Question

我目前正在用C创建一个闹钟。它的要点非常简单，我想：

获取所需闹钟时间的用户输入
将其与当地时间进行比较
以秒为单位递减计时器，并在达到闹钟时打印

然而，我遇到了一些麻烦，现在已经停留了一段时间。这是我到目前为止所做的：

int u_input(int* uhour, int* uminute, int* usecond);
void sc_alarm();

int main(int argc, char *argv[])
{
  int uhour, uminute, usecond;
  u_input(&uhour, &uminute, &usecond);
  sc_alarm(uhour, uminute, usecond);
}

int u_input(int* uhour, int* uminute, int* usecond)
{
  printf("Enter H M S:");
  scanf("%d %d %d", uhour, uminute, usecond);
}

void sc_alarm(int uhour, int uminute, int usecond)
{
  struct tm *tm;
  time_t ctime;
  ctime = time(NULL);
  tm = localtime(&ctime);

  int difference, alarm_total, local_time_total;
  int chour = tm->tm_hour;
  int cminute = tm->tm_min;
  int csecond = tm->tm_sec;

  difference = alarm_total - local_time_total;
  alarm_total = (uhour * 3600) + (uminute * 60) + usecond;
  local_time_total = (chour * 3600) + (cminute * 60) + csecond;

  do
  {
    printf("Time left: %d seconds\n",difference);       
    difference--;
    sleep(1);
  }
  while(difference > 0);

  printf("Alarm!\n");

}

Answer 1

您正在计算需要减去的值之前的差异。

替换

difference = alarm_total - local_time_total;
alarm_total = (uhour * 3600) + (uminute * 60) + usecond;
local_time_total = (chour * 3600) + (cminute * 60) + csecond;

带

alarm_total = (uhour * 3600) + (uminute * 60) + usecond;
local_time_total = (chour * 3600) + (cminute * 60) + csecond;
difference = alarm_total - local_time_total;

另外，你应该注意，如果你在Windows上，sleep以毫秒为单位获取它的参数，如果你在Linux或Unix上sleep接受它的参数在几秒钟内，您需要相应地修改sleep功能。

如何通过用户输入来实现闹钟？

1 个答案: