我采取了Nicola Bonelli和Johannes Schaub非常有帮助的答案，并将他们合并成一个解决方案，即恕我直言，更具可读性，清晰，不需要typeof扩展： template <class Type> class TypeHasToString { // This type won't compile if the second template parameter isn't of type T, // so I can put a function pointer type in the first parameter and the function // itself in the second thus checking that the function has a specific signature. template <typename T, T> struct TypeCheck; typedef char Yes; typedef long No; // A helper struct to hold the declaration of the function pointer. // Change it if the function signature changes. template <typename T> struct ToString { typedef void (T::*fptr)(); }; template <typename T> static Yes HasToString(TypeCheck< typename ToString<T>::fptr, &T::toString >*); template <typename T> static No HasToString(...); public: static bool const value = (sizeof(HasToString<Type>(0)) == sizeof(Yes)); }; 我用gcc 4.1.2查了一下。信贷主要是尼古拉Bonelli和约翰Schaub，所以给他们一个投票，如果我的回答可以帮助你:)... 展开详请