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Hi.
The default process is running of out stack. This is because you've
written a program in which one function calls another, which calls another,
which calls the first. Each time a function is called, an internal "call
stack" must keep track of the nestedness. If you've looped them to each
other (think of a snake eating its tail) eventually you run out of this
stack memory.
Instead, write a program like this:
void main() {
while (1) {
dosomestuff();
doadifferentthing();
whynottryathirdthing();
}
}
That's your main loop. The "while (1)" keeps it from ever finishing.
Each of the three "dosomestuff" functions must exit. So, e.g.:
void dosomestuff() {
fd(0); /* turn on motor 0 */
fd(1); /* turn on motor 1 */
}
Then when that finishes, control will return to main, which will continue
with the next function.
Fred
In lugnet.robotics.handyboard, John R. Edwards writes:
> I have written a small program for the Handyboard to test program flow in
> IC. It is supposed to print "Step1", "Step2", Step3" in sucession to the LCD
> in an endless loop until I reset the board. It works properly through 12
> cycles but on the 13th cycle, the program hangs up and "RUNTIME ERR 04" is
> printed out on the LCD. According to the HB Technical Reference Manual this
> is indicating a stack overflow error in running process but I am not running
> any processes that I know of. If anyone could help me out and tell me where
> my mistakes are I would be most thankful.
>
> John Edwards
>
> The following is my troubled program:
>
> /******************/
> /* Test_Loop.c */
> /******************/
>
>
>
> int n;
>
> void main() /* Push Start Button to Start */
> {
> while(!start_button());
> increment();
> while(start_button());
> }
>
> void increment() /* Add 1 to n each time through */
> { /* When n becomes larger than 3 */
> n = n + 1; /* n is reassigned the value of 1 */
> if(n < 1)
> {
> n = 1;
> }
> if(n > 3)
> {
> n = 1;
> }
> control();
> }
>
> void control() /* Reads Global Variable and */
> { /* Directs program to one of */
> if(n == 1) /* the Step# functions */
> {
> step1();
> }
> if(n == 2)
> {
> step2();
> }
> if(n == 3)
> {
> step3();
> }
> }
>
> void step1() /* Prints Step Number, Delays .1 second */
> { /* then Returns program to increment
> function */
> printf("\nStep1");
> sleep(.1);
> increment();
> }
>
> void step2() /* Prints Step Number, Delays .1 second */
> { /* then Returns program to increment
> function */
> printf("\nStep2");
> sleep(.1);
> increment();
> }
>
> void step3() /* Prints Step Number, Delays .1 second */
> { /* then Returns program to increment
> function */
> printf("\nStep3");
> sleep(.1);
> increment();
> }
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Message has 1 Reply:
 | | Re: IC problem
|
| Fred and Jim, Thank you for your quick responses. I can now see that I was trying to make a relatively simple piece of code very complicated. I rewrote the program using Fred's suggestions and wound up with the following simple, efficient program (...) (24 years ago, 5-Feb-01, to lugnet.robotics.handyboard)
|
Message is in Reply To:
| | IC problem
|
| I have written a small program for the Handyboard to test program flow in IC. It is supposed to print "Step1", "Step2", Step3" in sucession to the LCD in an endless loop until I reset the board. It works properly through 12 cycles but on the 13th (...) (24 years ago, 4-Feb-01, to lugnet.robotics.handyboard)
|
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