c++ template function overloading.

Question

Below are lines from "the c++ programming language"

template<class T > T sqrt(T );
template<class T > complex<T> sqrt(complex<T>);
double sqrt(double);
void f(complex<double> z )
{
s q r t (2 ); // sqrt<int>(int)
sqrt(2.0) ; // sqrt(double)
sqrt(z) ; // sqrt<double>(complex<double>)
}

I dont understand why sqrt(z) ; calls sqrt<double>(complex<double>) can any body please explain.

Author says, T sqrt<complex<T>> is more specialized than T sqrt <T> but there is a seperate declaration for template<class T > complex<T> sqrt(complex<T>); why not use that?

Answer 1

Well, the function used is the one you are talking about sqrt<double>(complex<double>) is an instance of the template template <class T> complex<T> sqrt(complex<T>).

Your misunderstanding was in the signification of the template instance and not in the overloading process.

Answer 2

In hindsight, it would have been easier if Bjarne would have written it as

template<class T> T sqrt(T);
template<class U> complex<U> sqrt(complex<U>);
double sqrt(double);
void f(complex<double> z )
{
    sqrt (2); // sqrt<int>(int)
    sqrt(2.0) ; // sqrt(double)
    sqrt(z) ; // sqrt<double>(complex<double>)
}

so you don't get confused by all the different T's. But the idea is simple; C++ finds the best match. There are three possible functions. The first two are perfect matches (no conversion needed) so the non-template version is ignored. Now, we have T=complex and U=double. Which version is chosen? Bjarne explains the second template is chosen here, because it's more specialized. This means that for any type U, there is a type T=complex<U> which makes the signatures of both templates identical.