在运行时保持对用户输入的响应,以重新启动选项
我已经编程了一个小型提醒应用程序,该应用程序每隔几秒钟发出哔哔声:
#include <windows.h>
#include <stdio.h>
int main()
{
int interval;
printf("How often shall I remind you?\n");
printf("Every x seconds: ");
scanf("%d", &interval);
if (interval < 0 || interval > 300)
{
printf("Invalid interval. Try 0 < interval < 300.");
return 1;
}
int numofbeeps = 3600/interval;
printf("Good. I will beep every %d seconds (that makes %d times) over the next 60 minutes.\n", interval, numofbeeps);
for(int ctr=0 ; ctr < numofbeeps ; ctr++)
{
Sleep(interval*1000);
Beep( 1000, 750 );
}
return 0;
}
是否有一种方法可以连续扫描用户输入(即使在 goto - 使用来重新启动提醒循环)?
I have programmed a small reminder application which beeps every few seconds as specified by the user:
#include <windows.h>
#include <stdio.h>
int main()
{
int interval;
printf("How often shall I remind you?\n");
printf("Every x seconds: ");
scanf("%d", &interval);
if (interval < 0 || interval > 300)
{
printf("Invalid interval. Try 0 < interval < 300.");
return 1;
}
int numofbeeps = 3600/interval;
printf("Good. I will beep every %d seconds (that makes %d times) over the next 60 minutes.\n", interval, numofbeeps);
for(int ctr=0 ; ctr < numofbeeps ; ctr++)
{
Sleep(interval*1000);
Beep( 1000, 750 );
}
return 0;
}
Is there a way to continuously scan for user input (even during sleep
times) and, upon a certain input, take a certain action (e.g. restart the reminder loop by overwriting a variable and some goto
-use)?
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