JAVASCRIPT- How to prevent object from reflecting changes to its all references

Question

This is the code:

var master = [];
var obj = {};

obj.name = "A";
master.push(obj);
console.log(master);

obj.name = "B";
master.push(obj);
console.log(master);

Following is the output on console:

The name property should be A at one place and B at another place

But both properties are displayed as B only.

How to avoid this overriding?

You have a reference, not a copy. – epascarello Mar 20 '17 at 14:28 — epascarello, Mar 20 '17 at 14:28

score 0 · Answer 1 · answered Mar 20 '17 at 14:31

That's what you wanted to do. First define a type, then instantiate 2 objects of it and at the end put them in an array and show the array with them and their value.

obj = function(name) {
  this.name = name;
}

var a = new obj("A");

var b = new obj("B");

var master = [a,b];

console.log(master);

JAVASCRIPT- How to prevent object from reflecting changes to its all references

1 Answers1