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#REDIRECT [[Pl2 CORRIG 4.DOC/fr]]
  
  +
CORRECTIONS of
  +
  +
EXERCISES NR°4
  +
  +
-----------------
  +
  +
  +
1) Exercise nr°1:
  +
--------------
  +
a) If TACR is worth %100 and the data register is worth 42, what will be
  +
the call frequency of a program on this TIMER?
  +
  +
With TACR = %100, we have a prescaler of 50, the data register
  +
is worth 42, so we get:
  +
  +
frequency = 245760 / 50 / 42 = about 117 calls per second (Delay Mode)
  +
  +
  +
Will the program be installed on TIMER A or TIMER B?
  +
  +
TACR manages the operating mode of TIMER A. (have you forgotten
  +
already ?)
  +
  +
  +
  +
b) How to stop a program under interruption installed
  +
on TIMER A ? :
  +
  +
It is enough to set the first 4 bits of the TACR register to 0
  +
  +
  +
c) We want to install a program under interruption that must execute
  +
about 245 times per second, give a possible value of the control
  +
and data registers to achieve this effect.
  +
  +
If the control register is worth %10, we have a prescaler of 10 and
  +
if the data register also worth 10, we get:
  +
  +
frequency = 245760 / 10 / 10 = 245.76 calls per second (Delay Mode)
  +
  +
  +
d) Can we install several programs under interruption in the
  +
same TIMER?
  +
  +
Of course NOT!
  +
Just look at how to install a routine
  +
on a TIMER, the first thing we do is set all the
  +
registers of the concerned TIMER to 0 to disable the routine that
  +
was already there....
  +
  +
  +
e) A program under interruption must necessarily execute in
  +
SUPERVISOR MODE.
  +
Find a reason for this ...
  +
  +
A simple reason is that at the end of the program (Before the RTE),
  +
we clear the concerned bit of the ISR register.
  +
And any modification of the MFP registries SHOULD be done in SUPERVISOR
  +
MODE...
  +
  +
  +
1) Exercise nr°2:
  +
--------------
  +
Here are the macros for TIMERS A and B:
  +
  +
  +
TIMERA MACRO $\1,$\2,$\3 ;MACRO with 3 parameters
  +
  +
; Zeroing the different bit vectors for TIMER A
  +
  +
and.b #%11011111,$FFFA13 ;IMRA
  +
and.b #%11011111,$FFFA0F ;ISRA
  +
and.b #%11011111,$FFFA0B ;IPRA
  +
and.b #%11011111,$FFFA07 ;IERA
  +
  +
; XBTIMER and installing our routine
  +
  +
pea \1 ;\1=LABEL of the start of the routine
  +
; to put under interruption
  +
move.w #\2,-(sp) ;\2=control register
  +
move.w #\3,-(sp) ;\3=data register
  +
move.w #0,-(sp) ;0 =TIMER A
  +
move.w #$1F,-(sp) ;XBTIMER
  +
trap #14 ;XBIOS
  +
adda.l #12,sp ;we reposition SP
  +
  +
; Setting to 1 the different bit vectors for TIMER A
  +
  +
or.b #%100000,$FFFA07 ;IERA
  +
or.b #%100000,$FFFA13 ;IMRA
  +
  +
ENDM ;End of the macro
  +
  +
  +
  +
  +
TIMERB MACRO $\1,$\2,$\3 ;MACRO with 3 parameters
  +
  +
; Zeroing the different bit vectors for TIMER B
  +
  +
and.b #%11111110,$FFFA13 ;IMRB
  +
and.b #%11111110,$FFFA0F ;ISRB
  +
and.b #%11111110,$FFFA0B ;IPRB
  +
and.b #%11111110,$FFFA07 ;IERA
  +
  +
; XBTIMER and installing the routine on TIMER B
  +
  +
pea \1 ;\1=LABEL of the start of the routine
  +
; to put under interruption
  +
move.w #\2,-(sp) ;\2=control register
  +
move.w #\3,-(sp) ;\3=data register
  +
move.w #1,-(sp) ;1 = TIMER B
  +
move.w #$1F,-(sp) ;XBTIMER
  +
trap #14 ;XBIOS
  +
adda.l #12,sp ;we reposition SP
  +
  +
; Setting to 1 the different bit vectors
  +
  +
or.b #%1,$FFFA07 ;IERB
  +
or.b #%1,$FFFA13 ;IMRB
  +
  +
ENDM ;END of the MACRO
  +
  +
  +
  +
2) Exercise nr°3:
  +
--------------
  +
Two solutions exist to stop TIMERS A and B.
  +
  +
The first consists of setting the first 4 bits of the TACR registers
  +
(For TIMER A) or TBCR (For TIMER B) to 0.
  +
  +
So we would write:
  +
  +
BCLR #0,$FFFA19 ;bit 0 of the TACR register
  +
BCLR #1,$FFFA19 ;bit 1
  +
BCLR #2,$FFFA19 ;bit 2
  +
BCLR #3,$FFFA19 ;bit 3
  +
  +
For TIMER A (In SUPERVISOR MODE!)
  +
  +
Or
  +
  +
BCLR #0,$FFFA1B ;bit 0 of the TBCR register
  +
BCLR #1,$FFFA1B ;bit 1
  +
BCLR #2,$FFFA1B ;bit 2
  +
BCLR #3,$FFFA1B ;bit 3
  +
  +
For TIMER B (In SUPERVISOR MODE!)
  +
  +
  +
There is also an XBIOS function that allows to block an
  +
interruption of the MFP 68901: It is the JDISINT function of code 26.
  +
  +
We just need to pass the IPL of the MFP that must be stopped.
  +
It is with this function that we will create our MACROS:
  +
  +
  +
For TIMER A (Level 13/6 in the MFP), we will write:
  +
  +
STOPTIMEA MACRO
  +
