I'm trying to use QString
as the key in a std::unordered_map
, however I get the error:
error C2280: 'std::hash<_Kty>::hash(const std::hash<_Kty> &)': attempting to reference a deleted function
I can't switch to QHash
because the value-type of the map is non-copyable. Is there any way to make this work?
Put the hash
implementation in a header, and make sure that you include that header everywhere the map is used.
A trivial implementation that forwards to qHash
should be sufficient:
#include <QHash>
#include <QString>
#include <functional>
namespace std {
template<> struct hash<QString> {
std::size_t operator()(const QString& s) const noexcept {
return (size_t) qHash(s);
}
};
}
Even though std::size_t
is larger than unsigned int
on common 64 bit platforms, and thus the hash doesn't change across its full length - this isn't a problem. The standard places no such requirement on an std::hash
implementation.
Let's not forget, though, that modifying anything in the std
namespace is generally undefined behavior and unnecessary.
TL;DR:
You're allowed to specialize certain type and variable templates, and only if the specialization depends on at least one user-defined type. You can't fully specialize for built-in nor C++ library types.
See Extending the namespace std for reference. There, we read:
It is undefined behavior to add declarations or definitions to namespace std or to any namespace nested within std, with a few exceptions noted below.
Mainly, it is allowed to specialize certain types in the std
namespace:
It is allowed to add template specializations for any standard library class template to the namespace std only if the declaration depends on at least one program-defined type and the specialization satisfies all requirements for the original template, except where such specializations are prohibited.
Note that what's legal is specializing types, i.e. classes. Functions and member functions? Not ever:
It is undefined behavior to declare a full specialization of any standard library function template, [... or] member function of a standard library class template, [... or] member function template of a standard library class or class template.
Another limited exception is for variable templates:
It is undefined behavior to declare a full or partial specialization of any standard library variable template, except where explicitly allowed.
Emphasis mine in all cases. As always, there are further details that one should get acquainted with.