task give up it's time tick in uc/os

Presumably after the task has executed it makes an API call to say "run me again 1ms after I last started" (in FreeRTOS this would be vTaskDelayUntil(), I don't know what the uCOS equivalent is). Once this function is called the calling task would be placed in a Blocked state and not consume any more processing time until its delay period expired - so what you want to happen will happen automatically.

I'm not even sure that uCOS will implement a time slicing scheme in any case as each task must have a unique priority - so timeslicing would not make any sense. Again this is different to FreeRTOS where you can assign unique priorities to each task, or allow tasks to share priorities - also you can choose to use time slicing or not.

Many methods are available. Simplest may be to do a delay for 1 ms. This assumes your tick rate OS_TICKS_PER_SECOND has enough resolution, at least

A second approach may be to have a timer generate an interrupt on a 1 ms interval. The ISR could then signal a semaphore (or about any other mechanism) OSSemPost(semaphore1) that the high priority task is pending on. Such as OSSemPend(semaphore1). Make sure to create the semaphore before trying to use it.

Scott

This is a very common job that a real time kernel must handle. If you want the high priority task to run in lock-step with the 1ms clock, then make some process that runs in interrupt-land set a semaphore on the 1ms clock (I'm not sure if Mucos uses "semaphore", "event", "flag" or some other terminology -- I'll leave that as an exercise for you). Then make the high-priority task pend on that semaphore.

The consequence of this will be that when the 1ms clock pops off the semaphore will be set, the high-priority task will run, then when it calls the OS to pend on the semaphore again it will block. Then other, lower priority, tasks will be able to run.

or use an OS_timer