Fast Loop in PICBASIC Pro

[skimask]

Hi guys,

I'm new to PICBasic!! I'm programming PIC16F877A using PICBASIC Pro & MicroStudio...

What's the FASTEST way to implement a simple loop with a loop counter (initially positive) which decrements & stops when it reaches zero...


Again, i'm talking about the FASTEST (in decrementing counter & repeating itself) way?!!!


Thx...

LOOP: For I=10 TO 1 STEP -1:NEXT I:GOTO LOOP

That's not so hard now is it? It's just like in the PBP manual, with a couple of very small changes.

[sayzer]

There are ten cars in front of you.

And someone is asking which one is the fastest.

Then you look at each one and say "that one".

But the fastest car you found is the fastest car among the ones in front of you.

There are other fastest cars around.

Now, "fastest" changes from one perspective to another.

The fastest for you may be the slowest for me.

Take Skimask's example:
LOOP: For I=10 TO 1 STEP -1:NEXT I:GOTO LOOP

if you run this with 4mhz, you get the "fastest" loop.

If you run the same loop at 20Mhz, you get the "fastest", too, but faster then
4mhz.

If you run it at 40Mhz, you still get "fastest" but faster then 20Mhz.

Now, you do the math.

-------------------------------------

[Bruce]

I VAR BYTE SYSTEM

LOOP1: ; 150 cycles. Not very fast
    For I=10 TO 1 STEP -1
    NEXT I
    
LOOP2: ; 90 cycles. A lot better
    I = 10
    REPEAT
     I=I-1
    UNTIL I=0

ASM   ; 31 cycles. Pretty fast
LOOP3
     MOVLW 0xA
     MOVWF I      ; I=10
INNER
     DECFSZ I,F   ; I=I-1. IF I=0 skip next instruction
     GOTO INNER  ; Loop until I=0
ENDASM

FINISH:
    GOTO FINISH

[sayzer]

 I VAR BYTE SYSTEM

LOOP1: ; 150 cycles. Not very fast
    For I=10 TO 1 STEP -1
    NEXT I
    
LOOP2: ; 90 cycles. A lot better
    I = 10
    REPEAT
     I=I-1
    UNTIL I=0
ASM   ; 31 cycles. Pretty fast
LOOP3
     MOVLW 0xA
     MOVWF I      ; I=10
INNER
     DECFSZ I,F   ; I=I-1. IF I=0 skip next instruction
     GOTO INNER  ; Loop until I=0
ENDASM

FINISH:
    GOTO FINISH

...and, there is also the expert approach!

You know, we kiss the hand of the experts here.



------------------------------------

[skimask]

...and, there is also the expert approach!

You know, we kiss the hand of the experts here.



------------------------------------

Or the keyboard of the experts...

However, I saw the original poster's first line
------------------
Hi guys,

I'm new to PICBasic!!
-----------------------
which says to me 'not fluent in any sort of asm', so he's probably not going to get much better than 90 cycles
JDG

[Bonaparte]

I VAR BYTE SYSTEM

LOOP1: ; 150 cycles. Not very fast
    For I=10 TO 1 STEP -1
    NEXT I
    
LOOP2: ; 90 cycles. A lot better
    I = 10
    REPEAT
     I=I-1
    UNTIL I=0

ASM   ; 31 cycles. Pretty fast
LOOP3
     MOVLW 0xA
     MOVWF I      ; I=10
INNER
     DECFSZ I,F   ; I=I-1. IF I=0 skip next instruction
     GOTO INNER  ; Loop until I=0
ENDASM

FINISH:
    GOTO FINISH

I will probably go for Loop 2 implementation..

(Assembly isn't an appealing choice!!)

However, is there any way to estimate the time required to implement this Loop (2). I need to calculate the time needed to decrement the counter, compare its current value to the end value & the repetition (go back to start of loop) of the body time...

I'm using a 20 MHz crystal as clock....

if it can't be estimated from BASIC code... can the .asm file generated by compiler in assembly help me estimate delay time in executing loop knowing the time for each assembly line code???

This is what i'm trying to do; generation of 40kHz pulses (25us)

Loop:

counter = 16

(set one o/p pin high
pause (delay) by 11 us
set o/p pin to low (zero)
puse (delay) by 10 us

decrement counter & compare it

repeat loop again

i need the total time of loop to be 25 microseconds....

