问题
I want to partially specialize an existing template that I cannot change (std::tr1::hash) for a base class and all derived classes. The reason is that I'm using the curiously-recurring template pattern for polymorphism, and the hash function is implemented in the CRTP base class. If I only want to partially specialize for a the CRTP base class, then it's easy, I can just write:
namespace std { namespace tr1 {
template <typename Derived>
struct hash<CRTPBase<Derived> >
{
size_t operator()(const CRTPBase<Derived> & base) const
{
return base.hash();
}
};
} }
But this specialization doesn't match actual derived classes, only CRTPBase<Derived>. What I want is a way of writing a partial specialization for Derived if and only if it derives from CRTPBase<Derived>. My pseudo-code is
namespace std { namespace tr1 {
template <typename Derived>
struct hash<typename boost::enable_if<std::tr1::is_base_of<CRTPBase<Derived>, Derived>,
Derived>::type>
{
size_t operator()(const CRTPBase<Derived> & base) const
{
return base.hash();
}
};
} }
...but that doesn't work because the compiler can't tell that enable_if<condition, Derived>::type is Derived. If I could change std::tr1::hash, I'd just add another dummy template parameter to use boost::enable_if, as recommended by the enable_if documentation, but that's obviously not a very good solution. Is there a way around this problem? Do I have to specify a custom hash template on every unordered_set or unordered_map I create, or fully specialize hash for every derived class?
回答1:
There are two variants in the following code. You could choose more appropriated for you.
template <typename Derived>
struct CRTPBase
{
size_t hash() const {return 0; }
};
// First case
//
// Help classes
struct DummyF1 {};
struct DummyF2 {};
struct DummyF3 {};
template<typename T> struct X;
// Main classes
template<> struct X<DummyF1> : CRTPBase< X<DummyF1> > {
int a1;
};
template<> struct X<DummyF2> : CRTPBase< X<DummyF2> > {
int b1;
};
// typedefs
typedef X<DummyF1> F1;
typedef X<DummyF2> F2;
typedef DummyF3 F3; // Does not work
namespace std { namespace tr1 {
template<class T>
struct hash< X<T> > {
size_t operator()(const CRTPBase< X<T> > & base) const
{
return base.hash();
}
};
}} // namespace tr1 // namespace std
//
// Second case
struct DummyS1 : CRTPBase <DummyS1> {
int m1;
};
//
template<typename T>
struct Y : T {};
//
typedef Y<DummyS1> S1;
namespace std { namespace tr1 {
template<class T>
struct hash< Y<T> > {
size_t operator()(const CRTPBase<T> & base) const
{
return base.hash();
}
};
}} // namespace tr1 // namespace std
void main1()
{
using std::tr1::hash;
F1 f1;
F2 f2;
F3 f3;
hash<F1> hf1; size_t v1 = hf1(f1); // custom hash functor
hash<F2> hf2; size_t v2 = hf2(f2); // custom hash functor
hash<F3> hf3; size_t v3 = hf3(f3); // error: standard hash functor
S1 s1;
hash<S1> hs1; size_t w1 = hs1(s1); // custom hash functor
}
回答2:
Instead of modifying std::tr1::hash you should make your own namespace and define there new structure hash which inherited from std::tr1::hash or is specialized for CRTPBase<Derived>.
template <typename Derived>
struct CRTPBase
{
size_t hash() {return 0; }
};
struct AA : CRTPBase <AA> {};
struct BB {};
//
namespace mynamespace {
template <typename Some, typename Dummy=char>
struct hash : std::tr1::hash<Some> {};
//
template <typename Derived>
struct hash<Derived,
typename boost::enable_if< std::tr1::is_base_of<CRTPBase<Derived>, Derived>, char>::type >
{
size_t operator()(const CRTPBase<Derived> & base) const
{
return base.hash();
}
};
} // namespace mynamespace {}
//
//
void ff()
{
using namespace mynamespace;
hash<AA> aa; // my hash
hash<BB> bb; // std::tr1::hash
}
来源：https://stackoverflow.com/questions/1032973/how-to-partially-specialize-a-class-template-for-all-derived-types