c++ automatic template deduction fails for template argument

Question

I would like to better understand why automatic template deduction (applied when compiled with g++ -std=c++17) works in the first three lines in main(), but fails in the fourth. Is there any chance it will be accepted by compilers in the near future?

template <typename P = void>
class A {
public:
    void f1() {}
};

template<typename C>
void g() {}


int main() {
    A<> a;       // works
    A aa;        // works
    g<A<>>();    // works
    g<A>();      // fails
    return 0;
}

Answer 1

It's just a matter of signature. Basically you're passing the wrong type.

Both A a and A<> a mean you want an instance of A with the default template parameter value, that is, you end up with A< void >.

The function g< C >() accepts a template parameter which happens to be a type, not another templated type. When you invoke it with A<>, you tell the compiler that you want to use "the instantiation" of the templated type A, which is valid. When you invoke it with A you tell the compiler you want to invoke g< C >() with C being a templated type which does not fit its signature.

If you declare/define g() like so template <typename <typename> TTemplatedType> g() it will accepts to be invoked like this g< A >() but g< A<> >() will fail because now it no longer wants something else than a templated type.

Answer 2

With C++17, template argument deduction is also performed when the name of a class template is used as the type of an object being constructed.

Nothing changes for explicit type inside template.