just a comment -- you need to say and write more
You haven't been very interactive here. And there are some questions outstanding. In particular, showing the assembly code output for the two cases which exhibit different behavior. That would be very helpful to see and likely would explain a lot. For now, there is just too little information from you to help nail this down to a root cause.
delta list/queue
Since you are looking for a cooperative scheduler and since I mentioned a delta queue or delta list to you, I thought I'd talk about that part. I'll assume that you can somehow debug and/or work out a method that allows the following function to be written:
long int last= uwTick;long int deltatime( void ) { long int curr= (long int) *(volatile unsigned long int *) &uwTick; long int diff= curr - last; last= curr; return diff;}If you can get something akin to that working correctly, providing values of 0 or larger (can be more than 1), then I can use that for a simple polling loop to provide what I think you are shooting for. It's more code. But it is more versatile.
Let's start with three tasks. The first task is to be executed every 19 milliseconds. The second task is to be executed every 15 milliseconds. And the third task, every 27 milliseconds. Also, I'll assume that all three are to be scheduled to run in the first millisecond, together. After that, they diverge and rejoin each other per their individual time periods, as appropriate.
void task1(void) { insert(0, 19); printf(" T1"); }void task2(void) { insert(1, 15); printf(" T2"); }void task3(void) { insert(2, 27); printf(" T3"); }Note that they are responsible for re-inserting themselves into the queue with the time delay they want, relative to the "current moment" they are executed. These routines may take much longer than 1 millisecond, if they want. That's okay. The scheduler won't miss a beat.
However, there may be some delay beyond the 1 millisecond period. If so, then that delay should impact deltatime shown above, causing it to yield a value larger than 1 because of the excess delay involved in running your tasks at the same instant of time.
Next is the list of the above tasks. I'm using a simple array that can be allocated statically, dynamically, or pretty much any way you please. The idea depends on the fact that you know, at compile time, the number of tasks. If that's not the case, if you need an extensible queue, then I'll leave the implementation of it up to you. But for my example case, I'll assume you know you have exactly three tasks.
I also need two more elements in the list, so a total of five in this example. One is for the queue HEAD and the other is for the queue TAIL. This wastes a little bit of data for simpler code. Again, you can do it differently. But this is how I'm handling it here.
typedef struct mytask_s { long int delta; int next; int prev; void (*run)(void);} mytask_t;mytask_t tasks[] = { { 0, 0, 0, &task1 }, /* T1 is tasks[0] */ { 0, 0, 0, &task2 }, /* T2 is tasks[1] */ { 0, 0, 0, &task3 }, /* T3 is tasks[2] */ { 0, 0, 0, 0 }, /* queue HEAD is tasks[3] */ { 0, 0, 0, 0 } /* queue TAIL is tasks[4] */};#define QHEAD ((int) (sizeof(tasks)/sizeof(tasks[0])-2))#define QTAIL ((int) (sizeof(tasks)/sizeof(tasks[0])-1))With the insert code here:
void insert(int taskID, long int delta) { int prev = QHEAD, next = tasks[QHEAD].next; for (; next != QTAIL && tasks[next].delta <= delta; prev = next, next = tasks[next].next) delta -= tasks[next].delta; tasks[taskID].next = next; tasks[taskID].prev = prev; tasks[taskID].delta = delta; tasks[next].prev = tasks[prev].next = taskID; if (next != QTAIL) tasks[next].delta -= delta; return;}There's not much more to all this:
