Subj : Re: function expression
To   : netscape.public.mozilla.jseng
From : Martin Honnen <Martin.Honnen@t-online.de>
Date : Sat Jan 17 2004 04:19 pm



Georg Maaß wrote:

> This works:
> 
> js> (function banana(){banana.taste='fruchtig';print(banana.taste);})();
> fruchtig
> 
> This fails:
> 
> js> function banana(){banana.taste='fruchtig';print(banana.taste);}();
> 94: SyntaxError: syntax error:
> 94: function banana(){banana.taste='fruchtig';print(banana.taste);}();
> 94: ................................................................^
> 
> Why is this not recognized as function expression? The compiler assumes 
> that this is a function declaration, but it is a function expression. If 
> this is invalid, how can I detect this from the spac, that this is invalid?

Looking at the ECMAScript edition 3 specification section 14 defines a 
Program to consist of SourceElements where a SourceElement is a 
Statement or a FunctionDeclaration.
When SourceElements is evaluated it is processed first for function 
declaration then the statements are evaluated.
A FunctionDeclaration is defined as
   function Identifier ( FormalParameterListopt ) { FunctionBody }
so the part
   function banana(){banana.taste='fruchtig';print(banana.taste);}
matches that and what remains is
   ();
and that doesn't follow the rules for any statement possible I think.
-- 

	Martin Honnen
	http://JavaScript.FAQTs.com/

.