The new operator

[Egbert Teeselink]

Hi guys,

I've searched around in books and on the web, but probably not good
enough because I haven't been able to find out what *exactly* the new
operator does; all sites and books explain it exclusively my example.

My underbelly feeling says that "new" creates a shallow copy of the
referenced object and then calls its constructor function, binding
"this" to the newly created copy, and returns the result of the
constructor function, if any, or its "this" object.

Does that make any sense? Is there any resource that I missed where
this is clarified? Thanks!

Egbert

[Thomas 'PointedEars' Lahn]

Re: The new operator

Egbert Teeselink wrote:

I've searched around in books and on the web, but probably not good
enough because I haven't been able to find out what *exactly* the new
operator does; all sites and books explain it exclusively my example.

^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^
Parse error.

My underbelly feeling says that "new" creates a shallow copy of the
referenced object and then calls its constructor function, binding
"this" to the newly created copy, and returns the result of the
constructor function, if any, or its "this" object.

That is not what happens, of course. There is no referenced non-Function
object to begin with.

Does that make any sense?

No.

Is there any resource that I missed where this is clarified? Thanks!

ECMAScript Language Specification, Edition 3 Final, section 11.2.2, and
implementation sources.


HTH

PointedEars
--
realism: HTML 4.01 Strict
evangelism: XHTML 1.0 Strict
madness: XHTML 1.1 as application/xhtml+xml
-- Bjoern Hoehrmann

[Joost Diepenmaat]

Re: The new operator

Egbert Teeselink <skrebbel@gmail .comwrites:

Hi guys,
>
I've searched around in books and on the web, but probably not good
enough because I haven't been able to find out what *exactly* the new
operator does; all sites and books explain it exclusively my example.
>
My underbelly feeling says that "new" creates a shallow copy of the
referenced object and then calls its constructor function, binding
"this" to the newly created copy, and returns the result of the
constructor function, if any, or its "this" object.

<newdoesn't copy anything. it creates a new "empty" object, sets its
prototype to the prototype property of the constructor, and then calls
the constructor with <thisset to the new object.

this may be helpful:

http://joost.zeekat.nl/constructors-considered-mildly-confusing.html

--
Joost Diepenmaat | blog: http://joost.zeekat.nl/ | work: http://zeekat.nl/

[Egbert Teeselink]

Re: The new operator

On Sat, 04 Oct 2008 16:53:37 +0200, Thomas 'PointedEars' Lahn
<PointedEars@we b.dewrote:

Egbert Teeselink wrote:

>I've searched around in books and on the web, but probably not good
>enough because I haven't been able to find out what *exactly* the new
>operator does; all sites and books explain it exclusively my example.

^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^
Parse error.

Woops, I meant "by example". And naturally not all sites and books, but
all that I came across.

>My underbelly feeling says that "new" creates a shallow copy of the
>referenced object and then calls its constructor function, binding
>"this" to the newly created copy, and returns the result of the
>constructor function, if any, or its "this" object.

>
That is not what happens, of course.

Of course not. Doh. Why did I even bother asking?

There is no referenced non-Function
object to begin with.

Ahh right, I thought you could basically do "new" on anything.

So rephrase, doing "new Something()" requires that Something is the name
of a function, and this function will be called with an empty object bound
to "this", right? So, it would be roughly equivalent to Something.call( {})
?

>Does that make any sense?

>
No.
>

>Is there any resource that I missed where this is clarified? Thanks!

>
ECMAScript Language Specification, Edition 3 Final, section 11.2.2, and
implementation sources.

Aye, the specs. Somehow never thought of those as documentation material.

Thanks,

Egbert

[Joost Diepenmaat]

Re: The new operator

"Egbert Teeselink" <e.nofruit.tees elink@student.b anana.tue.nlwri tes:

Ahh right, I thought you could basically do "new" on anything.

Just constructor calls.

So rephrase, doing "new Something()" requires that Something is the
name of a function, and this function will be called with an empty
object bound to "this", right? So, it would be roughly equivalent to
Something.call( {}) ?

It requires that Something refers to an object that is marked as being a
constructor. *Roughly* speaking this means you can call new on any
function written in javascript itself, and on some host objects that are
documented as being constructors. See the Ecma-262 specs, sections 8.6.2
and 13.

By the way: the name of the variable referring to the constructor is not
relevant:

var chain = { to: { constr: Something } };

new chain.to.constr ();

Will work just as well, and the resulting object cannot portably tell
the difference.

--
Joost Diepenmaat | blog: http://joost.zeekat.nl/ | work: http://zeekat.nl/

[Thomas 'PointedEars' Lahn]

Re: The new operator

Egbert Teeselink wrote:

On Sat, 04 Oct 2008 16:53:37 +0200, Thomas 'PointedEars' Lahn
<PointedEars@we b.dewrote:

>Egbert Teeselink wrote:
>There is no referenced non-Function object to begin with.

That is not quite correct. There is no referenced non-constructable object.

Ahh right, I thought you could basically do "new" on anything.
>
So rephrase, doing "new Something()" requires that Something is the name
of a function,

On/of a reference to a(ny) constructor, that is, a(ny) object that
implements the internal [[Construct]] method. Function objects do this,
several host objects (Option, Image, ActiveXObject etc.), too.

and this function will be called with an empty object bound to "this",
right? So, it would be roughly equivalent to Something.call( {})?

Correct, roughly.


PointedEars
--
realism: HTML 4.01 Strict
evangelism: XHTML 1.0 Strict
madness: XHTML 1.1 as application/xhtml+xml
-- Bjoern Hoehrmann

[Egbert Teeselink]

Re: The new operator

On Sat, 04 Oct 2008 16:44:37 +0200, Egbert Teeselink <skrebbel@gmail .com>
wrote:

Hi guys,
>
I've searched around in books and on the web, but probably not good
enough because I haven't been able to find out what *exactly* the new
operator does; all sites and books explain it exclusively my example.
>
My underbelly feeling says that "new" creates a shallow copy of the
referenced object and then calls its constructor function, binding
"this" to the newly created copy, and returns the result of the
constructor function, if any, or its "this" object.
>
Does that make any sense? Is there any resource that I missed where
this is clarified? Thanks!
>
Egbert

Thanks, Joost and Thomas, for your answers!

Egbert