15

On page load I am creating two Javascript Objects, objDemo1 and objDemo1Backup where the latter is simply an exact copy of the first.

e.g.

objDemo1 { 
    sub_1 = { something: 123, somethingElse: 321 },
    sub_2 = { something: 456, somethingElse: 654 }
}

I can modify the values in sub_ as well as add / delete new sub_'s but the only object I am editing is objDemo1. i.e. I never change objDemo1Backup

I have a reset button that when clicked will reset objDemo1 back to what it was when the page originally loaded (i.e. objDemo1 = objDemo1Backup). This is where I am having the issue..

How do I set objDemo1 to objDemo1Backup?

I have tried:

objDemo1 = objDemo1Backup;

and

objDemo1 = null;
var objDemo1 = objDemo1Backup;

...as well as similar variations but nothing seems to work. Any ideas?

  • Note: I can confirm that at the point of resetting, objDemo1Backup is exactly the same as it was when I created it and objDemo1 has changed.
  • My code is definetly hitting the "reset" functionality, where I've tried the objDemo1 = objDemo1Backup... I just cannot figure out the syntax to replace the object.
Adam Tomat
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    I'm going to assume that what is happening is that since both `objDemo1` and `objDemo1Backup` point to the same object, when you change one, both are changed. This may not seem intuitive, but it is the way javascript works. Google **"javascript clone object"** for more details. – Shmiddty Dec 18 '12 at 16:46
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    You're playing with object reference, not cloning it. You should clone your object instead of assign it using = – LittleSweetSeas Dec 18 '12 at 16:47

3 Answers3

11

I'm using angularjs and it took me some time to find out how to copy an object to another object. Normally you'll get an objects clone by calling clone or here in angular copy:

var targetObj = angular.copy(sourceObj);

This gives you a new cloned instance (with a new reference) of the source object. But a quick look into the docs reveals the second parameter of copy:

angular.copy(sourceObj, targetObj)

This way you can override a target object with the fields and methods of the source and also keep the target objects reference.

schmijos
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7

In JavaScript objects are passed by reference, never by value. So:

var objDemo, objDemoBackup;
objDemo = {
    sub_1: "foo";
};
objDemoBackup = objDemo;
objDemo.sub_2 = "bar";
console.log(objDemoBackup.sub_2);   // "bar"

To get a copy, you must use a copy function. JavaScript doesn't have one natively but here is a clone implementation: How do I correctly clone a JavaScript object?

var objDemo, objDemoBackup;
objDemo = {
    sub_1: "foo";
};
objDemoBackup = clone(objDemo);
objDemo.sub_2 = "bar";
console.log(objDemoBackup.sub_2);   // undefined
Community
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Halcyon
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-1

You can use Object.assign.

ObjectConstructor.assign(target: T, source: U): T & U

It takes up two parameters: target and source. When function completes, all internals of target object will be appended with source one.

ivanjermakov
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