为什么 std：：unordered_map 的 KeyEqual 不被其运算符==使用？

问：

在下面的代码中，我定义了模板参数和 for .我希望输出是，但实际上是.为什么会这样？是因为不用于比较地图吗？HashKeyEqualunordered_map1 1 1 11 1 0 1std::equal_to<Base*>==

#include <iostream>
#include <unordered_map>

using std::cout;
using std::endl;

class Base {
public:
    int x;
    Base(int _x) : x(_x) {}

    bool operator==(const Base& another) const {
        return x == another.x;
    }
    size_t hash() const {return x;}
};

template <>
struct std::hash<Base> {
    size_t operator()(const Base& r) const {
        return r.hash();
    }
};

template <>
struct std::hash<Base*> {
    size_t operator()(const Base *r) const {
        return r->hash();
    }
};

template <>
struct std::equal_to<Base*> {
    bool operator()(const Base *r1, const Base *r2) const {
        return (*r1) == (*r2);
    }
};

int main(int, char**){
    Base b1(1);
    Base b2(2);
    Base bb1(1);
    Base bb2(2);
    cout << (b1 == bb1) << endl;

    std::unordered_map<Base, int> map1;
    map1.emplace(b1, 1);
    map1.emplace(b2, 2);
    std::unordered_map<Base, int> map2;
    map2.emplace(bb1, 1);
    map2.emplace(bb2, 2);
    cout << (map1 == map2) << endl;

    std::unordered_map<Base*, int, std::hash<Base*>, std::equal_to<Base*>> map11;
    map11.emplace(&b1, 1);
    map11.emplace(&b2, 2);
    std::unordered_map<Base*, int, std::hash<Base*>, std::equal_to<Base*>> map22;
    map22.emplace(&bb1, 1);
    map22.emplace(&bb2, 2);
    cout << (map11 == map22) << endl;

    std::unordered_map<Base*, int, std::hash<Base*>, std::equal_to<Base*>> map33;
    map33.emplace(&b1, 1);
    map33.emplace(&b2, 2);
    cout << (map11 == map33) << endl;
}