编译器告诉我 `unordered_map` 中的键类型不匹配 operator==

Question

In the following program the compiler seems to be missing the operator== for the keys in the unordered map. The way I understand it is that std::variant automatically delegates == to it's fields (if the same index is set and false otherwise). Furthermore, I have a definition of operator== in the types A and B .

#include <variant>
#include <string>
#include <unordered_map>


struct Base {
public:
    virtual std::size_t get_hash() const = 0;
};

struct A : public Base {
    std::string s;
public:

    A(std::string str): s{str} {} 

    std::size_t get_hash() const override {
        return std::hash<std::string>{}(this->s);
    }

    bool operator==(const A a) {
        return a.s == this->s;
    }
};

struct B : public Base {
    int i;

public:
    B(int it): i{it} {}

    std::size_t get_hash() const override {
        return std::hash<int>{}(this->i);
    }

    bool operator==(const B b) {
        return b.i == this->i;
    }
};

namespace std {
    template<>
    struct hash<A> {
        size_t operator()(const A& a) const {
            return a.get_hash();
        }
    };

    template<>
    struct hash<B> {
        size_t operator()(const B& b) const {
            return b.get_hash();
        }
    };
}

int main(int argc, char* argv[]) 
{
    std::unordered_map<std::variant<A,B>, int> map;

    std::variant<A,B> key1{A("hi")};
    std::variant<A,B> key2{B(1)};

    map.insert({key1, 1});
    map.insert({key2, 2});
}    

Hence, I do not understand why the compiler complains about a lack of equality operator.

Answer 1

The operator== need to be const ,

also it's better to take a const T& instead of const T

struct A{
    // ...
    bool operator==(const A& a) const {
        return a.s == this->s;
    }
};
struct B{
    // ...
    bool operator==(const B& b) const {
        return b.i == this->i;
    }
};