C++ invoke explicit template constructor

c++, templates

Solution

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

Problem

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

Original source

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