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

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

function f(...)
print(type(...)) -- when f(1,2,3) is called, this statement equal to type(1,2,3). Lua allow this but it only use the first parameter.

end

f(1,2,3)




----- Original Message ----- 
From: "wj" <jiang_wang@skygp.com>
To: "Lua list" <lua@bazar2.conectiva.com.br>
Sent: Friday, January 12, 2007 2:55 PM
Subject: Re: '...' is a variable,who can explain


> function f(...)
> print(type(...))
> end
> 
> f(1,2,3)
> 
> dose work.
> 
> look this
> exp ::=  nil | false | true | Number | String | `...´ | function |    prefixexp | tableconstructor | exp binop exp | op exp 
> prefixexp ::= var | functioncall | `(´ exp `)´
> 
> ... is invalid out of function block.is right?
> ----- 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
>>
>