C++ template constructor

Question

There is no way to explicitly specify the template arguments when calling a constructor template, so they have to be deduced through argument deduction. This is because if you say:

Foo<int> f = Foo<int>();

The <int> is the template argument list for the type Foo, not for its constructor. There's nowhere for the constructor template's argument list to go.

Even with your workaround you still have to pass an argument in order to call that constructor template. It's not at all clear what you are trying to achieve.

You can create a templated factory function:

class Foo
{
public:
    template <class T> static Foo* create() // could also return by value, or a smart pointer
    {
        return new Foo(...);
    }
...        
};

As far as I understand, it's impossible to have it (because it would conflict with the default constructor - am I right?)

You are wrong. It doesn't conflict in any way. You just can't call it ever.

template<class...>struct types{using type=types;};
template<class T>struct tag{using type=T;};
template<class Tag>using type_t=typename Tag::type;

the above helpers let you work with types as values.

class A {
  template<class T>
  A( tag<T> );
};

the tag<T> type is a variable with no state besides the type it caries. You can use this to pass a pure-type value into a template function and have the type be deduced by the template function:

auto a = A(tag<int>{});

You can pass in more than one type:

class A {
  template<class T, class U, class V>
  A( types<T,U,V> );
};
auto a = A(types<int,double,std::string>{});