Is destructing assigning in JavaScript shallow or deep copy?

Question

I'm confused by the destructing assigning in JavaScript about shallow and deep copy. For example,

const obj = {key:{}, value:{}}
let {key} = obj
key = {msg: 'hello'}

Is the value of key in the sample above a shallow or deep copy of the key in obj ? What is the value of key should be?

Answer 1

 let {key} = obj

… is the same as saying:

let key = obj.key

It is a shallow copy.

---

 key = {msg: 'hello'}

… overwrites the reference to the object that also exists in obj.key with a reference to a new object.

This renders the previous line pointless as nothing was done with the value before it was overwritten.

Answer 2

It's shallow, though there's some nuance.

This:

let {key} = obj;

has exactly the same result as this:

let key = obj.key;

Eg, all it does is copy the value of obj.key into key , making both of them point to the same object — a shallow copy, if the word "copy" really applies. (Normally I think of a "shallow copy" as copying multiple properties, not just one, like Object.assign does.)

You can reach deeper into the object graph you're retrieving from by using nested destructuring:

 const obj = { a: { b: { c: { d: "hi", }, }, }, }; const {a: {b: {c}}} = obj; console.log(cd); // "hi"

but at the end of the day it's just like an assignment, so it's just grabbing the object reference (if the value is an object reference), not a copy of the object:

 const obj = { a: { b: { c: { d: "hi", }, }, }, }; const {a: {b: {c}}} = obj; console.log(cd); // "hi" console.log(obj.abc.d); // "hi" c.d = c.d.toLocaleUpperCase(); console.log(cd); // "HI" console.log(obj.abc.d); // "HI"

---

But , even if it were a deep copy, your code still wouldn't change anything in obj , because you're changing the value of what's in key (an object reference), not the state of the object that reference refers to. After let {key} = obj , key is completely disconnected from obj.key other than that they both refer to the same object. If you change the value in key , the variable , it has no effect on the object it refers to. If you're goign to do key =... , there's no point to let {key} = obj at all, just use let key .