"Peter Dimov" <pdimov@pdimov.com> writes:
Eric Niebler:
Can you write what the identity function object would look like in C++0x?
struct identity { template<class A> A operator()( A && a ) const { return std::forward<A>( a ); } };
I guess.
No. That will convert lvalues to rvalues. You also need the lvalue overload in order to preserve the reference-ness of lvalues. IIRC, in Andrei Alexandrescu's talk at ACCU 2008 he also claimed you need a const-reference overload: struct identity { template<class A> A operator()( A && a ) const { return std::move( a ); } template<class A> A& operator()( A & a ) const { return a; } template<class A> A const& operator()( A const & a ) const { return a; } }; I'm not sure we need the third overload, as the second will deduce A to be "const A". Unfortunately, the only compiler I have which handles rvalue references says that the first two are ambiguous when given an lvalue. Anthony -- Anthony Williams | Just Software Solutions Ltd Custom Software Development | http://www.justsoftwaresolutions.co.uk Registered in England, Company Number 5478976. Registered Office: 15 Carrallack Mews, St Just, Cornwall, TR19 7UL