Why is this sorting function returning undefined?

Question

The result that's logged to the console (answer) is the correct one, but it's coming out of the function as undefined. What is going on?

let sorted = [];

function reset(){
    sorted = [];
}

function sort(list) {
    let least = list[0];
    for (let i = 0; i < list.length; i++){
        if (list[i] < least ){
            least = list[i];
        }
    }
    sorted.push(least);
    list.splice(list.indexOf(least),1);
    if (list[0] !== undefined){
        sort(list)
    }
    else {
        let answer = sorted;
        reset();
        console.log(answer);
        return answer;
    }
}
let sampleArray = [3,2,1];
sort(sampleArray);

In your if branch you have no return, which means you get the default return value (undefined). — An0num0us
– An0num0us, Commented Apr 17, 2018 at 15:42

Mureinik · Accepted Answer · 2018-04-17 15:41:46Z

1

In the if branch you correctly call sort recursively, but just ignore the returned value instead of returning it. Just return it and you should be fine:

if (list[0] !== undefined) {
    return sort(list); // Here
}
else {
    // Rest of your code...

answered Apr 17, 2018 at 15:41

Mureinik

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Patrick Roberts · Accepted Answer · 2018-04-17 16:00:48Z

0

@Mureinik is correct about the problem. However, more to the point, if you want to make a sorted copy of the array in increasing order, you can easily do so using .slice() and .sort():

let sampleArray = [3, 1, 2];
// sorted array
let sortedArray = sampleArray.slice().sort((a, b) => a - b);

console.log(sortedArray);
// original array preserved
console.log(sampleArray);

(I'm getting a 500 error so here's an alternative demo for now)

answered Apr 17, 2018 at 16:00

Patrick Roberts

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