Defining a hash function in TR1 unordered_map inside a struct
According to this, it is possible to define an equality function in a TR1 unordered_map like this:
#incl开发者_JAVA技巧ude <tr1/unordered_map>
using namespace std;
using namespace std::tr1;
struct foo{
...
bool operator==(const foo& b) const{
return ..;
}
};
unordered_map<foo,int> map;
Is it possible to define the hash function the same way?
If you want to change the default hashing (or, more often, provide hashing for a type that isn't currently supported), you provide a specialization of std::tr1::hash<T>
for your key-type:
namespace std {
namespace tr1 {
template<>
struct hash<typename my_key_type> {
std::size_t operator()(my_key_type const &key) {
return whatever;
}
};
}
}
Note that specializing an existing template for a user-defined type is one of the rare cases where you specifically are allowed to write code in namespace std
.
The signature of the unordered_map class is this:
template<class Key,
class Ty,
class Hash = std::hash<Key>,
class Pred = std::equal_to<Key>,
class Alloc = std::allocator<std::pair<const Key, Ty> > >
class unordered_map;
Your example works because the default Pred, std::equal_to<>, by default checks for equality using operator==. The compiler finds your foo::operator== member function and uses that.
std::hash doesn't have any specialisation which will call a member function on your class, so you can't just add a member to foo with a custom hash. You will need to specialise std::hash instead. If you want that to call a member function on foo, go ahead. You'll end up with something like this:
struct foo
{
size_t hash() const
{
// hashing method here, return a size_t
}
};
namespace std
{
// Specialise std::hash for foo.
template<>
class hash< foo >
: public unary_function< foo, size_t >
{
public:
size_t operator()( const foo& f )
{
return f.hash();
}
};
}
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