LEGO MINDSTORMS Robots, Руководство пользователя, Страница: 181 / 226

Here is a simple example that uses two tasks. The first task shows "Hello" and "nurse" on the display. The second task just waits for the Run button to be pressed. When it is pressed, the first task
is stopped, the second task ends, and control returns to legOS.
#include "conio.h"
#include "direct-button.h"
#include "unistd.h"
#include "sys/tm.h"
pid-t pid;
int display_task(int argc, char
∗∗
argv) {
while(1) {
cputs("Hello");
lcd_refresh();
sleep(1);
cputs("nurse");
lcd_refresh();
sleep(1);
}
return 0;
}
int stop_task(int argc, char
∗∗
argv) {
msleep(200);
while (!PRESSED (button_state(), BUTTON_RUN))
;
kill(pid);
return 0;
}
int main() {
pid = execi(&display_task, 0, NULL, 0, DEFAULT_STACK_SIZE);
execi(&stop_task, 0, NULL, 0, DEFAULT_STACK_SIZE);
Page 201
tm_start();
return 0;
}
Let's start at the bottom and work our way up. The
main()
method uses
execi()
to start the two tasks:
display_task()
and
stop_task()
. The process ID of the display task is saved
away in the
pid
variable so that it can be stopped later. The last thing
main()
does is call
tm_start()
to start the task manager, which actually handles running the tasks.
The
stop_task()
waits for the Run button to be pressed. When the button is pressed,
stop_task()
calls
kill()
to stop the display process. What's that call to
msleep()
at the
beginning of
stop_task()?
When legOS first boots on the RCX, it's waiting for the Run button to be pressed. As soon as it is, your program is started. It's very possible that
stop_task()
will already be running before you have a chance to take your finger off the Run button. The call to
msleep()
simply delays a little while to give you a chance to release the Run button.
The display_task() is straightforward. It alternates between displaying "Hello" and "nurse" on the display, looping forever until it is killed by stop_task() .
Waiting (unistd.h)
For a cleaner way to wait for specific events, consider
wait_event()
: