错误：通过xxx作为xxx的'this'参数抛弃限定符 [英] error: passing xxx as 'this' argument of xxx discards qualifiers

问题描述

#include <iostream>
#include <set>

using namespace std;

class StudentT {

public:
    int id;
    string name;
public:
    StudentT(int _id, string _name) : id(_id), name(_name) {
    }
    int getId() {
        return id;
    }
    string getName() {
        return name;
    }
};

inline bool operator< (StudentT s1, StudentT s2) {
    return  s1.getId() < s2.getId();
}

int main() {

    set<StudentT> st;
    StudentT s1(0, "Tom");
    StudentT s2(1, "Tim");
    st.insert(s1);
    st.insert(s2);
    set<StudentT> :: iterator itr;
    for (itr = st.begin(); itr != st.end(); itr++) {
        cout << itr->getId() << " " << itr->getName() << endl;
    }
    return 0;
}

行内：

cout << itr->getId() << " " << itr->getName() << endl;

它给出一个错误：

../ main.cpp：35：错误：将'Student StudentT'作为'int StudentT :: getId（）'的'this'参数丢弃限定符

../main.cpp:35: error: passing 'const StudentT' as 'this' argument of 'int StudentT::getId()' discards qualifiers

../ main.cpp：35：error：传递'const StudentT'作为'std :: string的参数'StudentT :: getName（）'丢弃限定符

../main.cpp:35: error: passing 'const StudentT' as 'this' argument of 'std::string StudentT::getName()' discards qualifiers

此代码有什么问题？谢谢！

What's wrong with this code? Thank you!

推荐答案

std :: set 存储为 const StudentT 。因此，当您尝试使用 const 对象调用 getId（）时，编译器会检测到一个问题，在const对象上的一个非const成员函数是不允许的，因为非const成员函数使NO PROMISE不修改对象;因此编译器会做一个安全假设， getId（）可能会尝试修改对象，但同时，它也注意到对象是const;所以任何尝试修改const对象应该是一个错误。因此编译器生成错误消息。

The objects in the std::set are stored as const StudentT. So when you try to call getId() with the const object the compiler detects a problem, namely you're calling a non-const member function on const object which is not allowed because non-const member functions make NO PROMISE not to modify the object; so the compiler is going to make a safe assumption that getId() might attempt to modify the object but at the same time, it also notices that the object is const; so any attempt to modify the const object should be an error. Hence compiler generates error message.

解决方案很简单：将函数const作为：

The solution is simple: make the functions const as:

int getId() const {
    return id;
}
string getName() const {
    return name;
}

这是必要的，因为现在你可以调用 getId ）和 getName（）在const对象上：

This is necessary because now you can call getId() and getName() on const objects as:

void f(const StudentT & s)
{
     cout << s.getId();   //now okay, but error with your versions
     cout << s.getName(); //now okay, but error with your versions
}

实现运算符< as：

inline bool operator< (const StudentT & s1, const StudentT & s2)
{
    return  s1.getId() < s2.getId();
}

注意参数现在为 const 参考。

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