Zeljko Vrba <zvrba@globalnet.hr> writes:
David Abrahams wrote:
However, reliably taking the address of an overloaded function involves knowing the exact types in its signature. I'm not sure how you can do that.
I have googled a bit and found the following example in the Microsoft C++ reference: http://tinyurl.com/cljnn
int Func( int i, int j ); int Func( long l );
...
int (*pFunc) ( int, int ) = Func;
Do you see any problems with this?
Not at all. As I said, you need to know the exact signature. The problem is detecting that the function (or operator<) whose existence you detect has the exact signature you care about. Maybe it could be tested by passing a convertible-to-T POD struct through the function: // UNTESTED template <class T> struct convertible_to { operator T() const; }; namespace tester { typedef char (&yes) [1]; typedef char (&no) [2]; struct not_found {}; no is_found(not_found); yes i_found(...); not_found operator<(...); template <class A1, class A2> struct has_less1 { static bool const value = sizeof( is_found( convertible_to<A1>() < convertible_to<A2>() ) ) == yes; }; template <class T> T make(); template <class T, class U> no is(T&,U&); template <class T> yes is(T&,T&); template <class R, class A1, class A2, bool = has_less1<A1,A2>::value> struct has_less { static bool const value = sizeof( is(make<R>(), make<A1>() < make<A2>()) ) == yes; }; template <class R, class A1, class A2> struct has_less<R,A1,A2,false> { static bool const value = false; }; } The problem with this is that you still can't tell whether the operator< you found takes its arguments by reference or by value, and without that information, you can't do the cast you want reliably.
I guess it should be possible to take an address of an operator by replacing 'Func' with e.g. 'operator<'. I'll try it and report the result.
I don't think this problem can be solved. -- Dave Abrahams Boost Consulting www.boost-consulting.com