CODEWITHSUNDEEP
javascript
Aditya M P
Aditya M P

Reputation: 5337

Why does Object.prototype.toString.call() give different output than Object.prototype.toString()?

Here's something I found in some code I was reading, when I tested in the console:

Object.prototype.toString("foo"); // output: "[object Object]"

Object.prototype.toString.call("foo"); // output: "[object String]"

I think I may have a faint idea but I can't express in words... can anyone explain?

Upvotes: 1

Views: 92

Answers (2)

Matteo Tassinari
Matteo Tassinari

Reputation: 18584

The two calls are NOT equivalent.

The first call:

Object.prototype.toString("foo");

calls the toString method in the context of Object.prototype, with an additional "foo" parameter (unused), and Object.prototype is an Object, so the result is [object Object]

The second call:

Object.prototype.toString.call("foo");

calls the toString method in the context of "foo", and Object.prototype.toString builds an object from it (new String("foo")), so the result is [object String]

Upvotes: 4

Tibos
Tibos

Reputation: 27823

The first parameter of call is the object that will be this inside the function, not the first parameter of the function:

"use strict";
function test(a,b) {
  console.log(this, a, b);
};
var obj = {
  'func' : test
}
test(1,2) // outputs undefined 1 2
test.call(1,2); // outputs 1 2 undefined
obj.func(1,2) // outputs obj 1 2
obj.func.call(1,2) // outputs 1 2 undefined

Upvotes: 0

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