On Fri, Jul 31, 2009 at 11:52 AM, Simonson, Lucanus J<lucanus.j.simonson@intel.com> wrote:
Beman Dawes wrote:
On Thu, Jul 30, 2009 at 10:20 AM, Mathias Gaunard<mathias.gaunard@ens-lyon.org> wrote:
It would be very nice if Boost.Config provided the BOOST_NO_SFINAE_EXPR macro, that would be defined if the compiler does not support SFINAE applied to expressions. http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2008/n2634.html
GCC supports this feature since 4.4 both for C++03 and C++0x for example.
Yes, such a macro is needed
So we could do this instead of use mpl and?
template <typename T1, typename T2> enable_if_c<is_a_A<T1>::value && is_a_B<T2>::value>::type foo();
Also, it turns out that the ability to provide a default template parameter for function templates markedly clarifies code using enable_if. (I learned this Howard Hinnant, but I think he may have gotten the idea from Peter Dimov.)
This sounds interesting. Can you explain how?
The example from Howard is below. Note that the enable_if code is moved out of the function signature and into a default template parameter. --Beman #define requires(...) class = typename std::enable_if<(__VA_ARGS__)>::type template <class T> struct underlying { typedef typename remove_cv<typename remove_reference<T>::type>::type type; }; // forward with <type_traits> template <class T, class U, requires(!is_lvalue_reference<T> || is_lvalue_reference<T> && is_lvalue_reference<U>), requires(is_same<typename underlying<T>::type, typename underlying<U>::type>)> inline T&& forward(U&& t) { return static_cast<T&&>(t); }