Re: I have a problem with the priority of ecos
Gary Thomas <[email protected]>
| Newsgroups | gmane.os.ecos.devel |
|---|---|
| Message-ID | <[email protected]> |
On 04/12/2011 03:25 PM, moktar_bouain wrote:
>
> Hello ,
> I have a problem with the priority of ecos.
> I have the following configuration:
>
> #include<cyg/kernel/kapi.h>
>
> #include<stdio.h>
> #include<math.h>
> #include<stdlib.h>
> cyg_thread thread_s[2];
> char stack[2][4096];
> cyg_handle_t simple_threadA, simple_threadB;
> cyg_mutex_t cliblock;
> void taska(cyg_addrword_t data)
> {
> printf("TASKA \n");
> }
> void taskb(cyg_addrword_t data)
> {
> printf("TASKB \n");
>
> }
>
> void cyg_user_start(void)
> {
> printf("Entering twothreads' cyg_user_start() function\n");
>
> cyg_mutex_init(&cliblock);
>
> cyg_thread_create(10, taska, (cyg_addrword_t) 0,"Thread A", (void *)
> stack[0], 4096,&simple_threadA,&thread_s[0]);
> cyg_thread_create(0, taskb, (cyg_addrword_t) 1,"Thread B", (void *)
> stack[1], 4096,&simple_threadB,&thread_s[1]);
> }
>
> void main (cyg_addrword_t data)
> {
> for(;;)
> {
> cyg_thread_resume(simple_threadA);
> cyg_thread_resume(simple_threadB);
> }
> }
>
> when I execute this configuration:
> TASKB
> TASKB
> TASKB
> TASKB
> TASKB
> TASKB
> TASKB
> TASKB
> TASKB
> TASKB
> TASKB
>
> but when I changed the priority:
>
> cyg_thread_create(1, taska, (cyg_addrword_t) 0,"Thread A", (void *)
> stack[0], 4096,&simple_threadA,&thread_s[0]);
> cyg_thread_create(0, taskb, (cyg_addrword_t) 1,"Thread B", (void *)
> stack[1], 4096,&simple_threadB,&thread_s[1]);
> I find this false result
> TASKB
> TASKA
> TASKB
> TASKA
> TASKB
> TASKA
> TASKB
> TASKA
> TASKB
> TASKA
> TASKB
> TASKA
> TASKB
> TASKA
> TASKB
>
> Any help??
What scheduler are you using? (this controls how many priorities and what
type of priority mechanism is available)
Also, what makes you think that after a thread does a printf() call, it
should yield the CPU to the other thread? There is no reason for it and
in the simplest configuration printf() is just diag_printf() which is not
interrupt driven and there would never be a reason for one thread to lose
control of the CPU.
--
------------------------------------------------------------
Gary Thomas | Consulting for the
MLB Associates | Embedded world
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