AMDG Peng Yu wrote:
In line 51-55 of you attached file, it says the following.
template <typename T1, typename T2> Y<typename A::multiply_traits<T1, T2>::result_type> operator*(const Y<T1> &y, const T2 &t) { return Y<T1>(y.the_t() * t); }
This way requires that I have to explicitly say 'multiply_traits' is from the namespace A. I wanted to use the following instead, which does not need to specify which the namespaces T1 and T2 are. But the following code still results in the compiler error.
template <typename T1, typename T2> Y<BOOST_TYPEOF(T1() * T2())> operator*(const Y<T1> &y, const T2 &t) { return Y<T1>(y.the_t() * t); }
My question is not resolved. I'm attaching the source file with some comments to avoid any confusion. Would you please take another look at the problem?
Alright. If you don't want to refer to multiply_traits from namespace A, then put a generic multiply_traits somewhere else. When it is defined using Boost.Typeof, it will work for any types. It does not matter what namespace multiply_traits is defined in. Really. template<class T1, class T2> struct typeof_multiply { BOOST_TYPEOF_NESTED_TYPEDEF_TPL(nested, (make<T1>() * make<T2>())); typedef typename nested::type type; }; ... template<class T1, class T2> Y<typename typeof_multiply<T1, T2>::type> operator*(...); Really, all this does is to move the typeof calculation into a separate metafunction. In Christ, Steven Watanabe