c 函数指针

Question

编写一个程序，从用户那里接受总秒数。将此值与三个变量(小时、分钟、秒)的地址一起传递给名为 time() 的函数，该函数将计算小时、分钟和秒数。从 main() 打印此信息。

请帮助我如何修复我的代码以使该程序按预期方式工作。

   /* Adham Hamade
 .
  */

#include <stdio.h>
#include <conio.h>

//function prototype
void time(int &,int &,int &, int);


    int main()
   {
   //Variables
    int num;
    int hours;
    int minutes;
    int seconds;


    //reference number variables 
    int *h = &hours;
    int *m = &minutes; 
    int *s = &seconds;


    printf("Please enter number of seconds");
    scanf("%d",&num);

    time(h, m, s, num);

    printf("\n\nTime is %d hrs %d mins %d secs", hours, minutes, seconds);


 getch();
 return 0 ;   
 }

 void time(int &h,int &m ,int &s, int num)
 {
      int sec;
      int min;
      int hr;
      int t;


      hr = num / 3600 ;
      t = num %3600;
      min = t/60;
      sec = t%60;

      hr = &h;
      min = &m;
      sec = &s;


 }

Answer 1

C中没有引用调用，只有引用传递。
改变

void time(int &,int &,int &, int);  

到

void time(int *, int *, int *, int);  

在 time 的函数定义中，更改

 hr = &h;
 min = &m;
 sec = &s;

到

*h = hr;
*m = min;
*s = sec;