.

answered Oct 10, 2016 at 0:23

Pavel

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2 Comments

NightSkyCode Over a year ago

something is suppose to be inrementI(), quick typo!

Solomon Slow Over a year ago

@NightSkyCode, Looks like you're missing Pavel's point. The incrementI(i) function in your first example always returns i: It immediately returns 5 if i==5, otherwise it does some stuff, and then it returns i. It doesn't matter what "some stuff" is. It doesn't matter whether "some stuff" is a function call named something() or whether it is a recursive call to incrementI(). Either way, it does whatever, and then it ignores the result of doing whatever, and then it returns i.

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