Re: [boost] Variadic templates

From: Douglas Gregor (doug.gregor_at_[hidden])
Date: 2007-04-16 17:08:42

Hello,

On Apr 16, 2007, at 3:33 PM, Marco wrote:
> I gave a glance to the papers about variadic templates and I didn't  
> see
> examples using a construct such as f(args)... where args is a function
> parameter pack; so I guess that the following code it's illegal:

The specification of variadic templates is the authority on what can  
and can't be done with them. The compiler, and the examples in the  
papers, just help to illustrate some of the capabilities. In this  
case...

> template< typename T >
> T max( T x )
> {
>    return x;
> }
>
> template< typename T, typename... Args >
> // where are_same<T, Args...>
> T max( T x, T y, Args... args )
> {
>    return std::max( x, max( y, args... ) );
> }
>
>
> // #1
> template< typename T, typename... Args >
> // where are_same<T, Args...>
> T max_abs( T x, T y, Args... args )
> {
>    // this is the (probably) illegal construct:
>    return max( abs(x), abs(y), abs(args)... );
>    // I'd like that "abs(args)..." expands to "abs(arg1), ... ,abs 
> (argN)"
> }

Sure, that's fine.

> What's the rationale behind such a design choice ?

It's a great idea, so we did it :)

>
> I admit that a simple workaround exists:
>
> // #2
> template< typename T >
> T max_abs( T x )
> {
>    return abs(x);
> }
>
> template< typename T, typename... Args >
> // where are_same<T, Args...>
> T max_abs( T x, T y, Args... args )
> {
>     return std::max(abs(x), max_abs(y, args...) );
> }
>
> anyway version #1 is more straightforward, and maybe it's possible  
> to make
> up examples more complex than mine.

The implementations of Bind, Function, and Tuple involve much more  
complex uses of variadic templates.

>
> Another question: given the following functions:
>
> // #1
> template< typename T >
> void f(T ) {}
>
> // #2
> template< typename T, typename... Args >
> void f(T, Args... ) {}
>
>
> the function call f(1,2) is it ambiguous ?
> if it isn't, which is the function called ?
> #1 because it's more specialized ? (only a guess)

For f(1, 2), only #2 matches.
For f(1), both #1 and #2 match, but #1 is more specialized.

        - Doug