Question

我正在尝试学习C ++，并且遇到有关时间格式的小型转换程序的问题。

    #include <iostream>
    using namespace std;

    void conversion(int hours, int minutes) {
       if (hours > 12) {
           hours -= 12;
           minutes = minutes;
       }
       else
           hours = hours;
           minutes = minutes;

       cout << hours << ":" << minutes << endl;
       }

    void output(int hours, int minutes) {
        cout << hours << ":" << minutes;
    }

    int main() {
        int hours, minutes;
        cout << "Enter the hours: ";
        cin >> hours;
        cout << "Enter minutes: ";
        cin >> minutes;
        conversion(hours, minutes);
        output(hours, minutes);

    }

在main函数中，输出调用未接收小时和分钟的更新值。

Answer 1

这与时间格式无关。一个显示相同问题的小得多的程序是：

#include <iostream>
using namespace std;

void f(int i) {
   i = 20;
}

int main() {
    int i = 10;
    f(i);
    cout << "i is still " << i << endl;
}

问题是参数列表中的int i表示“按值传递”。该函数获取值的副本。当函数退出时，您对该副本所做的任何更改都将被丢弃。通过引用（void f(int& i)）或通过指针（void f(int *pi)）传递。在教科书中查找参考和指针。

您的功能中还存在许多问题：

void conversion(int hours, int minutes) {
   if (hours > 12) {
       hours -= 12;
       minutes = minutes;  // This line does nothing.
   }
   else
       hours = hours;      // This line does nothing.
       minutes = minutes;  // This line will get executed even if hours was originally
                           // >12.  (Only the hours=hours line is covered by the else).  

   cout << hours << ":" << minutes << endl;
   }

你可能想要：

void conversion(int& hours, int& minutes) {
   if (hours > 12) {
       hours -= 12;
   }
   cout << hours << ":" << minutes << endl;
}

24HR到12HR转换程序，输出中无法识别新值？

1 个答案: