Hello, i look for a portable way to determine the return-type of a member function: #include <boost/utility/result_of.hpp> #include <boost/type_traits/function_traits.hpp> class CMyClass { public: int method( void ) {return 5;} }; template< typename T > struct result_of {}; template< typename R, typename A > struct result_of< R(*)(A) > { typedef R type; }; template< typename R > struct result_of< R(*)(void) > { typedef R type; }; template< typename R, typename C > struct result_of< R(C::*)(void) > { typedef R type; }; template< typename R, typename C, typename A > struct result_of< R(C::*)(A) > { typedef R type; }; /*...some other specialisations...*/ /* (1) OK, but I must now return-type and signature */ //typedef int (CMyClass::*fnPointer_t)(void); /* (2) OK, but I must now return-type and signature */ //typedef typeof(int(CMyClass::*)(void)) fnPointer_t; /* (3) OK, compiler determine return-type and signature if CMyClass::method non-ambiguous */ typedef typeof(&CMyClass::method) fnPointer_t; int main( ) { fnPointer_t p = &CMyClass::method; /* (4) OK */ result_of< fnPointer_t >::type i; /* (5) OK, if CMyClass::method non-ambiguous */ result_of< typeof(&CMyClass::method) >::type i2; /* (6) Failure, type expected */ //result_of< p >::type i3; /* (7) OK */ result_of< typeof(p) >::type i4; /* (8) Failure, no function pointer support */ //boost::result_of< fnPointer_t >::type i5; /* (9) Failure, no function pointer support */ //boost::result_of< typeof(&CMyClass::method) >::type i6; /* (10) Failure, no function pointer support */ //boost::function_traits< fnPointer_t >::result_type i7; /* (11) Failure, no function pointer support */ //boost::function_traits< typeof(&CMyClass::method) >::result_type i8; } Which is a portable way? My prefered solution is number 5, but i dont now whether typeof(&CMyClass::method) is portable. Why function pointer are not supported by boost::result_of and boost::function_traits templates? Thanks and Greetings, Peter