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

From: Joel FALCOU (joel.falcou_at_[hidden])
Date: 2008-06-02 12:34:03

> 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<class T, class U> complex<T> operator+(const U& lhs, const 
> complex<T>& rhs); because they can cause ambiguities.
Is having the followign triplet lead to ambiguity ?

template<class T, class U>
complex<T> operator+(const U& lhs, const complex<T>& rhs);
template<class T, class U>
complex<T> operator+(const complex<T>& lhs, const U& rhs);
template<class T, class U>
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