move.w #13,-(sp) ;TIMER A ( IPL 13/6)
  +
move.w #26,-(sp) ;JDISINT
  +
trap #14 ;XBIOS
  +
addq.l #4,sp ;we reposition SP
  +
ENDM
  +
  +
For TIMER B (Level 8/6 in the MFP), we will write:
  +
  +
STOPTIMEB MACRO
  +
move.w #8,-(sp) ;TIMER B ( IPL 8/6 )
  +
move.w #26,-(sp) ;JDISINT
  +
trap #14 ;XBIOS
  +
addq.l #4,sp ;we reposition SP
  +
ENDM
  +
  +
  +
  +
3) Exercise nr°3:
  +
--------------
  +
Here is the listing of the program that varies the palette of colors
  +
under interruption.
  +
There was no difficulty.
  +
  +
  +
TEXT
  +
  +
INCLUDE "INIT_TOS.L" ;Setblock
  +
INCLUDE "MACROS.L" ;The MACROS
  +
INCLUDE "TIMERS.L" ;The file of MACROS
  +
;for the TIMERS
  +
  +
SETBLOCK ;initialisation
  +
  +
SUPER ;SUPERVISOR MODE
  +
  +
PRINTLINE CA ;text
  +
WAIT ;wait
  +
  +
TIMERA ROUTINE,200,%111 ;We install the routine
  +
;on TIMER A with the
  +
;data register=200
  +
;and control register
  +
;=%111
  +
  +
USER ;USER MODE for
  +
  +
KEEP 2000 ;return to DESKTOP with the program
  +
  +
; The routine under interruption that changes the colors:
  +
  +
ROUTINE cmpi #30,TESTE ;Do we have 30 in 'TESTE'?
  +
bgt DEUX ;If it's greater, go to 'DEUX'
  +
move.l #PAL1,$45A ;Address of the new palette in
  +
;'colorptr'
  +
add #1,TESTE ;Otherwise add 1 to 'TESTE'
  +
bclr #5,$FFFA0F ;Clear ISRA (bit 5=TIMER A)
  +
RTE ;RETURN FROM EXCEPTION
  +
  +
DEUX move.l #PAL2,$45A ;Address of the new palette in
  +
;'colorptr'
  +
add #1,TESTE ;Add 1 to 'TESTE'
  +
cmpi #60,TESTE ;Do we have 60 in 'TESTE' ?
  +
bne NON ;If NO, go to 'NON'
  +
move #0,TESTE ;Set 0 in 'TESTE'
  +
NON bclr #5,$FFFA0F ;Clear ISRA (bit 5=TIMER A)
  +
RTE ;RETURN FROM EXCEPTION
  +
  +
DATA
  +
  +
; Palette nr° 1
  +
  +
PAL1 DC.W $777,$700,$070,$000,$000,$777,$777,$777
  +
DC.W $777,$777,$777,$777,$777,$777,$777,$777
  +
  +
; Palette nr° 2
  +
  +
PAL2 DC.W $777,$700,$007,$000,$000,$777,$777,$777
  +
DC.W $777,$777,$777,$777,$777,$777,$777,$777
  +
  +
CA DC.B 27,'E','Routine under interruption (TIMER A )'
  +
DC.B ' that changes the',13,10,'color nr°3 of the'
  +
DC.B ' palette (System Variable $45A):',0
  +
  +
BSS
  +
  +
DS.B 500
  +
PILE DS.B 1 ;For SETBLOCK
  +
SAUV_SP DS.L 1 ;For SUPER and USER
  +
TESTE DS.W 1 ;For the routine under interruption
  +
  +
END
  +
  +
  +
  +
4) Exercise nr°4:
  +
--------------
  +
Here too, no major difficulty...
  +
  +
  +
TEXT
  +
  +
INCLUDE "INIT_TOS.L" ;SETBLOCK
  +
INCLUDE "MACROS.L" ;The MACROS
  +
INCLUDE "TIMERS.L" ;The TIMERS MACROS
  +
INCLUDE "MACROS_2.L" ;The MACRO HEXA
  +
  +
SETBLOCK ;initialisation
  +
  +
SUPER ;SUPERVISOR MODE
  +
  +
PRINTLINE LA ;text
  +
CCONOUT #13 ;13 +
  +
CCONOUT #10 ;10 = return to the line at
  +
;column 1
  +
WAIT ;wait
  +
  +
TIMERA PRG,50,%111 ;We install our routine
  +
;on TIMER A with the da-
  +
;ta register = 50 and the
  +
;control register=%111
  +
  +
; The loop that displays the values of the variable
  +
  +
BOUCLE HEXA VAL ;Displays the L-M of 'VAL' in HEXA
  +
CCONOUT #13 ;13 +
  +
CCONOUT #10 ;10 = return to the line at column 1
  +
WAIT ;wait
  +
cmpi.b #'Q',d0 ;Key = Q?
  +
beq RETOUR ;If YES, go to 'RETOUR'
  +
cmpi.b #'q',d0 ;Key = q?
  +
beq RETOUR ;If YES, go to 'RETOUR'
  +
jmp BOUCLE and we start again with 'BOUCLE'
  +
  +
  +
; The routine under interruption that just increments the L-M
  +
; in 'VAL'
  +
  +
PRG add.l #1,VAL ;Add 1 to the L-M in 'VAL'
  +
bclr #5,$FFFA0F ;Clear ISRA (bit 5=TIMER A)
  +
RTE ;RETURN FROM EXCEPTION
  +
  +
RETOUR USER ;USER MODE
  +
STOPTIMEA ;we stop TIMER A
  +
TERM ;and END!
  +
  +
DATA
  +
  +
LA DC.B 27,'E','Routine under interruption (TIMER A )'
  +
DC.B ' that increments the',13,10,'content of the di'
  +
DC.B 'splayed variable ([Q] to QUIT) ...',0
  +
  +
BSS
  +
  +
DS.B 500
  +
PILE DS.B 1 ;For SETBLOCK
  +
SAUV_SP DS.L 1 ;For SUPER and USER
  +
VAL DS.L 1 ;The variable to display
  +
  +
END
  +
  +
  +
--------------------
  +
  +
Well, the listings and the executable programs of these exercises
  +
can be found in the files:
  +
  +
1) TIMERS .L for the MACROS TIMERA,TIMERB,STOPTIMEA,STOPTIMEB
  +
  +
2) PALETTE.L for the program of changing the palette under
  +
interruption.
  +
and PALETTE.PRG
  +
  +
3) AFFICHE.L for the program that displays the L-M incremented under
  +
interruption.
  +
  +
and AFFICHE.PRG
  +
  +
  +
  +
  +
PIECHOCKI Laurent
  +
8, Impasse Bellevue Continued in the file:ANNEXE.DOC
  +
57980 TENTELING ----------

