>-----Original Message-----
>From: [EMAIL PROTECTED] [mailto:es-discuss-
>[EMAIL PROTECTED] On Behalf Of Douglas Crockford
...
>because you can think of the \ as being an abbreviation of function.
>
> \ name(a,b,c) {}
>
>Just don't start your function name with u.
>
Exactly, that's why I didn't mention this possibility. I don't think it would
be acceptable to have that \u ambiguity.
Smalltalk does not have a way (other than variable assignment) to name closures
for recursive reference and it rarely has proven to be a problem. If you
really want to pass a recursive lambda to a function g you could define it
using a const:
const fact = {|n| (n<=1) ? 1 : fact(n-1)* n};
g(fact)
or more perversely:
g({||const fact = {|n| (n<=1) ? 1 : fact(n-1)* n}; fact}() );
(the second statement of the outer lambda wouldn't be needed if const evaluates
to the value of its initialization expression)
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