相关疑难解决方法(0)

考虑以下系列的部分专业化：

template <typename T, typename Enable=void>
struct foo {
  void operator()() const { cout << "unspecialized" << endl; }
};

template <typename T>
struct foo<T, enable_if_t<
  is_integral<T>::value
>>{
  void operator()() const { cout << "is_integral" << endl; }
};

template <typename T>
struct foo<T, enable_if_t<
  sizeof(T) == 4
    and not is_integral<T>::value
>>{
  void operator()() const { cout << "size 4" << endl; }
};

template <typename T>
struct foo<T, enable_if_t<
  is_fundamental<T>::value
    and not (sizeof(T) == 4)
    and not is_integral<T>::value
>>{ …