Input/Output C++ 中的运算符重载

Question

卡在 ostream/istream 运算符重载 Q1。为什么我们使用 ostream& operator 作为朋友？ Q2。为什么我们在 ostream & operator << (ostream &out, const Complex &c) 中传递两个参数 Q3.为什么我们引用 Cout 和 in ？ istream & operator >> (istream &in, Complex &c).

参考：Overloading stream Oerator overloading- Geeks4Geeks

HERE IS THE CODE

#include <iostream> 
using namespace std; 

class Complex 
{ 
private: 
    int real, imag; 
public: 
    Complex(int r = 0, int i =0) 
    { real = r; imag = i; } 
    friend ostream & operator << (ostream &out, const Complex &c); 
    friend istream & operator >> (istream &in, Complex &c); 
}; 

ostream & operator << (ostream &out, const Complex &c) 
{ 
    out << c.real; 
    out << "+i" << c.imag << endl; 
    return out; 
} 

istream & operator >> (istream &in, Complex &c) 
{ 
    cout << "Enter Real Part "; 
    in >> c.real; 
    cout << "Enter Imaginary Part "; 
    in >> c.imag; 
    return in; 
} 

int main() 
{ 
Complex c1; 
cin >> c1; 
cout << "The complex object is "; 
cout << c1; 
return 0; 
}

Answer 1

A1。访问私有成员

A2。第一个参数是流，第二个是对象。 operator<< 和 operator>> 需要两个参数

A3。因为它们是在函数中修改的。这些函数从 resp 读取。写入流。

另外：

不要使用 using namespace std;

不要在构造函数体中初始化成员。使用构造函数初始化列表

Complex(int r = 0, int i =0) 
{ real = r; imag = i; }

应该是

Complex(int r = 0, int i = 0) : real(r), imag(i) {}

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