问题描述
我有一个抽象基类 Hashable ，可以散列的类派生自。现在我想扩展 std :: hash 到所有派生自 Hashable 的类。
下面的代码应该完全做到这一点。
#include< functional>
#include< type_traits>
#include< iostream>
$ b $ class Hashable {
public：
virtual〜Hashable（）{}
virtual std :: size_t Hash（）const = 0;
};
类派生：public Hashable {
public：
std :: size_t Hash（）const {
return 0;
}
};
//对std :: hash进行专门化以在Hashable或从
派生的任何类上进行操作// Hashable。
namespace std {
template< class C>
struct hash {
typename std :: enable_if< std :: is_base_of< Hashable，C> :: value，std :: size_t> :: type
operator（）（const C& object ）const {
return object.Hash（）;
}
};
}
int main（int，char **）{
std :: hash< Derived>散列器;
派生的d;
std :: cout<< （d）<<的std :: ENDL;
返回0;
}
上述代码与gcc 4.8.1完全一样，但是当我尝试使用gcc 4.7.2进行编译，我得到以下内容：
$ $ $ $ $ $ c ++ -std = c ++ 11 -o测试test_hash.cpp
test_hash.cpp：22：8：错误：重新定义'struct std :: hash< _Tp>'
在/usr/include/c++/4.7/functional:59包含的文件中：0，来自test_hash.cpp的
：1：
/usr/include/c++/4.7/bits/functional_hash.h:58:12：error：'struct std :: hash< _Tp>的前一个定义;'
/usr/include/c++/4.7/bits/functional_hash.h：在'struct std :: hash< Derived>'的实例化中：
test_hash.cpp：31：24：从这里需要
/usr/include/c++/4.7/bits/functional_hash.h:60:7：错误：静态断言失败：std :: hash不是专门用于这种类型的
任何人都可以想出一种方法来使任何类的 std :: hash 派生自 Hashable wi th gcc 4.7.2？
好像没有正确的方法来做我想做的事情。我已决定使用以下宏为每个派生类编写单独的特化：
//宏可方便地为一个从Hashable派生的类
#define DEFINE_STD_HASH_SPECIALIZATION（hashable）\
namespace std {\
template<> \
struct hash< hashable> {\
std :: size_t operator（）（const hashable& object）const {\
return object.Hash（）; \
} \
}; \
}
然后
//派生
的std :: hash特化。DEFINE_STD_HASH_SPECIALIZATION（Derived）;
I have an abstract base class Hashable that classes that can be hashed derive from. I would now like to extend std::hash to all classes that derive from Hashable.
The following code is supposed to do exactly that.
#include <functional>
#include <type_traits>
#include <iostream>
class Hashable {
public:
virtual ~Hashable() {}
virtual std::size_t Hash() const =0;
};
class Derived : public Hashable {
public:
std::size_t Hash() const {
return 0;
}
};
// Specialization of std::hash to operate on Hashable or any class derived from
// Hashable.
namespace std {
template<class C>
struct hash {
typename std::enable_if<std::is_base_of<Hashable, C>::value, std::size_t>::type
operator()(const C& object) const {
return object.Hash();
}
};
}
int main(int, char**) {
std::hash<Derived> hasher;
Derived d;
std::cout << hasher(d) << std::endl;
return 0;
}
The above code works exactly as expected with gcc 4.8.1, but when I try to compile it with gcc 4.7.2, I get the following:
$ g++ -std=c++11 -o test test_hash.cpp
test_hash.cpp:22:8: error: redefinition of ‘struct std::hash<_Tp>’
In file included from /usr/include/c++/4.7/functional:59:0,
from test_hash.cpp:1:
/usr/include/c++/4.7/bits/functional_hash.h:58:12: error: previous definition of ‘struct std::hash<_Tp>’
/usr/include/c++/4.7/bits/functional_hash.h: In instantiation of ‘struct std::hash<Derived>’:
test_hash.cpp:31:24: required from here
/usr/include/c++/4.7/bits/functional_hash.h:60:7: error: static assertion failed: std::hash is not specialized for this type
Can anybody think of a way to make this specialization of std::hash work for any class derived from Hashable with gcc 4.7.2?
It seems like there is no proper way to do what I wanted to do. I have decided to just write separate specializations for each derived class, using the following macro:
// macro to conveniently define specialization for a class derived from Hashable
#define DEFINE_STD_HASH_SPECIALIZATION(hashable) \
namespace std { \
template<> \
struct hash<hashable> { \
std::size_t operator()(const hashable& object) const { \
return object.Hash(); \
} \
}; \
}
and then
// specialization of std::hash for Derived
DEFINE_STD_HASH_SPECIALIZATION(Derived);
这篇关于将派生类的std :: hash专门化的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！