C++ 运算符重载错误：只读成员“”的赋值 object

Question

错误信息：

class.cpp: In function 'RationalNumber operator++(const RationalNumber&, int)':
class.cpp:27:28: error: assignment of member 'RationalNumber::numerator' in read-only object
  r.numerator = r.numerator+1;

为什么它是“在只读对象中”？

class.h文件

#ifndef CLASS_H
#define CLASS_H
#include <iostream>
using namespace std;
// #include "class2.h"

class RationalNumber
{
 private:
  int numerator,denominator;
 public:
    RationalNumber(int x,int y);
    RationalNumber();
    friend ostream& operator << (ostream& os,const RationalNumber& x);
    friend RationalNumber operator++ (const RationalNumber& r, int dummy);
    RationalNumber operator- (const RationalNumber& r) {
        RationalNumber r3;
        //cout << numerator << " " << r.numerator<<endl;
        r3.numerator = (numerator * r.denominator)-(r.numerator * denominator);
        r3.denominator = r.denominator * denominator;
        return r3;
    }

};
#endif

主.cpp

#include <iostream>
using namespace std;
#include "class.h"
// #include "class2.h"

int main()
{
    RationalNumber r1(21,7),r2(67,31),r3;
    r3 = r1 - r2;
    cout << r3;
    r1++;
    cout << r1;
}

有错误的cpp部分

RationalNumber operator++ (const RationalNumber& r, int dummy) 
{
    RationalNumber temp;
    temp = r;
    r.numerator = r.numerator+1;
    return temp;
}

Answer 1

错误信息：

class.cpp: In function 'RationalNumber operator++(const RationalNumber&, int)':
class.cpp:27:28: error: assignment of member 'RationalNumber::numerator' in read-only object
  r.numerator = r.numerator+1;

为什么是“在只读对象中”？

类.h文件

#ifndef CLASS_H
#define CLASS_H
#include <iostream>
using namespace std;
// #include "class2.h"

class RationalNumber
{
 private:
  int numerator,denominator;
 public:
    RationalNumber(int x,int y);
    RationalNumber();
    friend ostream& operator << (ostream& os,const RationalNumber& x);
    friend RationalNumber operator++ (const RationalNumber& r, int dummy);
    RationalNumber operator- (const RationalNumber& r) {
        RationalNumber r3;
        //cout << numerator << " " << r.numerator<<endl;
        r3.numerator = (numerator * r.denominator)-(r.numerator * denominator);
        r3.denominator = r.denominator * denominator;
        return r3;
    }

};
#endif

主程序

#include <iostream>
using namespace std;
#include "class.h"
// #include "class2.h"

int main()
{
    RationalNumber r1(21,7),r2(67,31),r3;
    r3 = r1 - r2;
    cout << r3;
    r1++;
    cout << r1;
}

cpp 部分有错误

RationalNumber operator++ (const RationalNumber& r, int dummy) 
{
    RationalNumber temp;
    temp = r;
    r.numerator = r.numerator+1;
    return temp;
}

Answer 2

正如所指出的，const 运算符表示您不能修改r 。 这是后缀 ++ 运算符的正确声明和实现：

...
    friend RationalNumber& operator++ (RationalNumber& r, int dummy);
...

RationalNumber& operator++ (RationalNumber& r, int dummy) 
{
    r.numerator += 1;
    return r;
}

然而，也正如所指出的，像这样使用增量运算符会让人感到困惑。 如果将分数增加 1 个整数而不是分子增加 1 似乎更有意义。比如 3/7 + 1 = 10/7
相反，你做的是 3/7 + 1/7 = 4/7

注意：我已经删除了临时变量，以便您返回原始的 object，因为这是运算符的目的，而不是创建新的 object。