Hi Lorenzo, Thanks for pushing this topic. Sorry to only replying now but it is hard to find some free time with 3 little children! I have looked a bit at the standard (ISO/IEC 14882, 2011). The relevant text is: 23.2.1, Table 96, 25.2.11. This is what I have found: if a and b are container<T> and t1 and t2 are T: - a==b: requires t1==t2 returning convertible to bool - a!=b: requires a==b - a<b: calls std::lexicographical_compare -> requires t1<t2 returning convertible to bool . a>b: requires b<a . a<=b: requires !(b<a) . a>=b: requires !(a<b) So in short: a==b and a!=b require t1==t2 and a<b, a>b, a<=b, a>=b require t1<t2 returning convertible to bool.
namespace std { // fwd decl so no need to include STL template< typename T, class Alloc > class vector; } // namespace std
Concerning your proposal to forward declare the containers, it is halas not allowed by the standard (17.6.4.2.1 Namespace std): "The behavior of a C++ program is undefined if it adds declarations or definitions to namespace std or to a namespace within namespace std unless otherwise specified." I imagine easily that it should behave as intended but it seems possible for an implementation to not like it.
template< typename T, class LhsAlloc, class RhsAlloc, typename Ret > struct has_equal_to < std::vector<T, LhsAlloc >, std::vector<T, RhsAlloc >, Ret > : has_equal_to<T, T, Ret> {};
The standard defines the operators only for identical allocators. operator== always returns bool and has_equal_to should return convertible to bool. This should be reflected in the declaration. Frédéric