enable_if using a constexpr bool test not working

Question

I have a maths function that I want to be able to accept either a double, or a array/vector/container of doubles, and behave slightly differently.

I am attempting to use SFINAE and type traits to select the correct function.

Here is a minimal example:

#include <iostream>
#include <vector>
#include <type_traits>

template <typename T>
constexpr bool Iscontainer()
{
    if constexpr (std::is_class<T>::value && std::is_arithmetic<typename T::value_type>::value) {
        return true;
    }
    return false;
}

// Function 1 (double):
template <typename T>
typename std::enable_if<std::is_arithmetic<T>::value>::type g(T const & t)
{
    std::cout << "this is for a double" << t << std::endl;
}

// Function 2 (vec), version 1:
template <typename T>
typename std::enable_if<IsContainer<T>()>::type g(T const & t)
{
    std::cout << "this is for a container" << t[0] << std::endl;
}

int main()
{
    std::vector<double> v {1, 2};
    std::array<double, 2> a {1, 2};
    double d {0.1};

    g<>(v);
    g<>(a);
    g<>(d);  // error here
}

I get a compile time error:

../main.cpp:8:47: error: ‘double’ is not a class, struct, or union type
     if constexpr (std::is_class<T>::value && std::is_arithmetic<typename     T::value_type>::value) {
                   ~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~

However, when I replace function 2 with:

// Function 2 (vec), version 2:
template <typename T>
typename std::enable_if<std::is_class<T>::value && std::is_arithmetic<typename T::value_type>::value>::type
g(T const & t)
{
    std::cout << "this is for a vector" << t[0] << std::endl;
}

It works.

My problem is I don't understand why the first version does not work.. And I prefer the readability of the first version.

Answer 1

The reason why it fails is simple. You do not invoke SFINAE, and when the compiler tries to evaluate the expressions it sees:

if constexpr (std::is_class<double>::value // this is fine it's false
   && std::is_arithmetic<typename double::value_type>::value // problem here!
)

The whole statement is evaluated, there is no short-circuit for the if. The closest solution to what you currently have is to explicitly split the if, so that the problematic part is discarded when T is not a class and the second check is nonsensical.

#include <iostream>
#include <vector>
#include <type_traits>

template <typename T>
constexpr bool IsVector()
{
    if constexpr (std::is_class<T>::value) {
        if constexpr (std::is_arithmetic<typename T::value_type>::value) {
            return true;
        }
    }
    return false;
}

// Function 1 (double):
template <typename T>
typename std::enable_if<std::is_arithmetic<T>::value>::type g(T const & t)
{
    std::cout << "this is for a double" << t << std::endl;
}

// Function 2 (vec), version 1:
template <typename T>
typename std::enable_if<IsVector<T>()>::type g(T const & t)
{
    std::cout << "this is for a vector" << t[0] << std::endl;
}

int main()
{
    std::vector<double> v {1, 2};
    double d {0.1};

    g<>(v);
    g<>(d);  // error here
}

Alternatively I'd suggest a using alias:

template <typename T>
using IsVector2 = std::conjunction<typename std::is_class<T>, std::is_arithmetic<typename T::value_type> >;

template <typename T>
typename std::enable_if<IsVector2<T>::value>::type g(T const & t)
{
    std::cout << "this is for a vector" << t[0] << std::endl;
}

You could also name it better. It doesn't really check whether T is a vector, or a container (after your edit). Your current definition is a bit loose as well.