Re: [Boost-users] Is there any class that can promote a type to the complex type?

Isn't that what the std::complex default constructor does? It does and I think it's a bad practice.

explicit complex(const complex<double>&); explicit complex(const complex<long double>&);

Why are they explicit ? Where is the risk to have non-explicit destructor ?

The standard states that "The effect of instantiating the template complex for any type other than float, double or long double is unspecified." (26.2) A point I wasn't aware of. What about enforcing this at the compile-time level using CT-assert ?

I think that you are correct about operators on two complex types. However, it is probably not a good idea to have operator overloads like the following template complex<T> operator+(const U& lhs, const complex<T>& rhs); because they can cause ambiguities. Is having the followign triplet lead to ambiguity ?

template complex<T> operator+(const U& lhs, const complex<T>& rhs); template complex<T> operator+(const complex<T>& lhs, const U& rhs); template complex<T> operator+(const complex<U>& lhs, const complex<T>& rhs); If it did, shouldn't SFINAE make the cut by restrictng U to types that verify : boost::is_floating_points<U>::value == true ? For me, it looks like complex can be reworked with current technologies to enforce its behavior and adds the missing pieces. -- Joel FALCOU Research Engineer @ Institut d'Electronique Fondamentale Université PARIS SUD XI France