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javascriptarraysmethods
Bill P
Bill P

Reputation: 95

JavaScript function containing array methods not returning desired result

I'm a JavaScript beginner and I've written the snippet below which in my mind should return an array: [5, 4, 3, 2, 1], but instead it returns [5]. I realize there are better ways to achieve the result, but what I'm trying to understand is why my solution doesn't work.

function reverseArray(array) {
  for (var i = 3; i >= 0; i--)
    array.concat(array.slice(i,i+1));
  for (var i = 3; i >= 0; i--)
    array.shift();
  return array;
};

var arrayValue = [1, 2, 3, 4, 5];
reverseArray(arrayValue);
console.log(arrayValue);
// JS Bin returns: [5]

The shift method appears to work, but the concat/slice code doesn't seem to alter the array. Why not? When I test the line as stand alone code it works fine. Any explanation would be greatly appreciated. Thanks!

Upvotes: 0

Views: 58

Answers (3)

artsiomshushkevich
artsiomshushkevich

Reputation: 23

There's answer how work concat and shift functions. Here's small snippet how to make this function easily by algorithmic approach if you don't want to create local copy of existing array.

function reverseArray(array) {
  var i = 0;
  var j = array.length - 1;

  while (i < j) {
    var tempArrayElement = array[i];
    array[i] = array[j];
    array[j] = tempArrayElement;

    i++;
    j--;
  }
}

Upvotes: 0

Bergi
Bergi

Reputation: 664277

What Mike said: array.concat(…) does not modify array, it creates a new array which is then ignored.

It seems you only want to append a single value - that's what push does:

function reverseArray(array) {
  for (var i = 3; i >= 0; i--)
    array.push(array[i]);
  for (var i = 3; i >= 0; i--)
    array.shift();
  return array;
}

You probably also want to use a variable value derived from array.length instead of the constant 3.

Upvotes: 2

Mike Cluck
Mike Cluck

Reputation: 32511

concat and slice creates a new array without changing the original. shift modifies the array it acts on.

Basically, you're generating a bunch of new sub arrays with concat but you aren't doing anything with them.

Upvotes: 2

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