我有一个抽象基类Hashable
,可以散列的类派生自。我现在想将std::hash
扩展到源自Hashable
。
以下代码应该就是这样做的。
#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;
}
上面的代码与gcc 4.8.1完全一样,但是当我尝试使用gcc 4.7.2编译它时,我得到以下内容:
$ 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
是否有人想到一种方法可以使std::hash
的这种专业化适用于使用gcc 4.7.2从Hashable
派生的任何类?
答案 0 :(得分:2)
似乎没有正确的方法来做我想做的事情。我决定使用以下宏为每个派生类编写单独的特化:
// 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(); \
} \
}; \
}
然后
// specialization of std::hash for Derived
DEFINE_STD_HASH_SPECIALIZATION(Derived);