为什么我不能添加带有类型查找的抽象层来剥离C ++中的引用? [英] Why can't I add a layer of abstraction with type lookup to strip references in C++?
问题描述
这是对以下问题的解答
已正确回答。但是我在下面尝试了自己的解决方案
which has been correctly answered. However I tried my own solution below
#include <type_traits>
template <typename T, typename enable = void> struct Traits {
static const bool value = false;
};
template <typename T> struct Traits<T,std::enable_if<std::is_reference<T>::value>> {
static const bool value = Traits<typename std::remove_reference<T>::type>::value;
};
struct Zip{};
template <> struct Traits<Zip,void> {
static const bool value = true;
};
template <typename E>
void Execute(E && e){
static_assert(Traits<E>::value);
}
int main(){
auto z = Zip();
Execute(z);
}
理论是,如果正确的专业化失败,那么下一个最专业的将是根据 T
是否作为参考进行匹配的一个。如果匹配,则删除引用,我们希望递归获得匹配。但这似乎不起作用。有没有办法解决这个问题,从而保持我的尝试精神?
the theory being that if the correct specialisation fails then the next most specialized will be the one that matches based on if T
is a reference. If this matches then the reference is stripped and we recurse hopefully getting a match. But this does not seem to work. Is there a way to fix this that keep the spirit of my attempt?
推荐答案
您正在滥用 std :: enable_if
。如所写,您的 Traits
结构专门针对 std :: enable_if
本身,而不是其结果。您需要访问嵌套的 :: type
类型别名:
You are misusing std::enable_if
. As written, your Traits
struct is specialized on std::enable_if
itself, not its result. You need to access the nested ::type
type alias:
typename std::enable_if<std::is_reference<T>::value>::type
// or
std::enable_if_t<std::is_reference<T>::value>
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