Question

有人可以帮助我回答这些问题吗？我很难理解运算符重载和链接。

如何在程序中为操作员+功能执行链接？
为什么操作员不能＆＃34;＆lt;＆lt;＆＃;作为成员函数重载？

参考operator +函数，它返回的Bank_Acct对象的名称是什么？

#include <iostream>
#include <string>

using namespace std;

const int SIZE = 10;

class Bank_Acct
{
public:
    Bank_Acct( );  // default constructor
    Bank_Acct(double new_balance, string Cname); // explicit value                                          
                                         // constructor
    void Print( ); //accessor function
    Bank_Acct & operator+(double amount); // mutator function

private:
    double balance;
    string name;
};

Bank_Acct::Bank_Acct()
{
    balance = 0;
    name = "NoName";
}

Bank_Acct::Bank_Acct(double amount, string Cname)
{
    balance = amount;
    name = Cname;

}

void Bank_Acct::Print()
{
    cout << endl << "Object name: " << name;
    cout << endl << "The new balance is: " << balance << endl;
}

Bank_Acct & Bank_Acct::operator+(double amount)
{
    balance += amount;
    return *this;
}

int main()

{ Bank_Acct my_Acct;

    cout.setf(ios::showpoint);
    cout.setf(ios::fixed);
    cout.precision(2);

    cout << "Original balance of my_Acct:" << endl;
    cout << "----------------------------" << endl;
    my_Acct.Print( );

    // the following statement contains chaining
    my_Acct + 18.75 + 14.35 + 10054.96;

    cout << endl << endl;

    cout << "The balance of my_Acct after addition to balance 3 times:" << endl;
    cout << "---------------------------------------------------------" << endl;
    my_Acct.Print();

    cout << endl << endl;

    return 0;
}

Answer 1

它返回修改对象的引用，因此您可以再次调用operator +。
你绝对可以重载＆lt;＆lt;作为成员函数，但如果您打算将其用作流操作符，则不会，因为左侧将是ostream而不是Bank_Acct
此是指向已调用成员函数的类实例的指针。

运算符重载和通道

1 个答案: