"Peter Dimov" <pdimov@pdimov.com> writes:
Anthony Williams:
"Peter Dimov" <pdimov@pdimov.com> writes:
Anthony Williams:
"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.
Try it.
I have, and it doesn't work.
With what compiler does it not work? Can you post the example program?
It's a pre-release compiler. I can't tell you which one as I'm under an NDA.
Besides, you can see in the signature: it returns "A" by value.
There's a special deduction rule for A&&, where A is a template parameter: when passed an lvalue of type T, A is deduced as T& and not as T. So in this case operator() returns a T& "by value".
I see that in 14.8.2.1p3. I'd not really intaken that before. Thanks. However, in this case I'm surely missing something else. Andrei's talk on Friday went on for a few slides about why the function above is insufficient, how you needed three overloads (A&&, A& and A const&), and with a sample implementation from Howard Hinnant that used a traits class to deduce the correct return value with only one overload. 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