如何在 C + + 0x 中组合散列值？

Well, just do it like the boost guys did it:

template <class T>
inline void hash_combine(std::size_t& seed, const T& v)
{
std::hash<T> hasher;
seed ^= hasher(v) + 0x9e3779b9 + (seed<<6) + (seed>>2);
}

I'll share it here since it can be useful to others looking for this solution: starting from @KarlvonMoor answer, here's a variadic template version, which is terser in its usage if you have to combine several values together:

inline void hash_combine(std::size_t& seed) { }


template <typename T, typename... Rest>
inline void hash_combine(std::size_t& seed, const T& v, Rest... rest) {
std::hash<T> hasher;
seed ^= hasher(v) + 0x9e3779b9 + (seed<<6) + (seed>>2);
hash_combine(seed, rest...);
}

Usage:

std::size_t h=0;
hash_combine(h, obj1, obj2, obj3);

This was written originally to implement a variadic macro to easily make custom types hashable (which I think is one of the primary usages of a hash_combine function):

#define MAKE_HASHABLE(type, ...) \
namespace std {\
template<> struct hash<type> {\
std::size_t operator()(const type &t) const {\
std::size_t ret = 0;\
hash_combine(ret, __VA_ARGS__);\
return ret;\
}\
};\
}

Usage:

struct SomeHashKey {
std::string key1;
std::string key2;
bool key3;
};


MAKE_HASHABLE(SomeHashKey, t.key1, t.key2, t.key3)
// now you can use SomeHashKey as key of an std::unordered_map

This could also be solved by using a variadic template as follows:

#include <functional>


template <typename...> struct hash;


template<typename T>
struct hash<T>
: public std::hash<T>
{
using std::hash<T>::hash;
};




template <typename T, typename... Rest>
struct hash<T, Rest...>
{
inline std::size_t operator()(const T& v, const Rest&... rest) {
std::size_t seed = hash<Rest...>{}(rest...);
seed ^= hash<T>{}(v) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
return seed;
}
};

Usage:

#include <string>


int main(int,char**)
{
hash<int, float, double, std::string> hasher;
std::size_t h = hasher(1, 0.2f, 2.0, "Hello World!");
}

One could certainly make a template function, but this could cause some nasty type deduction e.g hash("Hallo World!") will calculate a hash value on the pointer rather than on the string. This is probably the reason, why the standard uses a struct.

A few days ago I came up with slightly improved version of this answer (C++ 17 support is required):

template <typename T, typename... Rest>
void hashCombine(uint& seed, const T& v, Rest... rest)
{
seed ^= ::qHash(v) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
(hashCombine(seed, rest), ...);
}

The code above is better in terms of code generation. I used qHash function from Qt in my code, but it's also possible to use any other hashers.

I really like the C++17 approach from the answer by vt4a2h, however it suffers from a problem: The Rest is passed on by value whereas it would be more desirable to pass them on by const references (which is a must if it shall be usable with move-only types).

Here is the adapted version which still uses a fold expression (which is the reason why it requires C++17 or above) and uses std::hash (instead of the Qt hash function):

template <typename T, typename... Rest>
void hash_combine(std::size_t& seed, const T& v, const Rest&... rest)
{
seed ^= std::hash<T>{}(v) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
(hash_combine(seed, rest), ...);
}

For completeness sake: All the types which shall be usable with this version of hash_combine must have a template specialization for hash injected into the std namespace.

Example:

namespace std // Inject hash for B into std::
{
template<> struct hash<B>
{
std::size_t operator()(B const& b) const noexcept
{
std::size_t h = 0;
cgb::hash_combine(h, b.firstMember, b.secondMember, b.andSoOn);
return h;
}
};
}

So that type B in the example above is also usable within another type A, like the following usage example shows:

struct A
{
std::string mString;
int mInt;
B mB;
B* mPointer;
}


namespace std // Inject hash for A into std::
{
template<> struct hash<A>
{
std::size_t operator()(A const& a) const noexcept
{
std::size_t h = 0;
cgb::hash_combine(h,
a.mString,
a.mInt,
a.mB, // calls the template specialization from above for B
a.mPointer // does not call the template specialization but one for pointers from the standard template library
);
return h;
}
};
}

The answer by vt4a2h is certainly nice but uses the C++17 fold expression and not everyone is able to switch to a newer toolchain easily. The version below uses the expander trick to emulate a fold expression and works in C++11 and C++14 as well.

Additionally, I marked the function inline and use perfect forwarding for the variadic template arguments.

template <typename T, typename... Rest>
inline void hashCombine(std::size_t &seed, T const &v, Rest &&... rest) {
std::hash<T> hasher;
seed ^= hasher(v) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
(int[]){0, (hashCombine(seed, std::forward<Rest>(rest)), 0)...};
}

Live example on Compiler Explorer

The answer by Henri Menke works great, but if you treat warnings as errors with for example:

add_compile_options(-Werror)

GCC 9.3.0 will give this error:

Test.h:223:67: error: ISO C++ forbids compound-literals [-Werror=pedantic]
223 |     (int[]){0, (hashCombine(seed, std::forward<Rest>(rest)), 0)...};
|                                                                  ^
cc1plus: all warnings being treated as errors

We can update the code to avoid the error like this:

template <typename T, typename... Rest>
inline void hashCombine(std::size_t &seed, T const &v, Rest &&... rest) {
std::hash<T> hasher;
seed ^= (hasher(v) + 0x9e3779b9 + (seed << 6) + (seed >> 2));
int i[] = { 0, (hashCombine(seed, std::forward<Rest>(rest)), 0)... };
(void)(i);
}

You can use the rst C++ library that I developed to do that:

#include "rst/stl/hash.h"


struct Point {
Point(const int x, const int y) : x(x), y(y) {}


int x = 0;
int y = 0;
};


bool operator==(const Point lhs, const Point rhs) {
return (lhs.x == rhs.x) && (lhs.y == rhs.y);
}


namespace std {


template <>
struct hash<Point> {
size_t operator()(const Point point) const {
return rst::HashCombine({point.x, point.y});
}
};


}