boost::mem_fn erases the need for a separate mem_fn_ref. If http://www.boost.org/libs/bind/mem_fn.html fails to explain it to you, please ask more specific questions.
I thought I'd been pretty specific... guess not. Allow me to reiterate: Why doesn't std::mem_fun work with a container of boost::shared_ptr? I fiddled around with my compiler for a while (vc6), but ended up with a wildly unhelpful error message. Given the following lines of code (among others): --- typedef std::vector<boost::shared_ptr<std::string> > Strings; Strings strings; std::for_each( strings.begin(), strings.end(), std::mem_fun(std::string::size) ); --- my compiler puked up the following wildly useless error message: --- error C2664: 'mem_fun' : cannot convert parameter 1 from 'unsigned int (__thiscall std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >::*' to 'unsigned int (__thisca ll std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >::*' Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast --- Allow me to pull out those two types for easier viewing: unsigned int (__thiscall std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >::* unsigned int (__thiscall std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >::* Golly. Those 'two' types are about as related as you can get. So why won't std::mem_fun work through boost::shared_ptr's operator->? Here's the mem_fun that ships with VC6: --- template<class _R, class _Ty> class mem_fun_t : public unary_function<_Ty *, _R> { public: explicit mem_fun_t(_R (_Ty::*_Pm)()) : _Ptr(_Pm) {} _R operator()(_Ty *_P) const {return ((_P->*_Ptr)()); } private: _R (_Ty::*_Ptr)(); }; --- The relevant portion to our discussion is: "{return ((_P->*_Ptr()); }" which can be further boiled down to: return _P->_Ptr(); This looks an awful lot like it should work with a shared_ptr's operator->(). So why doesn't it? I admit that this is more of a C++ question than a boost one, but it's not too much of a tangent and I *am* curious. Mentioning mem_fun_ref in my previous post only served to muddy the waters. My apologies. --Mark Storer Senior Software Engineer Verity, Inc.