Almost Always Use Auto rule with templated constructor?

Question

I have recently adopted the pattern of Almost Always Auto in C++14, but have come across a case that I can't figure out how to write using the auto syntax: templated constructors.

Say I have the class:

class my_type{
    public:
        template<typename T>
        my_type(){/* ... */}
};

I tried:

auto var = my_type<float>{};

Which, of course, doesn't work because that presumes my_type is a template and not its constructor.

then how could I use auto syntax to initialize this variable?

Oh, you're asking how to select that ctor, got it. How would you write this without using `auto`? — Useless, Jul 22 '15 at 10:12
@Useless absolutely no idea, only just came across it now, actually. How would you do that? — RamblingMad, Jul 22 '15 at 10:17
@CoffeeandCode you can't, you need to deduce the template parameter from the arguments. If you could explain what that constructor is supposed to do, maybe we can suggest alternatives. — TartanLlama, Jul 22 '15 at 10:17
See for example [this question](http://stackoverflow.com/q/3960849/212858). Now, this has nothing to do with `auto`, since you can't do this anyway. — Useless, Jul 22 '15 at 10:18

score 1 · Answer 1 · answered Jul 22 '15 at 10:28

Although it isn't really about auto, there is a way to select templated ctors if you want to - you need to cheat, and give them an argument.

struct A {
private:
    template <typename T> struct TypeWrapper {};
    template <typename T> explicit A(TypeWrapper<T>) {}
public:
    template <typename T>
    static A make_A() {
        return A(TypeWrapper<T>{});
    }
};

int main() {
    auto i = A::make_A<int>();
    auto d = A::make_A<double>();
    auto r = A::make_A<A>();
    // etc. etc.
}

I really like this, but it would be better as a free function that is a `friend` of `A` — RamblingMad, Jul 22 '15 at 10:29

Almost Always Use Auto rule with templated constructor?

1 Answers1