CODEWITHSUNDEEP
javascript
박한빈
박한빈

Reputation: 61

Deleting a key for an object inside a replicated object also deletes existing objects

var origin = { a: 1, b: { c: 2, d: 3 } }; // origin object

var copy_obj = { ...origin } // or Object.assign({}, origin.b)

delete copy_obj.b.c; // delete copy object

console.log(origin) // { a: 1, b: { d: 3 } };

I'm studying the invariant nature of things.

However, the above example shows that you want to delete the b.c. element of copy_obj. But the origin is also deleted.

Why is this happening?

Upvotes: 6

Views: 64

Answers (1)

Subesh Bhandari
Subesh Bhandari

Reputation: 1122

Object.assign({}, origin.b) or spread syntax {...obj} performs a shallow copy. So the object stored in property b of origin is not cloned i.e it maintains a reference to the object being nested.

Use the following to make a deep clone.

JSON.parse(JSON.stringify(obj))


var origin = { a: 1, b: { c: 2, d: 3 } }; // origin object

var copy_obj = JSON.parse(JSON.stringify(origin));

delete copy_obj.b.c; 

console.log(origin)

Note: This only works for objects that can be serialized

For example:

// Object having functions
const obj= {
  a: 1,
  b: () => {return 2}
}


const copy = JSON.parse(JSON.stringify(obj));

// property `b` won't be copied as it is a function and can't be serialized.
console.log(copy)

Upvotes: 8

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