What is the purpose of `enum class` with a specified underlying type, but no enumerators?

Question

The std::align_val_t type is defined as:

namespace std
{
    enum class align_val_t : size_t
    {
    };
}

What is the purpose of such an empty enumeration?

What's the difference with a typedef?

Answer 1

It is used as a strongly typed integer, which disables implicit type conversions.

enum class strong_typed : size_t {};

typedef size_t defed;

int strong(strong_typed f);
int weak(defed f);

int main()
{
    weak(defed{5}); // works
    weak(5); // works

    strong(strong_typed{5}); // works
    strong(5);               // error cannot convert int to strong_typed
    strong(size_t{5});       // error cannot convert size_t to strong_typed
}

godbolt demo

you can get back the underlying type with std::to_underlying inside the function.

Strongly typed integers are useful when you have a function accepting many arguments of the same type and you want to disambiguate them to avoid swapping arguments by mistake, or invoking the wrong overload.

---

As for why it is used in std::align_val_t, according to Ted Lyngmo in the comments, it is likely to disambiguate operator new in the variadic case.

// called by new (size_t{5}) T{};
void* operator new  ( std::size_t count, size_t arg );

// called by new T{}; on a T with large alignment
void* operator new  ( std::size_t count, std::align_val_t al );