Javascript references to object

Question

I'm not sure how to explain this question, so I'll illustrate with code.

var bob = {
    "1": ["a", "b"]
}

var jim = bob[1]

jim.shift()

print(bob[1])

Running this with d8, I get an output of [b].

Notice how bob (the object I am referencing from jim) is being changed when I modify jim. I would like to have behavior where modifying jim does nothing to bob. That is, even after the shift(), I'd like bob[1] to still be [a,b] instead of [b]. I'm sure that this is a well-documented part of JS, but I wasn't sure how to search for it. Thanks for your help.

This question in one form or another has come up a lot lately, for some reason. — Pointy, Jul 16 '14 at 23:52
Google's javascript engine through the shell (same thing as node) — goodcow, Jul 16 '14 at 23:53
Look for "JS pass by reference or value", "JS extending objects", etc... — Shomz, Jul 16 '14 at 23:53
For reference: http://stackoverflow.com/questions/122102/what-is-the-most-efficient-way-to-clone-an-object — chris, Jul 16 '14 at 23:54
@chris: That seems to ask for a deep clone, and is about objects not specific to arrays. — Bergi, Jul 16 '14 at 23:58
Oh, I didn't know it had a different name. Do you have doc reference for it? — cookie monster, Jul 17 '14 at 00:03
@Bergi, I didn't see this question being specific to arrays either, just a coincidence that an array was used in the example. — chris, Jul 17 '14 at 00:04
@cookiemonster http://www.sandeepdatta.com/2011/10/using-v8-javascript-shell-d8.html — goodcow, Jul 17 '14 at 00:05

score 2 · Accepted Answer · answered Jul 16 '14 at 23:54

2

Make a copy of bob[1]

var bob = {
    "1": ["a", "b"]
}

var jim = bob[1].slice(0)

jim.shift()

print(bob[1])

answered Jul 16 '14 at 23:54

Patrik Oldsberg

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score 1 · Answer 2 · answered Jul 16 '14 at 23:53

1

Object values in JavaScript are references, not complete objects-as-values. When you assign bob[1] to jim, both jim and bob[1] reference the same object (the array). Changing the array via one reference does not affect the other reference; they're both pointing to the same (changed) array.

answered Jul 16 '14 at 23:53

Pointy

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How can I get the behavior that I want? Do I have to clone the object or something? Or is there a way to remove the reference? – goodcow Jul 16 '14 at 23:54
Yes, cloning the object is a solution. – Dested Jul 16 '14 at 23:54
2

@goodcow yes, you'd have to clone the object yourself. For arrays, you could use `jim = bob[1].slice(0);` - be aware that cloning objects is, in general, a hard problem to solve. You're also better off *avoiding* copying objects when possible, since it's clearly not free (in terms of performance). – Pointy Jul 16 '14 at 23:55
So generally, it's better to clone the fields of an object, not the object itself? – goodcow Jul 17 '14 at 00:04
@goodcow well the problem is that it's got the potential to get really messy. In this case, it's simple, so there's no problem. For more complicated objects, however, what to copy and how to copy it can be hard to figure out; there's no good general-purpose always-works solution, in other words. – Pointy Jul 17 '14 at 00:07

Javascript references to object

2 Answers2