为什么不std::unordered_map<tuple<int, int>, string>开箱即用?必须为tuple<int, int>例如定义散列函数是繁琐的
template<> struct do_hash<tuple<int, int>>
{ size_t operator()(std::tuple<int, int> const& tt) const {...} };
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构建一个以元组为键的无序映射(Matthieu M.)展示了如何自动执行此操作boost::tuple.有没有为c ++ 0x元组执行此操作而不使用可变参数模板?
当然这应该在标准:(
我有一个std::unordered_map<string, std::array<int, 2>>. emplace将值放入地图的语法是什么?
unordered_map<string, array<int, 2>> contig_sizes;
string key{"key"};
array<int, 2> value{1, 2};
// OK ---1
contig_sizes.emplace(key, value);
// OK --- 2
contig_sizes.emplace(key, std::array<int, 2>{1, 2});
// compile error --3
//contig_sizes.emplace(key, {{1,2}});
// OK --4 (Nathan Oliver)
// Very inefficient results in two!!! extra copy c'tor
contig_sizes.insert({key, {1,2}});
// OK --5
// One extra move c'tor followed by one extra copy c'tor
contig_sizes.insert({key, std::array<int, 2>{1,2}});
// OK --6
// Two extra move constructors
contig_sizes.insert(pair<const string, array<int, …Run Code Online (Sandbox Code Playgroud)