[mpl] how to find out whether a function is implemented
I hope the MPL label is right. At least the question has *some* relevance to metaprogramming. Basically, I need to find out whether a certain function F over a certain type is defined, and if yes, take its address. Input: the type T, and function NAME F. Output: pointer to F if Ret F(T, T& or T const&; or two arguments) is defined, 0 otherwise. The function can return anything, including void. Variations in input: - the function may take T, T&, or T const& - if possible, the method should work for user-defined operators (my C++ is a bit rusty in this area - is it at all possible to take an address of an overloaded operator?) I guess the F would then be named, e.g. 'operator<' ? If the general case is hard to work out, it's OK to have the solution just for Ret F(T const&) and Ret F(T const&, T const&). I hope the question is clear enough :) Thanks for your time and answers.
Zeljko Vrba <zvrba@globalnet.hr> writes:
I hope the MPL label is right. At least the question has *some* relevance to metaprogramming.
Basically, I need to find out whether a certain function F over a certain type is defined, and if yes, take its address.
I can help you with the first part: http://www.boost-consulting.com/boost/boost/detail/is_incrementable.hpp shows an example determining if ++x is defined. 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. -- Dave Abrahams Boost Consulting www.boost-consulting.com
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? 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.
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
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Zeljko Vrba