If:
var x = [1, 2, 3];
var y = [4, 5, 6];
var z = x;
and then if z[2] = y[0];
Why is it that console.log(x); is [1, 2, 4] and not [1, 2, 3]?
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Because you're referencing the same array.Dave Newton– Dave Newton2018-03-30 19:20:03 +00:00Commented Mar 30, 2018 at 19:20
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Memory references!Ele– Ele2018-03-30 19:20:07 +00:00Commented Mar 30, 2018 at 19:20
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stackoverflow.com/questions/518000/…geoidesic– geoidesic2018-03-30 19:23:43 +00:00Commented Mar 30, 2018 at 19:23
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3 Answers
When you do var z = x; you are no creating a new array, entirely separate from x, you are simply creating a reference to the original array. Hence, the change happens in both.
If you want to create a new object, you can use the new ES6 spread operator
var z = {...x};
Have a look at this answer for a more in-depth explanation of passing by reference and by value.
Cuz the 'z' variable is a pointer to the same array that 'x' point to.
1 Comment
Munim Munna
Please explain your answer so that everyone understands. Answers with explanations are helpful and appreciated by all.
An array in JavaScript is also an object and variables only hold a reference to an object, not the object itself. Thus both variables have a reference to the same object. So changing made through one variable reflected in other as well.
var x = [1, 2, 3];
var y = [4, 5, 6];
var z = x;
z[2]=y[0];
console.log(x);
var w=Object.assign([],x);
w[0]=y[1];
console.log(x);
console.log(w);Look at the example. If you want to change in a new variable and don't want reflected that change in original one than use Object.assign.
