Object.prototype.valueOf() method

Question

Object.prototype.valueOf.call(\"abc\")
{ \'0\': \'a\'
, \'1\': \'b\'
, \'2\': \'c\'
}
Object.prototype.valueOf.call(new String(\"abc\"))
{ \'0\': \'a\'
, \'1\': \'b\'
,          


        

Answer 1

Perhaps I've missed what the actual question is. Here is what ECMA-262 says:

15.2.4.4 Object.prototype.valueOf ( )

When the valueOf method is called, the following steps are taken:

1. Let O be the result of calling ToObject passing the this value as the argument.

2. If O is the result of calling the Object constructor with a host object (15.2.2.1), then

   a. Return either O or another value such as the host object originally passed to the constructor. The specific result that is returned is implementation-defined.

3. Return O.

In the expression:

Object.prototype.valueOf.call("abc")

Object.prototype.valueOf is being called with a string primitive as this. So at step 1, it is converted using the internal ToObject method.

If passed a value of Type String (which "abc" is) ToObject will return a String object.

Step 2 is irrelevant since the object is not a host object (it's a native object).

Step 3 returns the object created by toObject.

So test it:

var x = Object.prototype.valueOf.call("abc");

alert(typeof x); // object

typeof returns object because the resulting value is a String object (a nice quirk of the typeof operator). You can go further:

typeof x.match; // function
x.constructor;  // function String() {[native code]}
alert(x); // abc

All consistent with the returned value being (a reference to) a String object.