Firefox JS vs ECMA - am I missing something?

[Spike]

Q for the language lawyers!
Here's the code:

function f(p)
{
alert("f("+p+") with this="+this);
}

var a = [f];

a[0]("index");
-----------------------------
Shouldn't a[0]("index") invoke f with this===a, since a is the 'base'
of the reference a[0]?
It doesn't, it invokes f with (apparently) this===f.

whereas
var b = new Object;
b.m = f;
b.m("method");

invokes f with this===b.

I seem to be about 3 years late asking a question like this, but I'm
writing this Javascript compiler...

[Stevo]

Spike wrote:

Q for the language lawyers!
Here's the code:
function f(p)
{
alert("f("+p+") with this="+this);
}
var a = [f];
a[0]("index");

-----------------------------
Shouldn't a[0]("index") invoke f with this===a, since a is the 'base'
of the reference a[0]?

No. Why are you expecting it to use the base. In the call a[0]() it's
a[0] that represents this.

It doesn't, it invokes f with (apparently) this===f.

Same thing. a[0]===f

whereas
var b = new Object;
b.m = f;
b.m("method");

invokes f with this===b.

Same thing again. The object you're referencing here is b, so b is this.

[John G Harris]

On Sun, 2 Mar 2008 at 02:38:53, in comp.lang.javascript, Spike wrote:

>Q for the language lawyers!
Here's the code:

function f(p)
{
alert("f("+p+") with this="+this);
}

var a = [f];

a[0]("index");
-----------------------------
Shouldn't a[0]("index") invoke f with this===a,

<snip>
If you did

var z = a[0];
z.("index");

what do you think the 'this' value will be while the function is
executing ? Why would it be different when you do

a[0]("index");

instead ?

Remember that the 'this' value is calculated by the execution engine. It
uses a very simple rule and knows nothing about the complications in
your source code.

John
--
John Harris