重载运算符有理错误

本文介绍了重载运算符有理错误的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

所以我环顾四周，因为对于大多数C ++学生来说，这似乎是一个常见的家庭作业问题，但是我似乎找不到能回答我问题的人.我觉得我已经正确填写了代码，但是每次都会遇到相同的错误.

So I have looked around because this seems to be a common homework problem for most C++ students, but I can't seem to find one that will answer my issue. I feel that I have filled out the code correctly but I get the same error each time.

这是我的代码:

#include <iostream> using namespace std; class Rational { public: Rational() { num = 0; denom = 1; }; Rational(int n, int d) { num = n; denom = d; normalize(); } Rational(int n) { num = n; denom = 1; } int get_numerator() const { return num; } int get_denominator() const { return denom; } void normalize() { if ((num > 0 && denom < 0)||(num < 0 && denom < 0)) { num = -1 * num; denom = -1 * denom; } int gcdcheck = GCD(num,denom); num = num / gcdcheck; denom = denom / gcdcheck; } int Rational::GCD(int n, int d) { int temp; n = abs(n); d = abs(d); if (n > d) { // Do nothing everything is where it should be } else { temp = n; n = d; d = temp; } int factor = n % d; while (factor != 0) { factor = n % d; d = n; n = factor; } return d;//Return the value to normalize to simplify the fractions to simplist form } Rational operator+(Rational b) const { Rational add; //Addition of fractions (a*d/b*d + c*b/d*b) //Numerator = (a*d + c*b) add.get_numerator = b.get_numerator * denom + b.get_denominator * num; //Denomenator = (b*d) add.get_denominator = b.get_denominator * denom; add.normalize(); return add; } Rational operator-(Rational b) const { Rational sub; //Same as Addition just a minus sign //Numerator = (a*d + c*b) sub.get_numerator = b.get_numerator * denom + b.get_denominator * num; //Denomenator = (b*d) sub.get_denominator = b.get_denominator * denom; sub.normalize(); return sub; } Rational operator*(Rational b) const { //Multiply the numerators and denomenators Rational multi; multi.get_numerator = b.get_numerator * num; multi.get_denominator = b.get_denominator * denom; multi.normalize(); return multi; } Rational operator/(Rational b) const { //Division of fractions is done by the recipricol of one of the fractions Rational divi; divi.get_numerator = b.get_numerator * denom; divi.get_denominator = b.get_denominator * num; divi.normalize(); return divi; } //To avoid issues with rounding the compare functions will multiply instead to give clean whole numbers //This will be done by multiplying the denomenators by the opposite numerator bool operator==(Rational b) const { return ((b.get_numerator * denom == b.get_denominator * num)); } bool operator<(Rational b) const { return ((b.get_numerator * denom > b.get_denominator * num)); } double toDecimal() const { double result; result = static_cast<double> (num)/ static_cast<double> (denom); return result; } private: int num = 0; // default value is 0 int denom = 1; // default value is 1 }; ostream& operator<<(std::ostream& output, Rational& a) { if (a.get_denominator == 0) { output << "Divide by Zero"; } output << a.get_numerator << '/' << a.get_denominator; return output; }

我知道它的很多代码，并且我不希望有人通过它进行所有调试，我只是认为我可以将其全部发布，以防万一问题蔓延到比我认为的问题还远的地方.

I know its a lot of code and I don't expect someone to go through it all debugging I just thought I would post it all just in case the problem spans farther then where I think the issue is.

对于每个运算符，我都会得到相同的错误:

I get the same errors for each operator:

1:错误C3867:"Rational :: get_denominator":非标准语法；使用&"创建指向成员的指针

1: error C3867: 'Rational::get_denominator': non-standard syntax; use '&' to create a pointer to member

2:'*':错误C3867:'Rational :: get_denominator':非标准语法；使用&"创建指向成员的指针

2: '*': error C3867: 'Rational::get_denominator': non-standard syntax; use '&' to create a pointer to member

3:错误C3867:'Rational :: get_numerator':非标准语法；使用&"创建指向成员的指针

3: error C3867: 'Rational::get_numerator': non-standard syntax; use '&' to create a pointer to member

我查看了来自完成此问题的不同在线站点的代码，并尝试了他们的方法，但似乎不起作用.我添加了const和&到函数中的参数，我仍然遇到相同的问题.我是错误地调用一个函数还是初始化一个错误?

