Hi, The program below this email does not compile with gcc 4.1.2. g++ (GCC) 4.1.2 20061115 (prerelease) (Debian 4.1.1-21) Copyright (C) 2006 Free Software Foundation, Inc. This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. I understand that I should ask on the gcc mailing list first, as gcc says it is own bug. But I want to make sure it is not become the typeof package uses some too advanced compiler features. Can somebody let me know if the following code can be compiled by a more recent version of gcc or some other compiler? Thanks, Peng #include <boost/typeof/typeof.hpp> #include <iostream> namespace A { template <typename T> class X { public: X() { } X(T t) : _t(t) { } const T &the_t() const { return _t; } private: T _t; }; template <typename T1, typename T2> struct multiply_traits; template <typename T1, typename T2> struct multiply_traits<X<T1>, T2> { typedef X<T1> result_type; }; template <typename T1, typename T2> typename multiply_traits<X<T1>, T2>::result_type operator*(const X<T1> &x, const T2 &t) { return X<T1>(x.the_t() * t); } } namespace B { template <typename T> class Y { public: Y(T t) : _t(t) { } const T &the_t() const { return _t; } private: T _t; }; template <typename T1, typename T2> //Y<typename multiply_traits<T1, T2>::result_type> operator*(const Y<T1> &y, const T2 &t) { //Y<T1> operator*(const Y<T1> &y, const T2 &t) { Y<BOOST_TYPEOF(T1() * T2())> operator*(const Y<T1> &y, const T2 &t) { return Y<T1>(y.the_t() * t); } } int main () { A::X<int> x(2); B::Y<A::X<int> > y(x); std::cout << (x * 3).the_t() << std::endl; std::cout << (y * 5).the_t().the_t() << std::endl; }