2 routines with a switch

JAMB1234

Please take a look at my program and see if anyone can help please. When I switch IN0 from 1 to 0 the program will execute the next routine but not without running the previous routine one more time. Can anyone help me figure out how I can avoid running the last routine once before it goes on to the next routine.

Thanks

'{$STAMP BS2}
'{$PBASIC 2.5}

timecounterA VAR······· Word·············· ' Time Score of player A
timecounterB VAR······· Word·············· ' Time Score of Player B
DO

DO
DEBUG "PRESS BUTTONS TO STAGE" , CR
IF (IN0=1)THEN
GOTO PRO
ENDIF

IF (IN0=0)THEN
GOTO SUB
ENDIF

SUB:
DO
' Nothing
·LOOP UNTIL (IN3 = 1) AND (IN4 = 1)
HIGH 15
PAUSE 2000
HIGH 14
PAUSE 2000
HIGH 13
PAUSE 3000
HIGH 12
PAUSE 1000
LOW 15
LOW 13
LOW 14
LOW 12
HIGH 9
HIGH 10
HIGH 11
PAUSE 400
LOW 9
LOW 10
LOW 8
LOW 7
LOW 11
PAUSE 1

HIGH 8
HIGH 7
timecounterA = 0
timecounterB = 0
DO
PAUSE 1
IF (IN3 = 1) THEN
timecounterA = timecounterA + 1
ENDIF
IF (IN4 =1) THEN
timecounterB = timecounterB + 1
ENDIF
LOOP UNTIL (IN3 = 0) AND (IN4 = 0)
LOW 8
LOW 7
IF (timecounterA =0) THEN
DEBUG " Left Lane FOUL" , CR
DEBUG " Right Lane Wins" , CR, CR
HIGH 6
PAUSE 4000
LOW 6
ELSEIF (timecounterB = 0) THEN
DEBUG "· Right Lane FOUL" , CR
DEBUG·· "Left Lane Wins" , CR
HIGH 5
PAUSE 4000
LOW 5
ELSE
DEBUG "Left Lane time: ", DEC timecounterA/265,".",DEC2 timecounterA//1000,CR
DEBUG "Right Lane time: ", DEC timecounterB/265,".",DEC2 timecounterB//1000, CR, CR
IF (timecounterA < timecounterB) THEN
··· DEBUG "Left Lane is the winner!", CR
HIGH 15
PAUSE 2000
LOW 15
ELSEIF (timecounterB < timecounterA) THEN
DEBUG "Right Lane is the winner!", CR
HIGH 14
PAUSE 2000
LOW 14
ELSE
DEBUG "It's a Tie!", CR
ENDIF
DEBUG CR
DEBUG "To play again, Hold the", CR
DEBUG "BUTTONS DOWN AGAIN.",CR, CR
ENDIF
LOOP
DO
PRO:
DO
' Nothing
·LOOP UNTIL (IN3 = 1)
·HIGH 15
PAUSE 500
HIGH 14
PAUSE 500
HIGH 13
PAUSE 500
HIGH 12
PAUSE 1000
LOW 15
LOW 13
LOW 14
LOW 12
HIGH 9
HIGH 10
HIGH 11
PAUSE 400
LOW 9
LOW 10
LOW 8
LOW 7
LOW 11
PAUSE 1

HIGH 8
HIGH 7

timecounterA = 0
DO
PAUSE 1
IF (IN3 = 1) THEN
timecounterA = timecounterA + 1
ENDIF
LOOP UNTIL (IN3 = 0)
LOW 8
LOW 7
IF (timecounterA =0) THEN
DEBUG " FOUL " , CR
DEBUG " Right Lane Wins" , CR, CR
HIGH 6
PAUSE 4000
LOW 6
ELSE
DEBUG "Left Lane time: ", DEC timecounterA/260,".",DEC2 timecounterA//1000, " seconds ", CR
DEBUG "To play again, Hold the", CR
DEBUG "BUTTONS DOWN AGAIN.",CR, CR
ENDIF
LOOP

LOOP

sylvie369

Try changing this:
DO
DEBUG "PRESS BUTTONS TO STAGE" , CR
IF (IN0=1)THEN
GOTO PRO
ENDIF

IF (IN0=0)THEN
GOTO SUB
ENDIF

to this

DO
DEBUG "PRESS BUTTONS TO STAGE" , CR
IF (IN0=1)THEN
GOTO PRO
ELSE
GOTO SUB
ENDIF

JAMB1234

I just tried that but I would not go into the next routine at all.

