我正在努力学习类型特征的概念。我写了一些代码来测试我的理解:
#include <iostream>
#include <typeinfo>
#include <utility>
class Normal1 {};
class Normal2 {};
class Special {};
struct Normal_tag {};
struct Special_tag {};
template <typename T>
struct trait {
typedef Normal_tag Type;
};
template <>
struct trait<Special> {
typedef Special_tag Type;
};
template <typename T>
void handle_impl(T&& object, Normal_tag) {
std::cout << "normal called\n";
}
template <typename T>
void handle_impl(T&& object, Special_tag) {
std::cout << "special called\n";
}
// method 1: can't pass in rvalue
// template <typename T>
// void handle(T& object) {
// handle_impl(object, typename trait<T>::Type());
// std::cout << '\t' << typeid(T).name() << '\n'
// << '\t' << typeid(typename trait<T>::Type).name() << '\n';
// }
// method 2: always lvalue
// template <typename T>
// void handle(const T& object) {
// handle_impl(object, typename trait<T>::Type());
// std::cout << '\t' << typeid(T).name() << '\n'
// << '\t' << typeid(typename trait<T>::Type).name() << '\n';
// }
// method 3: try to use universal reference
template <typename T>
void handle(T&& object) {
// handle_impl(object, typename trait<T>::Type());
handle_impl(std::forward<T>(object), typename trait<T>::Type());
std::cout << '\t' << typeid(T).name() << '\n'
<< '\t' << typeid(typename trait<T>::Type).name() << '\n';
}
int main(int argc, char *argv[])
{
Normal1 n1;
Normal2 n2;
Special sp;
handle(sp); // This line
handle(n1);
handle(n2);
handle(Special());
handle(Normal1());
handle(Normal2());
return 0;
}
下面的输出不是我的预期,我希望特殊方法可以调用左值和右值参数:
normal called
7Special
10Normal_tag
normal called
7Normal1
10Normal_tag
normal called
7Normal2
10Normal_tag
special called
7Special
11Special_tag
normal called
7Normal1
10Normal_tag
normal called
7Normal2
10Normal_tag
我认为输出意味着类Special
用于实例化。但为什么我会Normal_tag
?为什么呼叫handle(sp);
表现得这样?
我希望通用引用能够处理左值和右值参数,这是一种不好的方法吗?
答案 0 :(得分:3)
当您致电handle(sp);
sp 为L值时,handle
模板中T
推断为Special&
,但您不知道#39; t具有Special&
template <>
struct trait<Special&> {
typedef Special_tag Type;
};
因此你将正常称为作为输出。