int main(int argc, char* argv[]) { /* initialize empty queue */ tasks[QHEAD].next = tasks[QHEAD].prev = QTAIL; tasks[QTAIL].next = tasks[QTAIL].prev = QHEAD; /* insert each of the three tasks into their initial positions */ /* this may be done in any order -- not just the one shown */ insert(0, 1); /* next tick, task 0 */ insert(1, 1); /* next tick, task 1 */ insert(2, 1); /* next tick, task 2 */ long int timestamp = 0; printf("time: %ld\n", timestamp); printqueue(timestamp); for (;;) { long int delta = deltatime(); if (delta < 1) continue; timestamp += delta; int next = tasks[QHEAD].next; if (next != QTAIL) do { long int residual = tasks[next].delta - delta; if (residual > 0) { tasks[next].delta = residual; break; } printf("\ntime: %ld: ", timestamp); do { /* remove the current task from the queue */ int nextup = tasks[next].next; tasks[nextup].prev = tasks[next].prev; tasks[tasks[next].prev].next = nextup; /* run the current task */ tasks[next].run(); /* move to the next possible task */ next = nextup; } while (next != QTAIL && tasks[next].delta == 0); /* there still may be some more time to account for */ /* in the newly recomputed queue -- so check on that */ delta = -residual; printf("\n"); printqueue(timestamp); } while (next != QTAIL && delta > 0); /* all time passed has been accounted for */ } printf("time: %ld\n", timestamp); printqueue(timestamp); return 0;}The printqueue code is just for diagnostics (along with some of the above code that uses printf):
void printqueue(long int timestamp) { for (int i = tasks[QHEAD].next; i != QTAIL; i = tasks[i].next) printf(" T%d: %ld\n", i+1, tasks[i].delta); return;}example run
If I set things up so that deltatime() just returns 1 all the time (the clock advances smoothly on the assumption that all of the tasks can be performed together in less than 1 millisecond) then I get the following output:
time: 0 T1: 1 T2: 0 T3: 0time: 1: T1 T2 T3 T2: 15 T1: 4 T3: 8time: 16: T2 T1: 4 T3: 8 T2: 3time: 20: T1 T3: 8 T2: 3 T1: 8time: 28: T3 T2: 3 T1: 8 T3: 16time: 31: T2 T1: 8 T2: 7 T3: 9time: 39: T1 T2: 7 T3: 9 T1: 3time: 46: T2 T3: 9 T1: 3 T2: 3time: 55: T3 T1: 3 T2: 3 T3: 21time: 58: T1 T2: 3 T1: 16 T3: 5time: 61: T2 T2: 15 T1: 1 T3: 5time: 76: T2 T1: 1 T3: 5 T2: 9time: 77: T1 T3: 5 T2: 9 T1: 5time: 82: T3 T2: 9 T1: 5 T3: 13time: 91: T2 T1: 5 T2: 10 T3: 3time: 96: T1 T2: 10 T3: 3 T1: 6time: 106: T2 T3: 3 T1: 6 T2: 6time: 109: T3 T1: 6 T2: 6 T3: 15time: 115: T1 T2: 6 T1: 13 T3: 2time: 121: T2 T1: 13 T3: 2 T2: 0time: 134: T1 T3: 2 T2: 0 T1: 17time: 136: T3 T2 T2: 15 T1: 2 T3: 10...For example, you'd expect to see that T1 executes at time 1, time 20, etc.
And that's what you see:
T1: 1, 20, 39, 58, 77, 96, 115, 134 ...T2: 1, 16, 31, 46, 61, 76, 91, 106, 121, 136 ...T3: 1, 28, 55, 82, 109, 136 ...All of the above are appropriate. And they do not drift.
(Note: If you look at the timestamp shown and then look at the ordered numbers in the queue displayed in the diagnostic immediately following the timestamp and any tasks that were run, you will see that the timestamp plus the first value provides you with the next timestamp. Also, the timestamp plus the first two values gives you the timestamp following that one. And the timestamp plus the all three values gives to the timestamp after that one. These are deltas relative to the current moment and they are ordered.)
side note 1
One nice thing is that you can allow each of the tasks to perform their own rescheduling, as I do, in the first few lines of code (if known then) or later. Just so long as it is performed before the task exits. That frees the scheduling code itself from having to worry about that part of it.
On the other hand, you could add another value to the task structure itself and let the scheduling code use that value, instead, and keep the rescheduling code out of the tasks, themselves.
