timer in CP/M+

[litwr]

I'm trying to find a way to use timer under CP/M+.  I've made the code which calls firmware (KL_TIME_PLEASE) but it doesn't work. 

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org &100
kl_time_please equ &bd0d
;; get address of routine to call to execute a firmware function
  ld hl,(1)
  ld de,&57
  add hl,de
  ld (firm_jump+1),hl
loop
  call firm_jump
  dw kl_time_please
  call PR00000    ;prints HL as decimal

;; make the end of line
  ld de,eol
  ld c,9
  call 5
  ld hl,cnt
  dec (hl)
  jr nz,loop

;; quit back to command-line
  ret

eol db 13,10,"$"
cnt db 20

;; execute a firmware function
firm_jump
  jp 0

PR00000 proc
        local PRD,PR0
        ld de,-10000
CALL PR0
        ld de,-1000
CALL PR0
ld de,-100
CALL PR0
ld de,-10
CALL PR0
ld A,L
PRD add a,$30
        ld e,a
        ld c,2
        jp 5   ;bdos

PR0 ld A,$FF
ld B,H
ld C,L
inc A
add HL,DE
jr C,$-4

ld H,B
ld L,C
JR PRD
        endp

What is wrong?  Or is there any other way to get timer data with Amstrad CPC6128 CP/M+ ? A lot of thanks for any help.

[AMSDOS]

I'm trying to find a way to use timer under CP/M+.  I've made the code which calls firmware (KL_TIME_PLEASE) but it doesn't work. 

Code Select Expand


org &100
kl_time_please equ &bd0d
;; get address of routine to call to execute a firmware function
  ld hl,(1)
  ld de,&57
  add hl,de
  ld (firm_jump+1),hl
loop
  call firm_jump
  dw kl_time_please
  call PR00000    ;prints HL as decimal

;; make the end of line
  ld de,eol
  ld c,9
  call 5
  ld hl,cnt
  dec (hl)
  jr nz,loop

;; quit back to command-line
  ret

eol db 13,10,"$"
cnt db 20

;; execute a firmware function
firm_jump
  jp 0

PR00000 proc
        local PRD,PR0
        ld de,-10000
   CALL PR0
        ld de,-1000
   CALL PR0
   ld de,-100
   CALL PR0
   ld de,-10
   CALL PR0
   ld A,L
PRD   add a,$30
        ld e,a
        ld c,2
        jp 5   ;bdos

PR0   ld A,$FF
   ld B,H
   ld C,L
   inc A
   add HL,DE
   jr C,$-4

   ld H,B
   ld L,C
   JR PRD
        endp

What is wrong?  Or is there any other way to get timer data with Amstrad CPC6128 CP/M+ ? A lot of thanks for any help.

I had a good look at this, and even use a different routine for printing the number (which was published in AA54), but had no luck with it. The Winape Assembler had a problem with this statement:

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jr c,$-4


in the PR0 routine, so I replaced it with jr c,pr0 which I'm presuming was my mistake, though the Assembler was telling me that &-4 was outside it's range ("&" is used instead of "$"), perhaps this line is doing something in your assembler that it's not meant to, though I'm only guessing now, everything else looks fine to me, though it's obviously not giving you the result you want.

[Executioner]

The WinAPE assembler shouldn't have a problem with jr c,$-4. I've just tested and it all seems fine and outputs FA as the relative address. $ is the current code location, so $-4 is the current code location minus 4. &-4 is the location #FFFC which would be outside the relative jump range of course.

[FloppySoftware]

I have translated the source code to ZSM and assembled with it.

When I execute the COM file, WinAPE hangs.

I don't know what's the problem (yet).

Edit: The following code compiled with MESCC runs perfectly under WinAPE:

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#include <mescc.h>
#include <printf.h>
main()
{
    patch();
    printf("HL = %04x\n", get_time());
}
#asm
patch
ld hl,(1)
ld de,057h
add hl,de
ld (firm_jump + 1),hl
ret
get_time
call firm_jump
defw 0bd0dh
ret
firm_jump
jp 0
#endasm
I think the problem is in the output subroutine.

[AMSDOS]

The WinAPE assembler shouldn't have a problem with jr c,$-4. I've just tested and it all seems fine and outputs FA as the relative address. $ is the current code location, so $-4 is the current code location minus 4. &-4 is the location #FFFC which would be outside the relative jump range of course.

Well the problem was occurring when I converted the "$" sign to "&". Perhaps the fault lies in what that means?
I haven't seen an assembler use "$" to represent a Hexadecimal number, but I guess it's possible.

[Executioner]

I haven't seen an assembler use "$" to represent a Hexadecimal number, but I guess it's possible.

$ is used to represent the current code location. For example,

org #1234

dw $


will generate the address #1234

org #1234

dw $ - 4

will generate the address #1230

[litwr]

I am terribly sorry for my stupid question.    The problem is very easy to solve.  I forgot to save HL before BDOS call.  So

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JP 5
in PR00000 should be replaced by

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push HL
call 5
pop HL

I was sure that BDOS (like AMSDOS, IBM PC DOS/BIOS, ...) preserves CPU registers.  Thanks for the help anyway.  The $-sign means the current code location and it also starts the hexadecimal number in pasmo-cross-assembler.

[FloppySoftware]

Ooops! 

Funny bugs (that we all do sometimes)