HELP PLEASE!!!

[mister_e]

Why not using the internal PWM module instead? it will run in background and you don't need to care about latency etc etc.

use something like this...

DEFINE OSC 20
TRISC=0
duty var word
duty = 250                                  ' duty value for 50% duty cycle
PR2 = 124                                   ' period for 40KHZ
T2CON = %00000100                           ' timer2 on, prescale 1:1
CCPR1L = duty>>2                            ' eight MSB of duty cycle value
CCP1CON = (duty.lowbyte&3)<<5 | %00001100   ' store  2 LSB of duty 
                                            '  & set PWM mode

START:
      PAUSE 1
      GOTO START

and look what happen on PORTC.2 pin.

Need to start it and stop it ? here's something using PORTB.0 as trigger source

DEFINE OSC 20
TRISC=0
duty        var word
PWMMode     var byte
StartStop   var bit

duty = 250                                  ' duty value for 50% duty cycle
PR2 = 124                                   ' period for 40KHZ
T2CON = %00000100                           ' timer2 on, prescale 1:1
CCPR1L = duty>>2                            ' eight MSB of duty cycle value
PWMMode = (duty.lowbyte&3)<<5 | %00001100   ' store  2 LSB of duty 
                                            '    & set PWM mode

START:
    IF PORTB.0=0 then            ' 
        startstop=startstop ^ 1  ' toggle between start/stop
        if startstop=1 then      '
            CCP1CON=PWMMODE      ' Start PWM
            ELSE                 '
                CCP1CON=0        ' Stop PWM
                PORTC.2=0        ' make sure the pin is low   
            ENDIF                '
        WHILE PORTB.0=0 : WEND   ' wait until trigger release
        PAUSE 100                ' debounce delay 
        ENDIF
    GOTO START

How to calculate those value easy? PicMultiCalc software available bellow for FREE
http://mister-e.org/pages/utilitiespag.html

ScreenShot

<img src="http://www.picbasic.co.uk/forum/attachment.php?attachmentid=1280&stc=1&d=116639098 4">

HTH

[paul borgmeier]

I agree with Steve (and everyone else) that the hardware PWM sounds best for what you described. But, if it will not work for you consider "unrolling" your loops like that shown below. You lose no time in looping but it does cost you more code space (16 repeats is not huge considering the amount of memory in the 877A).

DEFINE OSC 20
TRISB.0 = 0
PORTB = 0

PORTB.0 = 1   ' I = 1 (on for 12.6 uS)
@ nop
Pauseus 12
PORTB.0 = 0   ' (off for 12.4 uS)     
PAUSEUS 12

PORTB.0 = 1   ' I = 2
@ nop
Pauseus 12
PORTB.0 = 0     
PAUSEUS 12

;(more of the same here)

PORTB.0 = 1   ' I = 16
@ nop
Pauseus 12
PORTB.0 = 0     
PAUSEUS 12

Endhere:
goto Endhere
END

[mister_e]

could be...

DEFINE OSC 20
TRISB.0 = 0
PORTB = 0
pauseus 0 ' do nothing good but include PAUSEUS?C macro
TRISB.0 = 0
PORTB = 0
start:
asm
    CHK?RP PORTB
    local a=0
    while a<16        
        BSF PORTB,0
        nop
        PAUSEUS?C 12
        BCF PORTB,0
        nop
        PAUSEUS?C 12
a+=1
    endw
endasm
pause 100
goto start

now play with the a < x value, and look what happen to your code size.. interesting eh?

Now compare the code size of

PORTB.0 = 1   
@ nop
Pauseus 12
PORTB.0 = 0     
@ nop
PAUSEUS 12

copy and pasted 16 times, with the one generated with my solution... mmm interesting

Sometime MPASM directive are really handy!

EDIT: you could also use

@   local a=0
@   while a<16
        PORTB.0=1
        @ nop
        PAUSEUS 12
        PORTB.0=0
        @ nop
        PAUSEUS 12
@a+=1
@       endw

If you don't like the PAUSE?C, BSF, BCF option... both generate the same code.