访问类型成员

Question

在我的例子中,我有一个班级Foo<T>.在我的函数中,test我需要获取Foo正常类型的模板参数.首先我开始使用std::conditional但忘记了模板参数必须全部有效,无论选择哪一个.是为non-Foo类型创建类型特化的唯一方法吗？

例

#include <type_traits>

template <typename TYPE>
class Foo
{
public:
  using M = TYPE;
};

template <typename T>
void test(const T& a)
{
  // actually I would have used !is_foo<T>::value for the first arg
  // but this check is fine to minimise the example
  using MY_TYPE = typename std::conditional<
    std::is_same<T, int>::value,
    T,
    typename T::M>::type; // <---Error: error: type 'int' cannot be used prior to '::' because it has no members
}

int main()
{
  test(Foo<int>()); // MY_TYPE must be int
  test(int()); // MY_TYPE must be int
  return 0;
}

Answer 1

那么你可以UnFoo帮助你找到合适的类型:

template <typename T>
struct UnFoo {
    using type = T;
};

template <typename T>
struct UnFoo<Foo<T>> {
    using type = T;
};

template <typename T>
void test(const T& a)
{
  using MY_TYPE = typename UnFoo<T>::type; //maybe with a helper to get rid of typename
}

另一种选择是为其编写一个重载Foo<T>并将其委托给另一个函数,但这取决于你的实际test函数的作用.