Thanks

Lee Harker

One thing I would suggest just to save·some pain and suffering. It appears you are controlling a few things in sequence with pins 15, 13, 14, 12 and pins 9, 10, 8, 7. If at all possible you should change your hardware so that the devices that need to be in sequence are wired in sequence. That would make your program much easier to write and troubleshoot. Things like that should be in an easy to follow loop.

sam_sam_sam

JAMB1234

You could do this

DEBUG "PRESS BUTTONS TO STAGE" , CR
IF (IN0=1)THEN
GOTO PRO
ELSEIF (IN0=0)THEN
GOTO SUB
ENDIF
·

When I switch IN0 from 1 to 0 the program will execute the next routine but not without running the previous routine one more time.

Which routine is your default routine you could have the routine setup that if it loop so many time then it will do that routine

let say that your SUB routine is the default routine unless you hit that switch



· cntr = cntr + 1 *· Chg··············· · ' Chg·is your input PIN

·IF (cntr =·300 ) THEN GOTO·SUB

· ' cntr= How many loop it will run as long as (·Input =·1 )If cntr = 300··then it will run·PRO routine any thing less than 300 is same thing as a·do nothing loop it still checking the input pin state

This routine dose the same thing except it dose this

·' cntr= How many loop it will run as long as (·Input =·0 )·If cntr = 300 then it will run SUB routine any thing less than 300 is same thing as a·do nothing loop it still checking the input pin state

·DO

··· cntr = cntr + 1 * (1 - Chg)
··· IF (cntr = 300) THEN·GOTO PRO
·· I hope this help································· ·cntr can be any value

This an example of how you would this ___________________________________________________________________

·chg············ PIN···· 4
cntr··········· VAR···· Word

DO

·· cntr = cntr + 1 *· Chg·············'If input = 1 and·CNTR get·to a·count of 300
····· IF (cntr = 300 ) THEN·········· 'then it become true
······ GOTO· sub

·· cntr = cntr + 1 * (1 - Chg)····· 'If input =·0 and·CNTR get·to a·count of 300·
··· IF (cntr = 300) THEN···············'then it become true
······ GOTO· PRO
······ ENDIF
······ ENDIF
·····
·····

·LOOP

·Sub:
·HIGH 1
·PAUSE 500
·LOW 1
·RETURN

·PRO:
·HIGH 2
·PAUSE 500
·LOW 2
·RETURN

··___________________________________________________________________________________________

· DO


DO
DEBUG "PRESS BUTTONS TO STAGE" , CR
IF (IN0=1)THEN
GOTO PRO
ENDIF

IF (IN0=0)THEN
GOTO SUB
ENDIF

One problem that I see is that you have two DO and from what·I can see you are looping your hole routine
This make it hard to follow
One more thing it hard to follow your routine with all output pins not defined

LOW 15
LOW 13
LOW 14
LOW 12
HIGH 9
HIGH 10
HIGH 11
··Name ·Pin 15
· Name· Pin 14
· Name· Pin·13
· Name··Pin 12

and so on

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··Thanks for any··that you may have and all of your time finding them

·
·
·
·
Sam

Post Edited (sam_sam_sam) : 10/19/2008 2:26:57 AM GMT

Chris Savage

As a note this is a duplicate thread which is not allowed on the forums. I caught this too late to remove your other post without deleting the posts of those that tried to help but in the future remember that we will remove duplicate threads.

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Chris Savage
Parallax Engineering