有谁知道 bitset 的哈希函数使用的是什么算法，

这是来自网站:http://en.cppreference.com/w/cpp/utility/bitset/hash

#include <iostream>
#include <bitset>
#include <functional>

int main()
{
    std::bitset<4> b1(1);
    std::bitset<4> b2(2);
    std::bitset<4> b3(b2);
    std::bitset<4> b4(8);
    std::cout<<b4<<'\n';
    std::hash<std::bitset<4>> hash_fn;

    size_t h1 = hash_fn(b1);
    size_t h2 = hash_fn(b2);
    size_t h3 = hash_fn(b4);

    std::cout << h1 << '\n';
    std::cout << h2 << '\n';
    std::cout << h3 << '\n';
}

输出是

1000
4334672815104069193
16667047557902998627
2258353126044249582

http://en.cppreference.com/w/cpp/utility/bitset/hash

另外，为什么它不将这些位转换为 unsigend long 并生成哈希值？

最佳答案

作为noted by Igor ，C++标准没有指定算法，它only requires该哈希值仅取决于对象，并且在程序运行期间相同:http://eel.is/c++draft/hash.requirements

20.5.3.4 Hash requirements [hash.requirements] 1 A type H meets the Hash requirements if:

• (1.1) it is a function object type,
• (1.2) it satisfies the requirements of CopyConstructible and Destructible, and
• (1.3) the expressions shown in Table 29 are valid and have the indicated semantics.

2 Given Key is an argument type for function objects of type H, in Table 29 h is a value of type (possibly const) H, u is an lvalue of type Key, and k is a value of a type convertible to (possibly const) Key.

Table 29 — Hash requirements

• Expression Return type Requirement
• h(k) size_­t The value returned shall depend only on the argument k for the duration of the program. [ Note: Thus all evaluations of the expression h(k) with the same value for k yield the same result for a given execution of the program. — end note ] [ Note: For two different values t1 and t2, the probability that h(t1) and h(t2) compare equal should be very small, approaching 1.0 / numeric_­limits​::​max(). — end note ]
• h(u) size_­t Shall not modify u.

Gcc 的 libstdc++ bitset 实现使用 std::hash: https://github.com/gcc-mirror/gcc/blob/master/libstdc%2B%2B-v3/include/debug/bitset

#if __cplusplus >= 201103L
  // DR 1182.
  /// std::hash specialization for bitset.
  template<size_t _Nb>
    struct hash<__debug::bitset<_Nb>>
    : public __hash_base<size_t, __debug::bitset<_Nb>>
    {
      size_t
      operator()(const __debug::bitset<_Nb>& __b) const noexcept
      { return std::hash<_GLIBCXX_STD_C::bitset<_Nb>>()(__b._M_base()); }
    };
#endif

https://github.com/gcc-mirror/gcc/blob/1cb6c2eb3b8361d850be8e8270c597270a1a7967/libstdc%2B%2B-v3/include/std/bitset#L1561

  // DR 1182.
  /// std::hash specialization for bitset.
  template<size_t _Nb>
    struct hash<_GLIBCXX_STD_C::bitset<_Nb>>
    : public __hash_base<size_t, _GLIBCXX_STD_C::bitset<_Nb>>
    {
      size_t
      operator()(const _GLIBCXX_STD_C::bitset<_Nb>& __b) const noexcept
      {
        const size_t __clength = (_Nb + __CHAR_BIT__ - 1) / __CHAR_BIT__;
        return std::_Hash_impl::hash(__b._M_getdata(), __clength);
      }
    };

LLVM 的 libcxx 使用自己的 bitset 实现，异或所有单词:https://github.com/llvm-mirror/libcxx/blob/2c4b8af9aada61d83610330416eb8a39a8aa5494/include/bitset#L417

template <size_t _Size>
struct _LIBCPP_TEMPLATE_VIS hash<bitset<_Size> >
    : public unary_function<bitset<_Size>, size_t>
{
    _LIBCPP_INLINE_VISIBILITY
    size_t operator()(const bitset<_Size>& __bs) const _NOEXCEPT
        {return __bs.__hash_code();}
};

template <size_t _N_words, size_t _Size>
inline
size_t
__bitset<_N_words, _Size>::__hash_code() const _NOEXCEPT
{
    size_t __h = 0;
    for (size_type __i = 0; __i < _N_words; ++__i)
        __h ^= __first_[__i];
    return __h;
}

和 1 字位集的更简单变体:

inline
size_t
__bitset<1, _Size>::__hash_code() const _NOEXCEPT
{
    return __first_;
}

关于c++ - std::bitset 散列函数算法，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/43956125/