----- Original Message ----- From: "Kenny Riddile" <kfriddile@yahoo.com> To: <boost@lists.boost.org> Sent: Wednesday, January 27, 2010 6:14 PM Subject: Re: [boost] detecting a constructor with a specific signature
Daniel Frey wrote:
On 27.01.2010, at 15:52, Kenny Riddile wrote:
Ya, I was already using the "flag typedef" method, but was just wondering if something non-intrusive was feasible...maybe it isn't. A compiler with 0x support isn't an option I'm afraid.
No need for C++0x if you only need a special case. Requires a good compiler, though, GCC 4.3 is not good enough, 4.4+ is. The following compiles with GCC 4.4, -ansi -pedantic, replacing decltype with sizeof :)
#include <string> #include <iostream>
struct Foo { Foo( const std::string& ); };
struct Bar { Bar( const std::string& ); Bar( const std::string&, Foo& ); };
template< typename T > T make();
template< int > struct result { typedef double type; };
template< typename T > typename result< sizeof T( make< const std::string& >(), make< Foo& >() ) >::type select( int );
template< typename > char select( ... );
template< typename T > struct has_foo_ctor { enum { value = sizeof select< T >( 0 ) > 1 }; };
int main() { std::cout << has_foo_ctor< Foo >::value << std::endl; std::cout << has_foo_ctor< Bar >::value << std::endl; }
Regards, Daniel
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Alas, I'm using VC9. The first declaration of select() gives me the following error:
error C2564: 'T' : a function-style conversion to a built-in type can only take one argument
Hi, have you tried to double () template< typename T > typename result< sizeof T(( make< const std::string& >(), make< Foo& >() )) >::type select( int ); Vicente