Variable re-assignment of object reference leaves other object unaltered (no “transitive” assignment)

Question

I have an object that I want to replace.

var obj1 = { x: "a" };
var ref = obj1;
var obj2 = { y: "b" };

obj1 = obj2;

This results in ref being equivalent to { x: "a" }, but I want it to get changed as well to point to obj2 to get ref being equivalent to { y: "b" }.

I basically want to replace the reference to an object. So when I do obj1 = obj2, I want all references to obj1 to get replaced with a reference to obj2 — Chris, Jan 23 '17 at 19:54
@ValLeNain That's what I already thought, but you never know. There are often some workarounds I am still not aware of. Apparently not this time :) — Chris, Jan 23 '17 at 20:00
Closely related: [Why are changes in an object no longer reflected after reassignment?](/q/27476421/4642212). — Sebastian Simon, Jan 29 '23 at 13:10

score 3 · Answer 1 · edited May 23 '17 at 11:45

3

Not possible. JS passes objects by a copy of the reference, so in the step var ref = obj1 you're not actually assigning a reference pointer like you would in a language like C. Instead you're creating a copy of a reference that points to an object that looks like {x: 'a'}.

See this answer for other options you have: https://stackoverflow.com/a/17382443/6415214.

edited May 23 '17 at 11:45

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answered Jan 23 '17 at 19:54

Charles

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Md Johirul Islam · Answer 2 · 2017-01-23T20:10:41.327

You can try to change all the fields of obj1 with the fields of obj2

var obj1 = { x: 'a' };
var ref = obj1;
var obj2 = { y: 'b' };
obj1.x = obj2.y;
console.log(ref) // Print {x, 'b'}

If you want to add {y,'b'} you can follow the approach

obj1.y=obj2.y
console.log(obj1); prints {x: "b", y: "b"}
console.log(ref); prints {x: "b", y: "b"}

If you instead want to delete obj1.x you can do something like this

delete obj1.x;
console.log(ref) prints {y:'b'}

Variable re-assignment of object reference leaves other object unaltered (no “transitive” assignment)

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