c++ - 有人可以解释吗？我有一个幸运的猜测，为什么通过引用而不是值传递这个变量会搞砸我的程序？

Question

每当我将函数“double getsales (double &num)”更改为 double getsales (double num) 并适本地更改函数原型(prototype)时，程序就会出错。我不明白为什么我的幸运猜测修复了它，无论我尝试阅读多少有关引用变量的内容。

谁能解释一下？

#include <iostream>
#include <iomanip>

using namespace std;

double getsales (double &);
void findhighest (double, double, double, double);

int main()
{
    double northeast = 0;
    double southeast = 0;
    double northwest = 0;
    double southwest = 0;

    cout << "Enter NorthEast sales: $" ;
    cout << getsales(northeast) << endl;
    cout << "Enter SouthEast sales: $"; 
    cout << getsales(southeast) << endl;
    cout << "Enter NorthWest sales: $";
    cout << getsales(northwest) << endl;
    cout << "Enter SouthWest sales: $";
    cout << getsales(southwest) << endl;

    findhighest(northeast, southeast, northwest, southwest);

    return 0;
}

double getsales (double &num)
{
    do
    {
        if(!cin)
        {
            cin.clear();
            cin.ignore(100, '\n');
        }

        cin >> num;
        cout << "Number entered: ";


    }while(!cin || num <= 0);
    return num;
}

void findhighest (double ne, double se, double nw, double sw) 
{
    const char *who = "NorthEast";
    double high = ne;
    if(se > high)
    {
        who = "SouthEast";
        high = se;
    }
    if(nw > high)
    {
        who = "NorthWest";
        high = nw;
    }
    if(sw > high)
    {
        who = "SouthWest";
        high = sw;
    }

    cout << fixed << showpoint << setprecision(2) << endl;
    cout << who << "has the highest sale ($" << high << ")" << endl;
}

Answer 1

按值传递时，getsales 不会修改作为参数传递的原始数字。

因此，调用

findhighest(northeast, southeast, northwest, southwest);

将使用原始数字(为零)进行操作。

您应该使用修改后的值。