Javascript Array.push method issue

Question

I have the following code:

function build_all_combinations(input_array){
   array = [1,2,3]
   limit = array.length - 1
   combo = []

   for(i = 0; i<= limit; i++){
      splice_value = array.splice(0,1)
      push_value = splice_value[0]
      array.push(push_value)
      console.log(array) 
      combo.push(array) 
   }
   console.log(combo) 
}

Which outputs:

[2, 3, 1]
[3, 1, 2]
[1, 2, 3]
[[1, 2, 3], [1, 2, 3], [1, 2, 3]]

The last line should be: [[2, 3, 1],[3, 1, 2],[1, 2, 3]]

I'm obviously not grokking something about the way the array is working. Each individual array is correct, but when I go to push them to the combo array, something is failing along the way. What is it?

Are those variables defined somewhere else? what about i? it looks like it's global. Try making i local with var -> for(var i=0... — elclanrs
– elclanrs, Commented Jun 27, 2012 at 16:42

Peter Bratton · Accepted Answer · 2012-06-27 16:43:07Z

5

Each time you are pushing the same array into the combo array; ie, all the references are pointing to the same instance in memory. So when you update one, you've in reality updated them all.

If you want separate references, you'll need to create separate arrays.

answered Jun 27, 2012 at 16:43

Peter Bratton

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2 Comments

Noah Clark Over a year ago

I can use array.new() but then how do I set it equal to another array without setting the same instance in memory?

Pointy Over a year ago

@NoahClark easiest way to make a copy of array is var copy = old.slice(0);

jackwanders · Accepted Answer · 2012-06-27 17:04:19Z

3

shift and slice are your friends here:

var array = [1,2,3];
var combo = []
for(var i = 0; i<array.length; i++){
  array.push(array.shift());  // remove first element (shift) and add it to the end (push)
  combo.push(array.slice());  // add a copy of the current array to combo
}

DEMO

answered Jun 27, 2012 at 17:04

jackwanders

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1 Comment

Noah Clark Over a year ago

I've added this to the solution!

Jashwant · Accepted Answer · 2012-06-27 17:04:50Z

2

jordan002 has given the explanation for this behaviour.

Here's the workaround.

array = [1,2,3]
limit = array.length - 1
combo = []
for(i = 0; i<= limit; i++){
 splice_value = array.splice(0,1)
 push_value = splice_value[0];
 array.push(push_value);
 console.log(array);
 combo.push(array.concat([]));
}
console.log(combo);

You can store a copy in temp variable.

array = [1,2,3]
limit = array.length - 1
combo = []
for(i = 0; i<= limit; i++){
 splice_value = array.splice(0,1)
 push_value = splice_value[0];
 array.push(push_value);
 console.log(array);
 var temp = array.slice();  
 combo.push(temp);
}
console.log(combo)

Reference

edited Jun 27, 2012 at 17:04

answered Jun 27, 2012 at 16:48

Jashwant

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2 Comments

Noah Clark Over a year ago

This works! Would you be able to use array.new some how instead of concat? That seems more explicit about what you're doing.

Jashwant Over a year ago

I did the same few hours back :)

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