C++ invoke explicit template constructor

Question

Can you tell me how to invoke template constructor explicitly (in initializer list)? for example:

struct T { 
    template<class> T();
};

struct U {
    U() : t<void>() {} //does not work
    T t;
};

thanks

Answer 1

It's not possible. The Standard also has a note on this at 14.8.1/7

[Note: because the explicit template argument list follows the function template name, and because conversion member function templates and constructor member function templates are called without using a function name, there is no way to provide an explicit template argument list for these function templates. ]

Explanation: This says: Template arguments are passed in angle brackets after a function template name, such as std::make_pair<int, bool>. And constructors don't have a name of their own, but they abuse their class names in various contexts (so U<int>() means: Pass <int> to the class template U, and construct an object by calling the default constructor without arguments). Therefore, one cannot pass template arguments to constructors.

In your case, you are trying to pass template arguments in a member initializer. In that case, there's even more of a problem: It will attempt to parse and interpret t<void> as a base-class type and thinks you want to call the default constructor of a base class. This will fail, of course.

If you can live with it, you can work it around

struct T { 
    template<class U> T(identity<U>);
};

struct U {
    U() : t(identity<void>()) {}
    T t;
};

Given identity like it's defined in boost

template<typename T> struct identity { typedef T type; };

Within C++20 you can use std::type_identity as identity type.

• https://en.cppreference.com/w/cpp/types/type_identity