重载运算符有理错误 [英] Overloading Operator Rational Error
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;
}
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