THis is my code
let a = ["a","b","c"]
let b = "dd"
let c = a
c.push(b)
console.log(a);
console.log(c);But i want that when i log "a" it returns ["a","b","c"] And when i log "c" it returns ["a","b","c", "d"]
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Assigning an array reference to a different variable does not make a copy of the array. – Pointy Feb 20 '20 at 13:29
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Your statements are wrong: It does not log `["a","b","c"]` for `a` and neither `["a","b","c", "d"]` for `c`. – ASDFGerte Feb 20 '20 at 13:30
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1@ASDFGerte I think that last sentence means that that's what the OP *wants*. – Pointy Feb 20 '20 at 13:31
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Why would it ever log `["a","b","c", "d"]` when you're pushing `"dd"` to the array? – Heretic Monkey Feb 20 '20 at 13:32
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I would have said this is a dupe of [Why does changing an Array in JavaScript affect copies of the array?](https://stackoverflow.com/q/6612385/215552) – Heretic Monkey Feb 20 '20 at 13:33
2 Answers
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Array/object values are copied by reference instead of by value. So, if you do let c = a the c points to the same array object as a
You can use Spread Operator (Shallow copy) to make a new copy of the array.
let a = ["a","b","c"]
let b = "dd"
let c = [...a]
c.push(b)
console.log(a);
console.log(c);Be aware this does not safely copy multi-dimensional arrays.
let a = [["a"],["b"]]
let c = [...a]
let b = "dd"
c[0].push(b)
console.log(a);
console.log(c);
// [["a", "dd"], ["b"]]
// [["a", "dd"], ["b"]]
// They've both been changed because they share references
rMonteiro
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2It's helpful to explain *why* your solution achieves the desired result. – Pointy Feb 20 '20 at 13:31
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Try this to make new ref
let a = ["a","b","c"]
let b = "dd"
let c = [...a]
c.push(b)
console.log(a);
console.log(c);
Ahmed Kesha
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