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使用其基类的部分模板特化为 std::hash 散列一个对象

[英]Hash an object using its base class' partial template specialization for std::hash

I have a wrapper class for std::string that serves as base class for several others.我有一个std::string的包装类,作为其他几个的基类。 Instances of the subclasses will be used as keys in std::unordered_set so I need to provide a hash function for them.子类的实例将用作std::unordered_set键,因此我需要为它们提供散列函数。 Since the hash is only dependent on the std::string stored in the base class, I do not want to write a hash function for every subclass but rather use the one from the wrapper class.由于散列仅依赖于存储在基类中的std::string ,我不想为每个子类编写散列函数,而是使用包装类中的散列函数。

This is how I would like to solve the problem:这就是我想解决这个问题的方式:

#include <string>
#include <unordered_set>

class Wrapper {
public:
  std::string name;
  size_t _hash;
  explicit Wrapper(std::string str) : name(str), _hash(std::hash<std::string>()(name)) {}
  size_t hash() const { return _hash; }
};

class Derived : public Wrapper {};

namespace std {

template <> struct hash<Wrapper> {
  std::size_t operator()(const Wrapper &k) const { return k.hash(); }
};

template <typename T> struct hash<std::enable_if_t<std::is_base_of_v<Wrapper, T>>> {
  std::size_t operator()(const T &k) const { return k.hash(); }
};

} // namespace std

int main(void) {
  std::unordered_set<Wrapper> m1;
  std::unordered_set<Derived> m2;
}

This does not compile of course, since T cannot be deduced.这当然不能编译,因为无法推导出T Clang says: Clang 说:

20:30: error: class template partial specialization contains a template parameter that cannot be deduced; this partial specialization will never be used
20:20: note: non-deducible template parameter 'T'

And g++ says: g++ 说:

hash_subclass.cpp:21:30: error: template parameters not deducible in partial specialization:
 template <typename T> struct hash<std::enable_if_t<std::is_base_of_v<Wrapper, T>>> {
                              ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
hash_subclass.cpp:21:30: note:         'T'

I have found this solution, but I would like to avoid using a macro.我找到了这个解决方案,但我想避免使用宏。 Also, this goes against what I expect from inheritance.此外,这与我对继承的期望背道而驰。

Is there a solution for this?有解决方案吗? Can a subclass inherit its base class' specialization of std::hash ?子类可以继承其基类对std::hash吗?

Also, I'm not 100% sure about my use of std::enable_if and std::is_base_of .另外,我不是 100% 确定我使用std::enable_ifstd::is_base_of Could you tell me whether this would work assuming T could be deduced?如果可以推导出T你能告诉我这是否可行?

IRC, the problem with std::enable_if is that it does not work for classes with a single template parameter. IRC, std::enable_if的问题在于它不适用于具有单个模板参数的类。 Consequently, you cannot specialize std::hash by using std::enable_if .因此,您不能使用std::enable_if专门化std::hash

However, you can make your own hasher as follows:但是,您可以按如下方式制作自己的哈希器:

template <typename T, typename Enable = std::enable_if_t<std::is_base_of_v<Wrapper, T>>>
struct WrapperHasher {
   std::size_t operator()(const T& k) const { return k.hash(); }
};

And then use it as a second template argument of std::unordered_set :然后将其用作std::unordered_set的第二个模板参数:

std::unordered_set<Wrapper, WrapperHasher<Wrapper>> m1;
std::unordered_set<Derived, WrapperHasher<Derived>> m2;

But in your case, you can define a wrapper much more simply as:但在你的情况下,你可以更简单地定义一个包装器:

struct WrapperHasher {
   std::size_t operator()(const Wrapper& k) const { return k.hash(); }
};

And then write:然后写:

std::unordered_set<Wrapper, WrapperHasher> m1;
std::unordered_set<Derived, WrapperHasher> m2;

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