On Mon, 16 Apr 2007 23:08:42 +0200, Douglas Gregor <doug.gregor@gmail.com> wrote:
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 :)
Great! I had missed that patterned pack expansion works with function parameter packs too. I try to dare something more, now. Has it been taken in consideration to provide a typedef type pack defined throught a template parameter pack ? That is, something like: template< typename... Args > class { typedef typename Args::value_type value_types; // or the following more explicit syntax: // typedef... typename Args::value_type value_types; void f( value_types... args ); }; Regards, Marco Using Opera's revolutionary e-mail client: http://www.opera.com/mail/