错误：将“const…”作为“…”的“this”参数传递将丢弃限定符

Question

我有与这里、这里或这里相同的问题，除了我为参数和函数设置了const：

#include <unordered_map>
#include <functional>

namespace zzz {

struct identity_t {
    static const int ID_SIZE = 5;
    static const int LSB = 4; // = ID_SIZE - 1
    unsigned char id[ID_SIZE];
    inline bool operator< (const identity_t& rhs) const {
        for (int i = 0; i < ID_SIZE; i++) if (id[i] != rhs.id[i]) return (id[i] < rhs.id[i]);
        return false; // equal
    }
    unsigned char operator[] (const int& i) {return id[i];}
};

class hash_identity_t {
public:
    long operator()(const identity_t& x) const {
        return (long) x[identity_t::LSB];
    }
};

class equal_to_identity_t {
public:
     bool operator()(const identity_t& lhs, const identity_t& rhs) const {
         for (int i = 0; i < identity_t::ID_SIZE; i++) if (lhs[i] != rhs[i]) return false;
          return true;
     }
};

但我有同样的编译错误：

error: passing 'const zzz::identity_t' as 'this' argument of 'unsigned char zzz::identity_t::operator[](const int&)' discards qualifiers [-fpermissive]

Answer 1

使用std：：map的方法。

const mapped_type & at (const key_type &__k) const 

Answer 2

您正在尝试调用非const函数（identity_t::运算符[]）的const参数在const函数longhash_identity_t::运算符（constidentity_t

使操作符[]保持不变。

//--------------------------------------vvvvv
unsigned char operator[] (const int& i) const {return id[i];}