MultiCogSerialDebug object

Peter Verkaik · 2007-12-17 18:13

I wrote a MultiCogSerialDebug object, derived from SerialMirror, that uses
my Format object to print formatted messages.
There is only one print function, cprintf, that does the formatting and printing.
It acts as printf with one argument, but takes an additional boolean parameter, more, that allows
to print compound messages. While more == true, other COGs are prohibited from printing.
The attached picture shows the debug output for the test program.

regards peter

Post Edited (Peter Verkaik) : 12/17/2007 6:20:53 PM GMT

Mike Green · 2007-12-17 19:02

Thanks, Peter. This will be handy.

mirror · 2007-12-17 22:21

Hi Peter,

I had a look. Do you realise there'll be an instance of printbuf for every reference to this object. I don't think it will cause any problems as is, but you could think about moving printbuf to the DAT section for the same reason that all the other variables are in the DAT section.

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Harrison. · 2007-12-17 22:30

You can also take a shortcut when defining 'arrays' in the DAT sections.

The normal (messy) way:

DAT
myvar byte 0,0,0,0

The Propeller Tool way:

CON
   LENGTH = 4

DAT
myvar byte 0[noparse][[/noparse]LENGTH]

I made extensive use of this feature in my PropTCP stuff. I didn't know this was possible until I stumbled onto a post where someone posted some example code that utilized this ingenious feature.

Harrison

Peter Verkaik · 2007-12-17 23:03

Harrison,
Thanks for that pointer. It was in VAR because previous there was no lock.
Now that cprintf is the only print function and there is a lock, it can be moved
to DAT.
I also realized that the lock in method rxcheck is not required as there should
be only one cog that must handle incoming messages.
I have put a new version in the object exchange. And attached it here also.

regards peter

Fred Hawkins · 2007-12-18 03:04

Where does the 'Debug Terminal #1' window come from?

Paul Baker · 2007-12-18 03:15

BASIC Stamp IDE (Maybe it's in the Javalin Stamp's IDE as well, Im not sure)

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Paul Baker
Propeller Applications Engineer

Parallax, Inc.

Fred Hawkins · 2007-12-18 04:16

Well. Well. My mental map of the prop neighborhood has suffered a drastic reversal -- it's us that are on the wrong side of the tracks.

simonl · 2007-12-18 10:41

Hi Paul,

Any news on when the IDE will include the debug window?

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Cheers,

Simon

www.norfolkhelicopterclub.co.uk
You'll always have as many take-offs as landings, the trick is to be sure you can take-off again ;-)
BTW: I type as I'm thinking, so please don't take any offense at my writing style

Paul Baker · 2007-12-19 02:37

No specific timeframe I'm afraid, but it is #2 on the to-do llist and #1 is nearing completion.

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Paul Baker
Propeller Applications Engineer

Parallax, Inc.

simonl · 2007-12-19 10:11

Hey, thanks Paul.

You've got me curious now - what's #1 !?

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Cheers,

Simon

www.norfolkhelicopterclub.co.uk
You'll always have as many take-offs as landings, the trick is to be sure you can take-off again ;-)
BTW: I type as I'm thinking, so please don't take any offense at my writing style

Peter Verkaik · 2007-12-19 15:19

Here is a little problem.
My main code is

· buf[noparse][[/noparse]0] := $41
· buf[noparse][[/noparse]1] := $42
· buf[noparse][[/noparse]2] := 0
· debug.cprintf(string("1st char is %c\r"),buf[noparse][[/noparse]0],false)
· debug.cprintf(string("2nd char is %c\r"),buf[noparse][[/noparse]1],false)
· debug.cprintf(string("total string is %s\r"),@buf,false)
· debug.cprintf(string("1st char is %c\r"),$41,false)
· debug.cprintf(string("2nd char is %c\r"),$42,false)

The first 2 cprintf calls do not print the characters A and B
and the 3rd cprintf call does not print the string AB.
The last 2 cprintf calls do print the characters A and B.

The first 2 should have done also, as buf[noparse][[/noparse]0] == $41 and buf[noparse][[/noparse]1] == $42

What is going on here?
The picture shows the debug output.

regards peter

Post Edited (Peter Verkaik) : 12/19/2007 3:24:18 PM GMT

Peter Verkaik · 2007-12-19 17:53

This really is a serious problem.
The attached test program has main code

· 'send welcome message
· debug.str(string("MultiCogSerialDebug_test3 program"))
· debug.tx(13)
· buf[noparse][[/noparse]0] := $41
· if buf[noparse][[/noparse]0] == $41
··· debug.str(string("1st character loaded correctly"))
··· debug.tx(13)
· debug.str(string("Program finished"))
· debug.tx(13)
· repeat

It uses SerialMirror to display output.
The above outputs
MultiCogSerialDebug_test3 program
Program finished

Apparently buf[noparse][[/noparse]0] does not get $41 assigned.
However, disabling the first debug call,
the output becomes
1st character loaded correctly
Program finished

Could this be a compiler bug?

regards peter

Peter Verkaik · 2007-12-19 18:24

Problem solved.
Increasing the atnStack to 5 longs made the error disappear.

regards peter

Paul Baker · 2007-12-19 19:37

simonl said...
Hey, thanks Paul.

You've got me curious now - what's #1 !?

COM port management

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Paul Baker
Propeller Applications Engineer

Parallax, Inc.

mirror · 2007-12-20 22:59

Of course the object could also have been written as:

OBJ
  fmt: "Format"
  sm: "SerialMirror"
    
DAT
  lockCOG     byte  255
  semID       byte  0
 
VAR
  byte printfbuf[noparse][[/noparse]64] 
 
PUB start(rxpin, txpin, mode, baudrate, semaphore) : okay
  okay := sm.start(rxpin, txpin, mode, baudrate)
  semID := semaphore
  
PUB cprintf(format,arg,more)
  if lockCOG <> cogid
    repeat until not lockset(semID)
    fmt.sprintf(@printfbuf,string("[noparse][[/noparse]COG%1d]:"),cogid)
    sm.str(@printfbuf)
  lockCOG := cogid
  fmt.sprintf(@printfbuf,format,arg)
  sm.str(@printfbuf)
  if !more
    lockCOG := 255
    lockclr(semID)

MultiCogSerialDebug object

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