Why doesn't this recursive javascript function return the correct value

Question

Why does this function return undefined?

The interior function returns the correct value.

function arraySum(i) {

    // i will be an array, containing integers, strings and/or arrays like itself.
    // Sum all the integers you find, anywhere in the nest of arrays.

    (function (s, y) {
        if (!y || y.length < 1) {
            //console.log(s);
            // s is the correct value
            return s;
        } else {
            arguments.callee(s + y[0], y.slice(1));
        }
    })(0, i);
}

var x = [1, 2, 3, 4, 5];
arraySum(x);

OK, so I rewrote it with a named function and now it's working. function arraySum(i) { function acc (s, y) { if (!y || y.length < 1) { //console.log(s); // s is the correct value return s; } else { return acc(s + y[0], y.slice(1)); } } return acc(0,i); } var x = [1, 2, 3, 4, 5]; console.log(arraySum(x)); — hiester
– hiester, Commented Oct 6, 2013 at 2:31
I recommend not using arguments.callee or arguments.caller; it will be removed in ECMAScript 6, and is already not available in ECMAScript 5 Strict Mode. — Joe Simmons
– Joe Simmons, Commented Oct 6, 2013 at 2:39
"OK, so I rewrote it with a named function and now it's working" - Yes, but it works more because you added the missing return keyword in two places than because you introduced a function name. In any case, if you've figured out a solution you should post it as an answer - it's acceptable here at StackOverflow to answer your own questions. — nnnnnn
– nnnnnn, Commented Oct 6, 2013 at 2:42

chubbsondubs · Accepted Answer · 2013-10-06 02:27:05Z

1

Change it to

return arguments.callee( s + y[0], y.slice(1))

Or just use reduce :-) :

[1,2,3,4].reduce( function(sum, x) { return sum + x; }, 0 );

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce

answered Oct 6, 2013 at 2:27

chubbsondubs

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3 Comments

user2736012 Over a year ago

A return is also needed before the IIFE. return (function (s, y) {...

Joe Simmons Over a year ago

Why do you pass a second argument to reduce? It is unnecessary.

chubbsondubs Over a year ago

It's the initial value to pass to the first arg to the callback. It's for clarity.

thefourtheye · Accepted Answer · 2013-10-06 02:59:33Z

0

If what you said in the code's comment is true, this is what you need.

function arraySum(i) {

    // i will be an array, containing integers, strings and/or arrays like itself
    // Sum all the integers you find, anywhere in the nest of arrays.

    return (function (s, y) {
        if (y instanceof Array && y.length !== 0) {
            return arguments.callee(arguments.callee(s, y[0]), y.slice(1));
        } else if (typeof y === 'number') {
            return s + y;
        } else {
            return s;
        }
    })(0, i);
}

Output

var x = [1, 2, 3, 4, 5];
console.log(arraySum(x));
x = [1, 2, [3, 4, 5]];
console.log(arraySum(x));
x = [1, "2", 2, [3, 4, 5]];
console.log(arraySum(x));
x = [1, "2", [2, [3, 4, 5]]];
console.log(arraySum(x));

answered Oct 6, 2013 at 2:59

thefourtheye

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1 Comment

hiester Over a year ago

I should have removed those comments. That's the part of the quiz I was still working on.

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