var a = [], b = [];
a.push('a');
a.push('b');
a.push('c');
b = a;
b.push('d');
console.log(a);
Why is a = ["a", "b", "c", "d"] ?, how to avoid this cross reference when doing b=a; I would like just to have 2 separates objects, independants
and at the end a = ["a", "b", "c"], and b = ["a", "b", "c", "d"]
When you assign a to b, you are actually assigning a reference to a. So any change made to b will also affect a. To do what you really wanted, try:
b = [].concat(a);
which will make an actual copy of a.
In JS arrays are also objects (instance of Array object) and as such objects in JS are assigned by reference not by value, so when you do:
b = a
b is now assigned by reference to a, hence it will affect the a array.
To do what you want to do, you can do:
b = a.slice();
Here is an example:
var a = [], b = [];
a.push('a');
a.push('b');
a.push('c');
b = a.slice();
b.push('d');
console.log(a);
Result:
[a, b, c]
Because variables reference objects, they do not copy them.
You can use
var a = [], b = [];
a.push('a');
a.push('b');
a.push('c');
b = a.slice();
b.push('d');
console.log(a);
Because assignment does not copy variable content. (See the related Is this a variable scoping bug in NodeJS or do I just need more sleep question.)
For arrays, you could use concat to create a (shallowly) copied array:
a = [1,2,3]
b = [].concat(a)
b.push(4)
... or sliceas the other answers say.
Once you do b = a both a and b point to the same array in memory; you need to push the elements to both a and b separately, or once you push elements to a you can loop through it and push the same to b, e.g.
for ( var i = 0; i < a.length; i++ ){ b.push( a[i] ); }