I have looked at code from different online sites that have done this problem and tried their methods but it doesn't seem to work. I have added const and & to the parameters in the functions and I still get the same issues. Am I calling a function incorrectly or initializing one wrong?

推荐答案

代码中有多个问题.这是更正的代码.

You have multiple problems in the code. Here is the corrected code.

您返回的值不是引用.

在类中定义函数时，无需指定全名

缺少函数调用的()

在代码末尾有一些注释.

There are some comments on the code at the end.

#include <iostream> #include <cmath> using namespace std; class Rational { public: Rational() { num = 0; denom = 1; }; Rational(int n, int d) {` num = n; denom = d; normalize(); } Rational(int n) { num = n; denom = 1; } int& get_numerator() { return num; } int& get_denominator() { return denom; } void normalize() { if ((num > 0 && denom < 0) || (num < 0 && denom < 0)) { num = -1 * num; denom = -1 * denom; } int gcdcheck = GCD(num, denom); num = num / gcdcheck; denom = denom / gcdcheck; } int GCD(int n, int d) { int temp; n = abs(n); d = abs(d); if (n > d) { // Do nothing everything is where it should be } else { temp = n; n = d; d = temp; } int factor = n % d; while (factor != 0) { factor = n % d; d = n; n = factor; } return d;//Return the value to normalize to simplify the fractions to simplist form } Rational operator+(Rational b) const { Rational add; //Addition of fractions (a*d/b*d + c*b/d*b) //Numerator = (a*d + c*b) add.get_numerator()= b.get_numerator() * denom + b.get_denominator() * num; //Denomenator = (b*d) add.get_denominator() = b.get_denominator() * denom; add.normalize(); return add; } Rational operator-(Rational b) const { Rational sub; //Same as Addition just a minus sign //Numerator = (a*d + c*b) sub.get_numerator() = b.get_numerator() * denom + b.get_denominator() * num; //Denomenator = (b*d) sub.get_denominator() = b.get_denominator() * denom; sub.normalize(); return sub; } Rational operator*(Rational b) const { //Multiply the numerators and denomenators Rational multi; multi.get_numerator() = b.get_numerator() * num; multi.get_denominator() = b.get_denominator() * denom; multi.normalize(); return multi; } Rational operator/(Rational b) const { //Division of fractions is done by the recipricol of one of the fractions Rational divi; divi.get_numerator() = b.get_numerator() * denom; divi.get_denominator() = b.get_denominator() * num; divi.normalize(); return divi; } //To avoid issues with rounding the compare functions will multiply instead to give clean whole numbers //This will be done by multiplying the denomenators by the opposite numerator bool operator==(Rational b) const { return ((b.get_numerator() * denom == b.get_denominator() * num)); } bool operator<(Rational b) const { return ((b.get_numerator() * denom > b.get_denominator() * num)); } double toDecimal() const { double result; result = static_cast<double> (num) / static_cast<double> (denom); return result; } private: int num = 0; // default value is 0 int denom = 1; // default value is 1 }; ostream& operator<<(std::ostream& output, Rational& a) { if (a.get_denominator() == 0) { output << "Divide by Zero"; } output << a.get_numerator() << '/' << a.get_denominator(); return output; }

对代码的一些注释...返回引用，尤其是返回给私有成员真的很糟糕.我建议您创建一个set函数.

Some comments on the code... Returning a reference, especially to a private member is really bad. I suggest you to create a set function.

因此基本上像以前一样保留get函数

so basically keep the get function as before

int get_denominator() const { return denom; }

并创建一个新的函数来设置值

and create a new function to set value

int set_denominator(int in) { denom = in; }