`this` becomes window in super constructor even using `bind`

Question

I'm new to Javascript. When I did some objective oriented programming, I want to invoke constructor of super class in a sub-class:

function P(p1, p2) {
  alert(this);
  this.v1 = p1;
  this.v2 = p2;
}

var middleObj = function() {};
middleObj.prototype = P.prototype;

var C = function(p1, p2) {
    P.bind(this, p1, p2);
    P();
}
C.prototype = new middleObj();
C.prototype.constructor = C;

var objChild = new C(1,2);
alert(objChild instanceof P);
alert(objChild.v2);

If I call P() directly in C(), this turns to window, and I can understand that. However, even if I bind this to C(), still I can not get the proper value of this. As a result, the second alert will show undefined because v2 is not initialized properly.

If you look more closely at the answers in the linked dupetarget, you'll see that they always use the return value of `bind` (and they explain why). — T.J. Crowder, Aug 19 '17 at 11:01

Suren Srapyan · Accepted Answer · 2017-08-19T11:00:24.580

1

P.bind() will return a function, you need to assign it to your P again. Like this

 P = P.bind(this, p1, p2)

But in this case I suggest you to use just P.call(this, p1, p2);

function P(p1, p2) {
  alert(this);
  this.v1 = p1;
  this.v2 = p2;
}

var middleObj = function() {};
middleObj.prototype = P.prototype;

var C = function(p1, p2) {
    P.call(this, p1, p2);
}
C.prototype = new middleObj();
C.prototype.constructor = C;

var objChild = new C(1,2);
alert(objChild instanceof P);
alert(objChild.v2);

edited Aug 19 '17 at 11:00

answered Aug 19 '17 at 10:58

Suren Srapyan

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Thanks for clarifying. It is solved – John Doe Aug 19 '17 at 11:08

`this` becomes window in super constructor even using `bind`

1 Answers1