我选择了Java中的哈希表的概念，所以我意识到，对于用于定制类的通用哈希集合容器，必须为散列函数和相应的相等函数提供定义。
在Java中，这意味着重写方法
int hashCode（）和
boolean equals（Object o）。 接受5个模板参数（你能相信吗？），它看起来像一个怪物，并让我的头旋转。
感谢它，如果可以给出一个简单的例子，当前的玩具类：
class一些{ public： int a; };其中hash函数只返回a的值，并且相等测试函数返回true iff成员'a'的值是相同的。
谢谢
解决方案步骤1：为您的类型重载 operator == ：
bool operator ==（const Some& x，const Some& y） { return xa == ya; $ / code>第二步：专用 std :: hash $ b
namespace std { template<> code> struct hash< Some> { typedef一些argument_type; typedef size_t result_type; size_t operator（）（const Some& x）const { return x.a; } }; }第3步：简单测试：
int main（） { std :: unordered_set< Some>测试; test.insert（一些{42}）; }第四步：获利！
I picked the concepts of hash tables in Java, so I was aware that for a generic "hash set" container to work for custom classes, one must provide definition for a hash function and an corresponding equality function.
In Java, this would mean overriding method
int hashCode()and
boolean equals (Object o).
I was expecting the same logic in c++'s STL, but was having trouble understanding the syntax. Specifically, std::unordered_set<> accepts 5 template arguments (can you believe that?), which looks like a monster and makes my head spin.
So I would be appreciate it if one could give a simple example for the current toy class:
class Some{ public : int a; };Of which the hash function simply returns the value of a, and the equality test functions returns true iff the values of member 'a' are the same.
Thanks
解决方案Step 1: Overload operator== for your type:
bool operator==(const Some& x, const Some& y) { return x.a == y.a; }Step 2: Specialize std::hash for your type:
namespace std { template<> struct hash<Some> { typedef Some argument_type; typedef size_t result_type; size_t operator()(const Some& x) const { return x.a; } }; }Step 3: A simple test:
int main() { std::unordered_set<Some> test; test.insert(Some{42}); }Step 4: Profit!
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