Add two arrays without using the concat method

Question

Here is a sample of what I would like to do

function test(r){
  var arr = ['d','e','f'];
  r.push(arr);
  /*
  More Code
  */
  return r;
}
var result = test(['a','b','c']);
alert(result.length);//I want this to alert 6

What I need to do is pass in an array and attach other arrays to the end of it and then return the array. Because of passing by reference I cannot use array.concat(array2);. Is there a way to do this without using something like a for loop to add the elements one by one. I tried something like r.push(arr.join()); but that did not work either. Also, I would like the option of having objects in the arrays so really the r.push(arr.join()); doesn't work very well.

These days, you can use an iterable, and join arrays logically, without any data copy. Check out this code. — vitaly-t
– vitaly-t, Commented Oct 22, 2024 at 10:33

qw3n · Accepted Answer · 2017-02-25 05:44:20Z

62

>>> var x = [1, 2, 3], y = [4, 5, 6];
>>> x.push.apply(x, y) // or Array.prototype.push.apply(x, y)
>>> x
[1, 2, 3, 4, 5, 6]

Alternatively using destructuring you can now do this

//generate a new array
a=[...x,...y];
//or modify one of the original arrays
x.push(...y);

edited Feb 25, 2017 at 5:44

qw3n

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answered Mar 9, 2011 at 1:21

emulbreh

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4 Comments

meouw Over a year ago

Interesting use of apply to push all members of an array onto another array +1

qw3n Over a year ago

I had seen the apply method before, but never really understood it (note to self need to study up on apply), but it certainly makes a concise way of doing what I wanted. How does the second method differ from the first one though? Using the Array.prototype instead of x.push.apply.

emulbreh Over a year ago

Both are equivalent. Calling x.push(y) means "call the function x.push with scope x and the arguments [y]. In this case x.push resolves to Array.prototype.push, thus: Array.prototype.push.apply(x, [y]). I only mentioned the second form as it's sometimes preferable to only reference x once.

Barbz_YHOOL Over a year ago

x.push.apply(x, y) solved it for me, especially the second x even though I don't understand why it's like that

mVChr · Accepted Answer · 2011-03-09 01:29:39Z

2

function test(r){
  var _r = r.slice(0), // copy to new array reference
      arr = ['d','e','f'];

  _r = _r.concat(arr); // can use concat now

  return _r;
}
var result = test(['a','b','c']);
alert(result.length); // 6

answered Mar 9, 2011 at 1:29

mVChr

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1 Comment

meouw Over a year ago

Wrong, the array you return is not the modified array that was passed to the function, it is a copy

Community · Accepted Answer · 2017-05-23 11:46:13Z

1

This is emulbreh's answer, I'm just posting the test I did to verify it. All credit should go to emulbreh

// original array
var r = ['a','b','c'];

function test(r){
  var arr = ['d','e','f'];
  r.push.apply(r, arr);

  /*
  More Code
  */
  return r;
}
var result = test( r );
console.log( r ); // ["a", "b", "c", "d", "e", "f"]
console.log( result === r ); // the returned array IS the original array but modified

edited May 23, 2017 at 11:46

CommunityBot

11 silver badge

answered Mar 9, 2011 at 1:43

meouw

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