如何在不指定枚举的情况下将SFINAE用于枚举？

Question

Using answer from Substitution failure is not an error (SFINAE) for enum I tried to write a code that would get enum value from a class, and if this enum value is not found it would have a fallback value. As I'm beginner with templates, after couple of hours I gave up and found a "solution" using macros :(

Is there a way to do the same thing without macros and without copying the code for every possible enum value?

This is what I came up with:

struct foo
{
    enum FooFields
    {
        enumFoo,
        enumHehe
    };
};

struct bar
{
    enum BarFields
    {
        enumHehe = 2
    };
};

#define GETENUM_DEF(testedEnum) \
template<class T> \
struct get_ ## testedEnum{\
  typedef char yes;\
  typedef yes (&no)[2];\
\
  template<int>\
  struct test2;\
\
  template<class U>\
  static int test(test2<U::testedEnum>*){return U::testedEnum;};\
  template<class U>\
  static int test(...){return -1;};\
\
  static int value(){return test<T>(0);}\
};

GETENUM_DEF(enumFoo)
GETENUM_DEF(enumHehe)

int main() {

    std::cout<<get_enumFoo<foo>::value()<<std::endl; //returns 0;
    std::cout<<get_enumFoo<bar>::value()<<std::endl; //returns -1;

    std::cout<<get_enumHehe<foo>::value()<<std::endl; //returns 1;
    std::cout<<get_enumHehe<bar>::value()<<std::endl; //returns 2;

    return 0;
}

Answer 1

C++ requires that you define a get_someField for every field you want to get, but you'd have to do that with or without macros.

Playing with SFINAE is part of what is known as template meta-programming. What you are doing is effectively detecting whether the expression T::enumFoo is valid, and returning that value if it is, else -1 .

To detect whether expression is valid, we can do something like this:

#include <type_traits>

// You need void_t to avoid a warning about the lhs of the comma operator 
// having no effect. C++ 17 has std::void_t
template<class...> using void_t = void;

template<class T, class = void>
struct get_enumFoo
{
    static constexpr int value = -1;
};

template<class T>
struct get_enumFoo<T, void_t<decltype(T::enumFoo)>>
{
    static constexpr int value = T::enumFoo;
};

template<class T, class = void>
struct get_enumHehe
{
    static constexpr int value = -1;
};

template<class T>
struct get_enumHehe<T, void_t<decltype(T::enumHehe)>>
{
    static constexpr int value = T::enumHehe;
};

Using it (your example):

#include <iostream>

int main() {
    std::cout << get_enumFoo<foo>::value << std::endl; //returns 0;
    std::cout << get_enumFoo<bar>::value << std::endl; //returns -1;

    std::cout << get_enumHehe<foo>::value << std::endl; //returns 1;
    std::cout << get_enumHehe<bar>::value << std::endl; //returns 2;
}

It works as follows:

We define a case for SFINAE to fall back on in the case that the expression T::enumFoo is invalid:

template<class T, class = void>
struct get_enumFoo
{
    static constexpr int value = -1;
};

We then define the case that sees if T::enumFoo is valid:

template<class T>
struct get_enumFoo<T, void_t<decltype(T::enumFoo)>>
{
    static constexpr int value = T::enumFoo;
};

If T::enumFoo is invalid, then the decltype(T::enumFoo) is an invalid expression, so SFINAE kicks in and we fall back to our previous case.

If T::enumFoo is valid, then decltype(T::enumFoo) is some type, but void_t<decltype(T::enumFoo)> is void. So we are specializing get_enumFoo<T, void> to have the field value = T::enumFoo .

Try it online

---

To further reduce the boilerplate behind adding new get_field traits, you could define a base class:

namespace detail {
    struct get_foo_defaultValue
    {
        static constexpr int value = -1;
    };
}

Then the base case would be

template<class T, class = void>
struct get_enumFoo
    : detail::get_foo_defaultValue
{};