OpenEuphoria

1. Multitasking - need help

• Posted by X_Files Mar 21, 2012
• 1488 views

Hi. I'm trying to learn multitasking in Euphoria. I have 2 questions:

Here is my example (from manual):

include std\os.e 
 
atom t1_running = 0 
 
procedure task1() 
  for i = 1 to 10 do 
    printf( 1, " Task (%2d) - sec: %7.4f\n", { i, time() } ) 
    task_yield() 
  end for 
  t1_running = 0 
end procedure 
 
atom t1 = task_create( routine_id( "task1" ), "" ) 
 
task_schedule( t1, 3 ) 
t1_running = 1 
 
while t1_running = 1 do 
  sleep( 4 ) 
  printf( 1, "MAIN PROG  - sec: %7.4f\n", { time() } ) 
  task_yield() 
end while 

Output is:

MAIN PROG  - sec:  4.2340 
 Task ( 1) - sec:  4.2340 
MAIN PROG  - sec:  8.2340 
 Task ( 2) - sec:  8.2340 
 Task ( 3) - sec:  8.2340 
 Task ( 4) - sec:  8.2340 
MAIN PROG  - sec: 12.2340 
 Task ( 5) - sec: 12.2340 
 Task ( 6) - sec: 12.2340 
 Task ( 7) - sec: 12.2340 
MAIN PROG  - sec: 16.2340 
 Task ( 8) - sec: 16.2340 
 Task ( 9) - sec: 16.2340 
 Task (10) - sec: 16.2340 
MAIN PROG  - sec: 20.2340 

1. First time task is executed once. Why?

2. Same example, but I change task scheduler to task_schedule( t1, { 3, 3.2 } ) Output is:

MAIN PROG  - sec:  4.2500 
 Task ( 1) - sec:  4.2500 
MAIN PROG  - sec:  8.2500 
 Task ( 2) - sec:  8.2500 
MAIN PROG  - sec: 12.2500 
 Task ( 3) - sec: 12.2500 
MAIN PROG  - sec: 16.2500 
 Task ( 4) - sec: 16.2500 
MAIN PROG  - sec: 20.2650 
 Task ( 5) - sec: 20.2650 
MAIN PROG  - sec: 24.2650 
 Task ( 6) - sec: 24.2650 
MAIN PROG  - sec: 28.2650 
 Task ( 7) - sec: 28.2650 
MAIN PROG  - sec: 32.2650 
 Task ( 8) - sec: 32.2650 
MAIN PROG  - sec: 36.2650 
 Task ( 9) - sec: 36.2650 
MAIN PROG  - sec: 40.2650 
 Task (10) - sec: 40.2650 
MAIN PROG  - sec: 44.2650 

I see that task is executed always when time (first parameter in sequence) passes and is not depend on second parameter. What is second parameter?