从位集到BITSET上提振无序（井）图 [英] Unordered (hash) map from bitset to bitset on boost

问题描述

我想使用的缓存，通过升压转换器的实施 unordered_map ，从来，dynamic_bitset 到来，dynamic_bitset 。的问题，当然，是有从位集没有默认哈希函数。它似乎并不像一个概念性的问题，但我不知道如何制定出的技术性。我应该怎么办呢？

I want to use a cache, implemented by boost's unordered_map, from a dynamic_bitset to a dynamic_bitset. The problem, of course, is that there is no default hash function from the bitset. It doesn't seem to be like a conceptual problem, but I don't know how to work out the technicalities. How should I do that?

推荐答案

我发现了一个意想不到的解决方案。原来提升有一个选项的#define BOOST_DYNAMIC_BITSET_DONT_USE_FRIENDS 。当这个定义，私有成员包括 m_bits 成为公共的（我认为这是有需要处理旧的编译器或东西）。

I found an unexpected solution. It turns out boost has an option to #define BOOST_DYNAMIC_BITSET_DONT_USE_FRIENDS. When this is defined, private members including m_bits become public (I think it's there to deal with old compilers or something).

所以，现在我可以用@ KennyTM的回答，改了一下：

So now I can use @KennyTM's answer, changed a bit:

namespace boost {
    template <typename B, typename A>
    std::size_t hash_value(const boost::dynamic_bitset<B, A>& bs) {            
        return boost::hash_value(bs.m_bits);
    }
}

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