使用运算符重载时,没有运算符“<<”匹配这些操作数

Nod*_*eJS 1 c++ overloading ostream

以下尝试运算符重载时发生错误:

#include<iostream>
#include<string>
#include<ostream>
using namespace std;

class Dollar
{
private:
    float currency, mktrate, offrate;
public:
    Dollar(float);
    float getDollar() const;
    float getMarketSoums() const;
    float getofficialSoums() const;
    void getRates();

    // In the following function I was trying to overload "<<" in order to print all the data members:
    friend void operator<<(Dollar &dol, ostream &out)
    {
        out << dol.getDollar() << endl;
        out << dol.getMarketSoums() << endl;
        out << dol.getofficialSoums() << endl;
    }
};

Dollar::Dollar(float d)
{
    currency = d;
}

float Dollar::getDollar() const
{
    return currency;
}

float Dollar::getMarketSoums() const
{
    return mktrate;
}

float Dollar::getofficialSoums() const
{
    return offrate;
}

void Dollar::getRates()
{
    cin >> mktrate;
    cin >> offrate;
}

int main()
{
    Dollar dollar(100);
    dollar.getRates();

    // In this line I am getting the error. Could you please help to modify it correctly?
    cout << dollar;

    system("pause");
    return 0;
}
Run Code Online (Sandbox Code Playgroud)

Rai*_*op7 5

您必须将std::ostreamobject 作为第一个参数传递给插入运算符,<<而不是作为第二个参数,只要您以这种方式调用它:

friend void operator << (ostream &out, Dollar &dol);
Run Code Online (Sandbox Code Playgroud)
  • constant reference只要此函数只是打印而不打算修改对象的成员,您就应该将对象传递给插入运算符:

    friend void operator << (ostream &out, const Dollar& dol);
    
    Run Code Online (Sandbox Code Playgroud)
  • 所以通过引用传递,避免多份拷贝,const避免无意修改。

  • 如果你想调用它以你想要的方式工作,你可以这样做:

    friend void operator<<(const Dollar &dol, ostream &out){
        out << dol.getDollar() << endl;
        out << dol.getMarketSoums() << endl;
        out << dol.getofficialSoums() << endl;
    }
    
    Run Code Online (Sandbox Code Playgroud)

主要例如:

    operator << (dollar, cout); // this is ok
    dollar << cout; // or this. also ok.
Run Code Online (Sandbox Code Playgroud)

如您所见,我颠倒了调用插入运算符的顺序以匹配上面的签名。但我不推荐这样做,只是为了更多地了解它应该如何工作。

  • 并且第二个参数应该声明为 `const`,因为 `operator&lt;&lt;` 不应该修改正在打印的对象 (2认同)