[英]How can i call a operator overloaded member function(or use the operator) from another operator overloading member function in the same class?
我正在用c ++編寫代碼來處理復數。 我也在練習運算符重載。 所以我重載*
(乘法運算符),現在我想在我的重載/
(除法運算符)中使用重載運算符,但是當我使用*
它的顯示錯誤時。 這是代碼:
#include <iostream>
#include <cmath>
using namespace std;
class Imaginary
{
public:
//constructors
Imaginary(double a,double b):x(a),y(b){}
Imaginary():x(0.0),y(0.0){}
//setter methods for x and y
void Setx(double x) { this->x = x; }
void Sety(double y) { this->y = y; }
//getter methods for x and y
double Getx(){return this->x;}
double Gety(){return this->y;}
//overloaded operators
Imaginary operator+(Imaginary&);
Imaginary operator-(Imaginary&);
Imaginary operator*(Imaginary&);
Imaginary operator~();
Imaginary operator/(Imaginary&);
void print();
private:
double x;
double y;
};
Imaginary Imaginary::operator+(Imaginary &i){
Imaginary ti;
ti.Setx(this->x+i.x);
ti.Sety(this->y+i.y);
return ti;
}
Imaginary Imaginary::operator-(Imaginary &i){
Imaginary ti;
ti.Setx(this->x-i.x);
ti.Sety(this->y-i.y);
return ti;
}
Imaginary Imaginary::operator*(Imaginary &i){
Imaginary ti;
ti.Setx((this->x*i.x) - (this->y*i.y));
ti.Sety((this->y*i.x)+(this->x*i.y));
return ti;
}
Imaginary Imaginary::operator~(){
int y;
y = this->y;
this->y = -y;
return *this;
}
Imaginary Imaginary ::operator/(Imaginary &i){
Imaginary numerator,denominator,ti;
//i want to use here the overloaded *(multiplacation) operator
numerator = (*this) * (~i);//showing error
denominator = (*this) * (~i);//showing error
ti.Setx(numerator.Getx()/denominator.Getx());
ti.Sety(numerator.Gety()/denominator.Getx());
return ti;
}
void Imaginary::print(){
cout<<x;
if (y>0)
cout<<"+i"<<y<<endl;
else if (y<0)
cout<<"-i"<<abs(y)<<endl;
}
int main()
{
Imaginary res;
Imaginary z1(2,3);
Imaginary z2(1,-1);
z1.print();
z2.print();
/*res = z1+z2;
cout<<"Addition:-\n";
res.print();
res = z1-z2;
cout<<"Subtraction:-\n";
res.print()*/
res = z1*z2;
cout<<"Multiplication:-\n";
res.print();
res = z1/z2;
cout<<"Division:-\n";
res.print();
return 0;
}
錯誤信息如下: -
D:\Games\Cheese\main.cpp|66|error: no match for 'operator*' (operand types are 'Imaginary' and 'Imaginary')|
請有人告訴我如何糾正它。
在這種情況下,錯誤不是最好的錯誤。 您的operator*
定義為
Imaginary Imaginary::operator*(Imaginary &i)
這要求右手邊是左值。 當你這樣做
(*this) * (~i)
在operator/
, (~i)
返回作為右值的Imaginary
。 您不能將該rvalue綁定到lavue引用,因此不會考慮您的重載並且您會收到編譯器錯誤。
解決這個問題的最簡單方法是采用const &
而不是
Imaginary Imaginary::operator*(const Imaginary &i)
兩件事情
首先,
(*this)*(~i);
是右側( operator~()
)是臨時的表達式。 如果該參數聲明為const
,則operator*()
只能接受臨時作為引用參數( Imaginary &
)。
operator*()
兩邊最好指定為const
,因為 - 一般來說 - 乘以兩個值不會改變任何一個(並產生不同的結果)。 其他運算符也是如此(例如operator/()
)。
解決方案是將operator*()
(成員函數)的規范更改為
Imaginary Imaginary::operator*(const Imaginary &i) const;
完成后,調用operator*()
的另一種方法是替換
numerator = (*this) * (~i);
同
numerator = this->operator*(~i);
或者甚至(因為這發生在一個成員函數中)
numerator = operator*(~i); // this-> is implied
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