Either way is fine.
side note 2
Also take note that the above method does NOT require looking at all the tasks. Only the first task in the queue is examined. The reason is that the insertion routine ensures that tasks after the first task in the queue only have the time remaining after the task before it. So in the above example, when I first inserted all three to start at time 1, the queue would have a delta of 1 for the first entry in the queue, a delta of 0 for the second, and a delta of 0 for the third. That's because when the first task starts, it's 1 value will have expired and then the scheduler will run all three of them at once (anything after the first queue entry that has a delta of 0 is run at the same time as the first entry is run.) This saves looping over all the tasks all the time. Only the first entry needs to be examined, because it is known a priori that all the rest in the queue is to be executed at the same time, or else later. The deltas keep track of by how much.
complex tasks
Finally, although I started this out by using a fixed-size array with a known set of tasks, the fact is that the above works perfectly well if the array is instead the maximum number of tasks and you have fewer and varying numbers of tasks.
For example, to illustrate just how useful this scheme can be, in one measurement system where I applied the above idea, I knew that I had 100 milliseconds per measurement to be emitted precisely. The system spec was 10 measurements per second.
However, each measurement consisted of a varying number of decay curves, their \$\tau\$ or duration being predicted by the prior measurement. And each decay reading period had to be preceded by an illumination period that was also variable in duration. In the worst case, I knew I needed 10 milliseconds of dead time to allow the prior decay curves to settle out enough so that I could make a measurement offset reading, as well.
So the total cycle consisted of N [illumination + decay cycles + short-delay] cycles, followed by a dead period waiting for offset to arrive, followed by an offset measurement, followed by a measurement calculation cycle and DAC update, followed by a very short dead period, before the next cycle began again.
Each measurement cycle task would schedule two things: itself 100 milliseconds from `now' and would set up a remaining time counter for the current cycle and then schedule the first [illumination + decay cycles + short-delay] cycle to occur almost immediately.
Each [illumination + decay cycles + short-delay] cycle would start the illumination and then schedule the decay ADC reading task, as well as scheduling the next [illumination + decay cycles + short-delay] cycle, if the remaining time counter permitted it. Otherwise, it would schedule the [offset delay + offset ADC reading series + calculation/DAC update] task at an exact period before the next measurement cycle.
Once the DAC update happened, there would only be one task left -- the next cycle to start exactly 100 milliseconds after the last one.
So things can be quite complex, if you need them to be. This is a versatile and yet also relatively simple system that can perform almost any complex task needed. It does all this without requiring all the complexity and difficulties of a pre-emptive scheduler.
summary
Of course, you still must be sure that the tasks to be completed can be completed in the time you allow. Nothing will solve the problem where your tasks require more time that is available!!!
Also, none of the above solves your problem with the timer. Until we see the assembly code for both cases and some more details, perhaps, I think the root cause remains to be seen. But if you can solve that problem, perhaps consider the above as the next step. It's easy and yet quite flexible.
Take a moment and think about your future. Not just this project, but future ones too. Having a good conceptual process is worth its weight in gold. Is this worth the effort to fully understand? I think YES it is. It's not complicated and it can solve very complex problems and do it with simplicity and ease. Does it solve every possible need? Of course not! Pre-emptive systems offer a whole new array of valuable features that you cannot get here. But they also bring along many difficulties, too.
This is a tool like any other. But I can definitely say that this is a tool worth learning about and deepening into your soul. Pay the price. It will give back many dividends in exchange.
Dr. Comer
All of the above comes from work by Dr. Comer. If you want to read more about this example, and more, see his first textbook from 1983 on XINU:
You can also read more in his later editions. But to be frank about it, this first edition is really his very best on this topic. The second edition requires more "work" in reading.
Dr. Douglas Comer also has introduced another important concept that is only discussed in detail in his first 1983 book show above. That is the idea of memory marking. I cannot recommend more highly learning about both these ideas -- the delta list and memory marking from his first edition on XINU. The 1983 book is a gold mine. It's also very accessible and easy to read and understand. I cannot more highly recommend that book.