比 std::reference_wrapper 更好地模拟 C++ 引用的类型
我正在设计一个类，它的行为类似于 C++ 引用，但做了一些额外的工具性工作(簿记).
I am designing a class that behaves like a C++ references but does some extra instrumental stuff (bookkeeping).
最初我认为 std::reference_wrapper 将是一个很好的模型.但过了一段时间我意识到 std::reference_wrapper 的行为，即使在原则上，也不能作为 C++ 引用，因为赋值重新绑定了内部指针.
Initially I thought that std::reference_wrapper<T> would be a good model. But after a while I realized that std::reference_wrapper<T> doesn't behave, even in principle, as a C++ reference because assignment rebinds the internal pointer.
double a = 5.;
double b = 3.;
double& ref = a;
ref = b;
assert(&ref != &b); // ref is not bound to b
std::reference_wrapper<double> refwpr = a;
refwpr = b;
assert(&refwpr.get() == &b); // ref wrapper was rebound to b
我当然可以为我自己的类更改该行为(而不是重新绑定)，但我认为可能已经存在模拟引用的类.例如类似于std::real_reference".
I can of course change that behavior for my own class (not to rebind) but I though that maybe a class that emulates a reference is already out there. For example something like "std::real_reference<T>".
是否有一个类(标准或非标准)可以更接近地模拟引用?
我认为它也很有用，例如在人们看到的许多地方std::vector<std::reference_wrapper<T>> 作为 std::vector<T&> 的替代方案出现，但这是一种误导，因为语义是不同，但可以通过一些修改来实现std::vector.
I think it can be useful too, for example in many places one sees
std::vector<std::reference_wrapper<T>> presented as an alternative to std::vector<T&> but this is misleading because the semantics is different but could be achieved with some modifications std::vector<std::real_reference<T>>.
我现在认为，如果一个完美"的引用包装器是可能的，那么引用可以只实现指针，我们就不需要T&但只是有一个库功能 reference 在所有方面都与 T& 完全一样......
I now think that if a "perfect" reference wrapper were possible, then references could be just implemented with pointers and we wouldn't need T&
but just simply have a library feature reference<T> that works exactly like T& in all respects....
... 这反过来意味着该语言可以允许您在类型上重载" & 属性以生成自定义类型而不是语言引用.例如 使用 mytype&= custom_reference_emulator_of_mytype.
... which in turn means that the language could allow you to "overload" the & attribute over types to produce a custom type instead of a language reference. For example using mytype& = custom_reference_emulator_of_mytype.
推荐答案
template <typename T>
class Tref {
std::tuple<T &> t;
public:
template <typename... Args>
Tref(Args&&... args): t(std::forward<Args>(args)...) {}
operator T &() { return std::get<0>(t); }
decltype(&std::get<0>(t)) operator&() { return &std::get<0>(t); }
};
Tref<double &> t(a);
t = b;
assert(&t != &b);
std::reference_wrapper 所做的是在模板删除引用时进行绑定.您不需要构建一个类来满足您的需求，只需使用普通引用即可.(我可能描述不清楚.我的英语很差.)
What std::reference_wrapper does is binding when the template erase the reference. You needn't build a class to meet your needs, and just use normal reference. (I may not describe clearly. My English is poor.)
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