Question

通过引用传递似乎不起作用。输出的第二行应该使用引用，而是输出与之前相同的内容。

double passByValue(double, double);
double passByRef(double&, double&);

int main(){    
    double firstNumber=0, secondNumber=0;
    char quit =' ';
    while (quit != 'y'){

        cout <<"Please enter your first number: ";
        cin >> firstNumber;

        cout << "\nPlease enter your second number: ";
        cin >> secondNumber;

        cout <<firstNumber <<"\t" << secondNumber << "\t" << passByValue(firstNumber, secondNumber)<<endl;
        cout <<firstNumber <<"\t" << secondNumber << "\t" << passByRef(firstNumber ,secondNumber )<<endl;

        cout <<"Do you want to quit? y/n";
        cin >> quit;
    }
}

double passByValue(double first, double second){
    first +=5;
    second +=5;
    double sum =first + second;
    return sum;
}

double passByRef(double &a, double &b){
    a +=5;
    b +=5;
    double sum = a + b;
    return sum;
}

Answer 1

试试这个：

double sum = passByValue(firstNumber, secondNumber);
cout << firstNumber <<"\t" << secondNumber << "\t" << sum <<endl;
sum = passByRef(firstNumber, secondNumber);
cout << firstNumber <<"\t" << secondNumber << "\t" << sum <<endl;

这可确保您看到因调用该函数而对firstNumber和secondNumber所做的更改。

Answer 2

代码是正确的，你的解释不是。
在致电firstNumber后尝试显示secondNumber和passByReference，您应该知道：

0     0     10  
0     0     10  
5     5

问题在于，当您拨打passByValue时，这些行

first +=5;
second +=5;

实际上没有增加firstNumber和secondNumber：first和second是副本。
另一方面，当你通过引用传递时，它们不是5，

为什么Pass By Ref不起作用

2 个答案: