Why is the template specialization not chosen?

问题

I wrote the following code:

#include <iostream>
#include <string>
#include <type_traits>

template<typename, typename = void>
struct is_incrementable : std::false_type {};

template<typename T>
struct is_incrementable<T, decltype( ++std::declval<T&>() )> : std::true_type {};

int main()
{
    std::cout << is_incrementable<std::string>::value << std::endl;
    std::cout << is_incrementable<int>::value << std::endl;
}

When I run it, I get 0 0. But I expected 0 1.

Any ideas?

回答1:

For std::string, the primary template is chosen and the specialization is considered. But decltype(++std::declval<T&>()) is ill-formed, so it is not considered and the primary template (non-specialized template) is used, which results in 0.

If you use int, it gets a bit more complicated. The primary template gets chosen by the compiler, as always, and then the specialization is considered (that's because specialization are always considered better matches). The specialization is <int, int&> for int, but it doesn't match the non-specialized template <int, void> (the void is the default template argument), so the specialization is ignored because it doesn't match.

So, the types of default template parameter have to match, or else the specialization is not taken into account, as a specialization is only taken when every template argument matches the specialization.

Just append a void() at the end to make the specialization match for the second template parameter, as the left expression is discarded and the type of void() is void, which matches the primary template's second template parameter.

template<typename T>
struct is_incrementable<T, decltype( ++std::declval<T&>(), void() )> : std::true_type {};

In C++17, you would use std::void_t for that.

来源：https://stackoverflow.com/questions/44858395/why-is-the-template-specialization-not-chosen