Is it possible to mix SFINAE and template specialisation?

Question

Here is what I am roughly trying to achieve:

// the declaration
template<typename... Args>
struct ArgsEstimate;

// specialisation for string, SFINAE would be overkill
template<typename... Args>
struct ArgsEstimate<std::string&, Args...> {
    static const std::size_t size = 64 + ArgsEstimate<Args...>::size;
};

// specialisation for arithmetic types
template<typename AirthmeticT,
         typename std::enable_if<std::is_arithmetic<AirthmeticT>::value>::type* = nullptr,
         typename... Args>
struct ArgsEstimate<AirthmeticT, Args...> {
    static const std::size_t size = sizeof(AirthmeticT) + ArgsEstimate<Args...>::size;
};

// specialisation for pointer types
template<typename PtrT,
         typename std::enable_if<std::is_pointer<PtrT>::value>::type* = nullptr,
         typename... Args>
struct ArgsEstimate<PtrT, Args...> {
    static const std::size_t size = 32 + ArgsEstimate<Args...>::size;
};

The problem is that this code gives a compilation error "template parameters not deducible in partial specialization" at the points I have done enable_if. A static_assert inside the struct won't work either since there will be redefinition.

I know, I can probably do this with SFINAE and function overloading alone. However, for cases like just std::string, using SFINAE is an overkill.

So I was wondering if there is clean way of mixing template specialisation and SFINAE.

bolov bolov · Accepted Answer · 2020-06-23T05:10:45

Direct answer to your question

You can, but you really can't. Your case is complicated by variadic template arguments.

// specialisation for arithmetic types
template<class AirthmeticT, class... Args>
struct ArgsEstimate<
    AirthmeticT,
    std::enable_if_t<std::is_arithmetic_v<AirthmeticT>>,
    Args...>
{
    static const std::size_t size = sizeof(AirthmeticT) + ArgsEstimate<Args...>::size;
};

This works... sort of. You just need to make sure the second parameter is always void:

ArgsEstimate<int, void, /* ... */> ok; // will use the integer specialization

ArgsEstimate<int, int, int> wrong; // oups, will use the base template.

This is impractical.

C++20 concepts

Concepts elegantly solve this:

// specialisation for arithmetic types
template<class T, class... Args>
    requires  std::is_arithmetic_v<T>
struct ArgsEstimate<T, Args...>
{
    static const std::size_t size = sizeof(T) + ArgsEstimate<Args...>::size;
};

The pre-concepts solution

What you need to do is to split your class into two classes. One that defines the size just for 1 argument. Here you can use SFINAE. And the other one that summs them:

template <class T, class Enable = void>
struct ArgEstimate {};

// specialisation for string, SFINAE would be overkill
template<>
struct ArgEstimate<std::string&>
{
    static const std::size_t size = 64;
};

// specialisation for arithmetic types
template<class T>
struct ArgEstimate<T, std::enable_if_t<std::is_arithmetic_v<T>>>
{
    static const std::size_t size = sizeof(T);
};

// specialisation for pointer types
template <class T>
struct ArgEstimate<T*>
{
    static const std::size_t size = 32;
};

// the declaration
template<class... Args> struct ArgsEstimate;

template<class T>
struct ArgsEstimate<T>
{
    static const std::size_t size = ArgEstimate<T>::size;
};

template<class Head, class... Tail>
struct ArgsEstimate<Head, Tail...>
{
    static const std::size_t size = ArgEstimate<Head>::size + ArgsEstimate<Tail...>::size;
};

And if you have C++17 you can use fold expression to simplify the sum:

template<class... Args>
struct ArgsEstimate
{
    static const std::size_t size = (... + ArgEstimate<Args>::size);
};

Also just wanted to point out that you don't need SFINAE for pointers:

// specialisation for pointer types
template <class T, class... Args>
struct ArgsEstimate<T*, Args...> {
    static const std::size_t size = 32 + ArgsEstimate<Args...>::size;
};

Is it possible to mix SFINAE and template specialisation?

1 Answers

Direct answer to your question

C++20 concepts

The pre-concepts solution