Deduce a class enum for generic types

Question

Consider this simple, but complete example:

#include <iostream>

enum class Foo
{
    A,
    B
};

template <Foo F>
struct X{};

template <Foo F>
constexpr Foo deduce_foo(X<F>&& arg)
{
    return F;
}

int main()
{
    if ( deduce_foo(X<Foo::B>{}) == Foo::A )
        std::cout << "A";
    else
        std::cout << "B";
    std::cout << std::endl;
}

which deduces the enum class correctly (function deduce_foo(...)).

Now, I want to make it more generic, substituting X by a template T, i.e:

template <Foo F, typename T>
constexpr Foo deduce_foo(T<F>&& arg) { ... }

But compiler (g++ 7.3.0) gives me this error:

error: ‘T’ is not a template constexpr Foo deduce_foo(X&& arg)

What Im doing wrong?

Your `typename T` isn't actually a "typename". It's a template, not the name of a specific type. — Drew Dormann, Mar 05 '18 at 16:05
Possible duplicate: https://stackoverflow.com/questions/213761 — Drew Dormann, Mar 05 '18 at 16:06
It should be `template class C> constexpr Foo deduce_foo(C&& arg) { ... }` — Jarod42, Mar 05 '18 at 16:07

score 3 · Accepted Answer · answered Mar 05 '18 at 16:10

3

you need to declare that T is a template, try:

template <Foo F, template<Foo> class T>
constexpr Foo deduce_foo(T<F>&& arg)
{
    return F;
}

answered Mar 05 '18 at 16:10

Alan Birtles

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Deduce a class enum for generic types

1 Answers1