.splice() - how to protect oryginal array?

Question

var a = [1, 2, 3];
var b = a;      // b = [1, 2, 3];

console.log(a); // 1, 2, 3
console.log(b); // 1, 2, 3

b.splice(0, 1);
console.log(b); // 2, 3
console.log(a); // 2, 3 <----------- WHY?

I just needed to copy my oryginal "a" array because I want to stay it as [1, 2, 3] forever. How to pop first element from "b" array without touching the oryginal one? Thanks

`var b = a;` isn't a copy ....they share same array reference — charlietfl, Jan 31 '16 at 02:13

score 1 · Accepted Answer · answered Jan 31 '16 at 02:15

Your code just needs one small fix:

var a = [1, 2, 3];
var b = a.slice();

I'm not sure of the specifics, but when you are assigning arrays or objects to another variable, the array/object is not copied by value, but rather by reference. .slice method duplicates all elements in the array to the new element rather than just saving a reference to the old one.

.splice() - how to protect oryginal array?

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