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Clone Arrays with JavaScript

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Believe it or not, there are reasons we use JavaScript frameworks outside of animations and those sexy accordions that people can't do without. The further you get into high-powered JavaScript applications (assuming you're creating true web applications, not websites), the more the need for basic JavaScript functionalities; i.e. JavaScript utilities that have nothing to do with DOM. One of those basic utilities is the ability to clone an array. Quite often I see developers iterating over array items to create their clone; in reality, cloning an array can be as easy as a slice!

The JavaScript

To clone the contents of a given array, all you need to do is call slice, providing 0 as the first argument:

var clone = myArray.slice(0);

The code above creates clone of the original array; keep in mind that if objects exist in your array, the references are kept; i.e. the code above does not do a "deep" clone of the array contents. To add clone as a native method to arrays, you'd do something like this:

Array.prototype.clone = function() {
	return this.slice(0);
};

And there you have it! Don't iterate over arrays to clone them if all you need is a naive clone!

ydkjs-3.png

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Discussion

  1. faruzzy

    This is a great tip! so worth it!

  2. Interesting tip!

  3. Just to note, I use .slice() all the time without passing in ‘zero’ and have never seen an issue. So even though the MDN docs do not say it is optional, it appears to work just fine with no arguments :)

  4. Mark

    It is also much faster, especially for large arrays. Here’s a benchmark I put together:

    http://jsperf.com/loop-vs-slice-copy/3

  5. What is the differences between clonearr=originalarr and clonearr=originalarr.slice(0) ?

  6. jonathan de montalembert

    I don’t understand, why not use clone = myArray?

  7. Stefan

    Mesuutt and Jonathan: doing cloneArr = originalArr will only copy the pointer of the array and not clone it:

    var a = [1,2,3]; // a.length == 3
    b = a; // a.length == 3 and b.length == 3
    b.push(4); // a.length == 4 and b.length == 4 
    
    
    • thanks @stefan, I understand exactly now :)

  8. There’s a subtle bug that can result from adding something to the prototype of Array. If you iterate over arrays with using for...in construct:

    var a = [3,7,9];
    for (var i in a) {
      console.log(i);
    }
    

    Note that the for loop will also iterate over the case when i == “clone”!
    To be safe, you must iterate by using the length property:

    for (var i=0; i<a.length; i++) {
      console.log(i);
    }
    
  9. Eric H.

    Take note: if your array is filled with objects, those objects remain linked:

    var aAll = [{x:0, y:0}, {x:1, y:1}],
    	aAll1 =aAll.slice(0); 
    console.log(aAll[0].x, aAll1[0].x); // 0, 0
    aAll1[0].x += 10
    console.log(aAll[0].x, aAll1[0].x); // 10, 10
    aAll1[0] = {};
    console.log(aAll[0].x, aAll1[0].x); // 10, undefined
    
  10. MikeOne

    So what if I need to create a clone of an array filled with objects? I cannot have the objects to be by reference in the clone. Iteration my only option?

  11. Scott S. McCoy

    No one noticed this empties the array in question? That’s not exactly a clone.

    • Scott S. McCoy

      It’s really lame you cannot delete or edit comments here.

      Use slice() not splice().

  12. Maybe it’s worth mentioning in a comment, this is very similar to how Python copies lists (arrays)

    my_list = [1, 2, "some text"];
    
    // To create a refrence
    my_ref = my_list
    
    // To create a copy
    my_copy = my_list[:]
    
  13. Adding Array.prototype.clone as mentioned in this article is a terrible idea… as it’s only a shallow clone, and if another developer on your team will use your clone method, they can trip up over this.

  14. Hi, thx for solution, exactly what Iwas looking for

  15. Hi.
    I know this post is kinda old, but we were on a discussion about it around here, and we prepared these tests.
    http://jsperf.com/array-content-copy/2
    many interesting thing around here!
    for instance, array.slice is the fastest on google chrome, while in firefox, array.concat was the fastest.
    Another interesting thing is the way the browser optimizes the most verbose test, which is not the slowest one!

  16. a handy way to get a deep copy is:

    var clone = JSON.parse( JSON.stringify( myArray ) );
    
  17. I added this method, using JSON’s parse/stringify methods to that perf.
    It works, but this does not have the best of the performances:

    http://jsperf.com/array-content-copy/7

  18. Massimo
    var clone = myArray.slice(0);
    

    don’t make a clone. Is the same of

    clone=myArray;
    

    every change on clone is reflected on myArray. To make a clone you must copy every element singly.

  19. sean

    not according to this test

    
    

    // bad clone
    var a = [1,2,3,4,5];
    var b = a;
    a.push(99);
    console.log(a); // [1,2,3,4,5,99]
    console.log(b); // [1,2,3,4,5,99]

    // good clone
    var x = [1,2,3,4,5];
    var y = x.slice(0);
    x.push(77);
    console.log(x); // [1,2,3,4,5,77]
    console.log(y); // [1,2,3,4,5]

  20. Yep, that’s not true, Massimo.
    try this out

    var a1= [1, 2, 3, 4];
    var a2= a1.slice(0);
    a2[2]= 'X';
    console.log(a1); // [1, 2, 3, 4] 
    console.log(a2); // [1, 2, "X", 4] 
    
  21. Andy

    This will only work if your array has simple values in it. If your array contains objects (or other arrays!!!) it will not copy them.

    For example

    var a = [ [ 1 ] ];
    var b = a.slice( 0 );
    a[0] === b[0] // true!
    

    Be careful! here is a function you can use if you know your arrays are nested: http://blog.andrewray.me/how-to-clone-a-nested-array-in-javascript/

    • Hi Andy.

      That’s not entirely the case!
      What happens here, is that objects and arrays are passed as reference, and even though, what you were comparing there was their value at [0]!
      If you simply run this, you will understand:

      a[0]= 'x';
      console.log(b[0]); // [1]
      

      By changing a[0], you are NOT altering b[0].
      BUT, if you had done this, instead:

      a[0][0] = 'a';
      console.log(b[0][0]); // "a"
      

      This time it changed the array on “b”, but because it was a reference to the same array/object, therefore, any change into one, will be applied to the other.

      So, yes, we must be always aware of objects being passed from some place to another, in any case :)
      In the example you gave, b is a copy of a, but as a copy, it also copied the pointers/references to the original Array stored on each position.

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