Germán Diago wrote:

Hello. I would like to know if it's possible to obtain the base type from this expression:

struct MyStruct { string field1; };

decltype(&MyStruct::field1) is a pointer to member. I would like to obtain type string instead, removing the pointer to member. Is it possible?

Something like the following? I know it's clumsy... but this works in g++ 4.0.1. #include <iostream> #include <string> /*------------------------------- MemberType --*/ template <typename DEFAULT> struct MemberType { typedef void type; }; template struct MemberType { typedef TYPE type; }; template <typename T> MemberType<T> getMemberType(const T& instance) { return MemberType<T>(); } /*----------------------------- Example Class --*/ class Example { public: Example(const std::string& value): field1(value) {} std::string field1; }; int main(int argc, char *argv[]) { Example example("foo"); typedef typeof(getMemberType(&Example::field1)) field1_MT; field1_MT::type extract(example.field1); std::cout << "Got " << extract << std::endl; return 0; }