Use boost::hash_value to define std::hash in C++11 - c++

Question

Is there an easy way to do the following with C++11 & Boost:
use the standard definitions of std::hash whenever available from <functional>
use boost::hash_value to define std::hash in those cases where std::hash is missing but boost::hash_value is available in <boost/functional/hash.hpp>.
For example:
std::hash<std::vector<bool>> should come from the standard library,
std::hash<std::vector<unsigned>> should be implemented with boost::hash_value.

Answer 1

The first idea that comes to mind is to use SFINAE and try std::hash<> if possible and otherwise use boost::hash_value(), like this:
#include <string>
#include <functional>
#include <type_traits>
#include <boost/functional/hash.hpp>
struct my_struct_0 {
std::string s;
};
template <typename T>
struct has_std_hash_subst { typedef void type; };
template <typename T, typename C = void>
struct has_std_hash : std::false_type {};
template <typename T>
struct has_std_hash<
T,
typename has_std_hash_subst<decltype( std::hash<T>()(T()) ) >::type
> : std::true_type {};
template <typename T>
static typename std::enable_if<has_std_hash<T>::value, size_t>::type
make_hash(const T &v)
{
return std::hash<T>()(v);
}
template <typename T>
static typename std::enable_if<(!has_std_hash<T>::value), size_t>::type
make_hash(const T &v)
{
return boost::hash_value(v);
}
int main()
{
make_hash(std::string("Hello, World!"));
make_hash(my_struct_0({ "Hello, World!" }));
}
Unfortunately, there is always a default specialization of std::hash that triggers static_assert failure. This may not be the case with other libraries but it is the case with GCC 4.7.2 (see bits/functional_hash.h:60):
/// Primary class template hash.
template<typename _Tp>
struct hash : public __hash_base<size_t, _Tp>
{
static_assert(sizeof(_Tp) < 0,
"std::hash is not specialized for this type");
size_t operator()(const _Tp&) const noexcept;
};
So the above SFINAE approach doesn't work — static_assert in there is a show-stopper. Therefore, you cannot really determine when std::hash is available.
Now, this does not really answer your question but might come handy — it is possible to do this trick the other way around — check for Boost implementation first and only then fall back to std::hash<>. Consider the below example that uses boost::hash_value() if it is available (i.e. for std::string and my_struct_0) and otherwise uses std::hash<> (i.e. for my_struct_1):
#include <string>
#include <functional>
#include <type_traits>
#include <boost/functional/hash.hpp>
struct my_struct_0 {
std::string s;
};
struct my_struct_1 {
std::string s;
};
namespace boost {
size_t hash_value(const my_struct_0 &v) {
return boost::hash_value(v.s);
}
}
namespace std {
template <>
struct hash<my_struct_1> {
size_t operator()(const my_struct_1 &v) const {
return std::hash<std::string>()(v.s);
}
};
}
template <typename T>
struct has_boost_hash_subst { typedef void type; };
template <typename T, typename C = void>
struct has_boost_hash : std::false_type {};
template <typename T>
struct has_boost_hash<
T,
typename has_boost_hash_subst<decltype(boost::hash_value(T()))>::type
> : std::true_type {};
template <typename T>
static typename std::enable_if<has_boost_hash<T>::value, size_t>::type
make_hash(const T &v)
{
size_t ret = boost::hash_value(v);
std::cout << "boost::hash_value(" << typeid(T).name()
<< ") = " << ret << '\n';
return ret;
}
template <typename T>
static typename std::enable_if<(!has_boost_hash<T>::value), size_t>::type
make_hash(const T &v)
{
size_t ret = std::hash<T>()(v);
std::cout << "std::hash(" << typeid(T).name()
<< ") = " << ret << '\n';
return ret;
}
int main()
{
make_hash(std::string("Hello, World!"));
make_hash(my_struct_0({ "Hello, World!" }));
make_hash(my_struct_1({ "Hello, World!" }));
}
Hope it helps.
UPDATE: Perhaps you could use the hack described here as pointed out by #ChristianRau and make the first SFINAE approach work! Though it is very dirty :)

Answer 2

My answer might not be correct, but I will try to explain why I think that the answer is no.
I don't think that std::hash<T> and boost:hash<T> can be used interchangeably, so I've tried hiding object creation (even if this is not perfect solution), and return their result, which is size_t. Method should be of course chosen at compile time, so function dispatch is what comes to my mind, sample code:
template <typename T>
size_t createHash(const T& t, false_type)
{
return boost::hash<T>()(t);
}
template <typename T>
size_t createHash(const T& t, true_type)
{
return std::hash<T>()(t);
}
template<typename T>
size_t createHash(const T& t)
{
return createHash<T>(t, std::is_XXX<T>::type());
}
int main()
{
vector<unsigned> v; v.push_back(1);
auto h1 = createHash(v);
cout << " hash: " << h1;
//hash<vector<unsigned> > h2;
}
The idea of this code is simple: if you can construct type of type std::hash<T>, choose second implementation, if not - choose first one.
If the first implementation is chosen, code compiles without a problem, you can check it by using fe. std::is_array<T>::type() in a wrapper function, which is of course not true, so boost::hash implementation will be choosed. However, if you use a trait which will return true_t for a vector<unsigned>, like fe. std::is_class<T>::type() then the compiler will report "C++ Standard doesn't provide...", which is a result of a static_assert.
For this to work, we would need to force compiler to return true_t if a type is really constructible (it doesn't fail static_assert) and false_t if it doesn't. However, I don't think there is a possibility to do that.