Passing Arguments in javascript

Question

I want to understand the following behaviour because the explanation on this site javascript garden is not enough for me.

It would be much appreciated if you could give me a clear explanation about the questions which are in the inline comments.

Here the example:

function Foo() {}

Foo.prototype.method = function(a, b, c) {
    console.log(this, a, b, c);
};

Foo.method = function() {
    Function.call.apply(Foo.prototype.method, arguments);
};


Foo.prototype.method(1,2,3) // Foo { method=function()} 1 2 3 //this output is obvious
Foo.method(1,2,3)  // Number {} 2 3 undefined // I want understand why the first argument is a number and the last one is undefined

I think you are asking about what's discussed here - stackoverflow.com/questions/7459769/… — Saket
– Saket, Commented Sep 19, 2011 at 16:47

Felix Kling · Accepted Answer · 2011-09-19 17:07:37Z

7

Function.call.apply(Foo.prototype.method, arguments);

is the same as

Foo.prototype.method.call(arguments[0], arguments[1], arguments[2], ...);

which, in your case, is the same as:

Foo.prototype.method.call(1, 2, 3);

This means, inside Foo.prototype.method, this will refer to 1, but as this always has to reference an object (in non-strict environment), 1 is converted to a Number object.

The last value is undefined because you are effectively only passing 2 and 3 (two arguments) to the method (instead of three).

So in the end, the code is doing something similar to this:

var obj = new Number(1);
obj.method = Foo.prototype.method;
obj.method(2,3);

edited Sep 19, 2011 at 17:07

answered Sep 19, 2011 at 16:49

Felix Kling

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1 Comment

Christian C. Salvadó Over a year ago

Note that on strict mode, this is no longer implicitly converted to number, e.g.: (function () {'use strict'; return typeof this; }).call(1) == 'number'

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