C++ 专用 function 模板别名语法

Question

I have

class ClassA {};
class ClassB {};

auto func_a() -> ClassA {
    return ClassA(); // example implementation for illustration. in reality can be different. does not match the form of func_b
}

auto func_b() -> ClassB {
    return ClassB(); // example implementation for illustration. in reality can be different. does not match the form of func_a
}

I want to be able to use the syntax

func<ClassA>() // instead of func_a()
func<ClassB>() // instead of func_b()

(this is as part of a bigger template)

but I don't know how to implement this template specialization for the function alias. Help? What is the syntax for this?

[Edit] The answers posted so far do not answer my question, so I'll edit my question to be more clear.

The actual definition of func_a and func_b is more complex than just "return Type();". Also they cannot be touched. So consider them to be

auto func_a() -> ClassA {
    // unknown implementation. returns ClassA somehow
}
auto func_b() -> ClassB {
    // unknown implementation. returns ClassB somehow
}

I cannot template the contents of func_a or func_b. I need the template for func<ClassA|ClassB> to be specialized for each one of the two and be able to select the correct one to call

Answer 1

You might do something like (c++17):

template <typename T>
auto func()
{
    if constexpr (std::is_same_v<T, ClassA>) {
        return func_a();
    } else {
        return func_b();
    }
}

Alternative for pre-C++17 is tag dispatching (which allows customization point):

// Utility class to allow to "pass" Type.
template <typename T> struct Tag{};

// Possibly in namespace details
auto func(Tag<ClassA>) { return func_a(); }
auto func(Tag<ClassB>) { return func_b(); }

template <typename T>
auto func() { return func(Tag<T>{}); }

Answer 2

You can do it like this:

#include <iostream>
#include <type_traits>

class A 
{
public:
    void hi()
    {
        std::cout << "hi\n";
    }
};

class B
{
public:
    void boo()
    {
        std::cout << "boo\n";
    }
};

template<typename type_t>
auto create()
{
    // optional : if you only want to be able to create instances of A or B
    // static_assert(std::is_same_v<A, type_t> || std::is_same_v<B, type_t>);

    type_t object{};
    return object;
}

int main()
{
    auto a = create<A>();
    auto b = create<B>();

    a.hi();
    b.boo();

    return 0;
}