Re: '...' is a variable,who can explain

["jason zhang" <jzhang@...>]

No, type() does'nt accept variable argument.

----- Original Message ----- 
From: "David Burgess" <dburgess@gmail.com>
To: "Lua list" <lua@bazar2.conectiva.com.br>
Sent: Friday, January 12, 2007 2:00 PM
Subject: Re: '...' is a variable,who can explain


It does raise the question of whether

  type(...)

should return a value?

db

On 1/12/07, jason zhang <jzhang@sunrisetelecom.com.cn> wrote:
>
>
> function ::= function funcbody
>  funcbody ::= `(´ [parlist1] `)´ block end
> parlist1 ::= namelist [`,´ `...´] | `...´
>
> ... means vararg function.
>
>
>
> ----- Original Message -----
> From: wj
> To: lua@bazar2.conectiva.com.br
> Sent: Friday, January 12, 2007 1:24 PM
> Subject: '...' is a variable,who can explain
>
>
> look at this code:
>
> --codes begin
>
> function joinFunc(f1,f2)
>   return function (...)
>            return f1(f2(...)) --...treated as a variable! is it maze!
>    end
> end
>
> f1 = joinFunc(print,math.abs)
>
> f1(-2) --out:2
>
> --codes end
>
> why the
>