Revision as of 15:07, 17 December 2023

                           -----------------
                            CORRECTIONS of 

                            EXERCISES NR°4
                             
                           -----------------


1) Exercise nr°1: 

  --------------
 a) If TACR is worth %100 and the data register is worth 42, what will be
    the call frequency of a program on this TIMER?
   
    With TACR = %100, we have a prescaler of 50, the data register
    is worth 42, so we get:

    frequency = 245760 / 50 / 42 = about 117 calls per second (Delay Mode)



    Will the program be installed on TIMER A or TIMER B?

    TACR manages the operating mode of TIMER A. (have you forgotten
    already ?) 



 b) How to stop a program under interruption installed
    on TIMER A ? :

    It is enough to set the first 4 bits of the TACR register to 0


 c) We want to install a program under interruption that must execute
    about 245 times per second, give a possible value of the control
    and data registers to achieve this effect.

    If the control register is worth %10, we have a prescaler of 10 and
    if the data register also worth 10, we get:

    frequency = 245760 / 10 / 10 = 245.76 calls per second (Delay Mode)



 d) Can we install several programs under interruption in the
    same TIMER?
    
    Of course NOT!
    Just look at how to install a routine
    on a TIMER, the first thing we do is set all the 
    registers of the concerned TIMER to 0 to disable the routine that
    was already there....


 e) A program under interruption must necessarily execute in
    SUPERVISOR MODE.
    Find a reason for this ...

    A simple reason is that at the end of the program (Before the RTE),
    we clear the concerned bit of the ISR register.
    And any modification of the MFP registries SHOULD be done in SUPERVISOR
    MODE...


1) Exercise nr°2: 

  -------------- 
  Here are the macros for TIMERS A and B:


TIMERA MACRO $\1,$\2,$\3 ;MACRO with 3 parameters

Zeroing the different bit vectors for TIMER A
         and.b     #%11011111,$FFFA13  ;IMRA
         and.b     #%11011111,$FFFA0F  ;ISRA
         and.b     #%11011111,$FFFA0B  ;IPRA
         and.b     #%11011111,$FFFA07  ;IERA
XBTIMER and installing our routine
         pea       \1                  ;\1=LABEL of the start of the routine
                                       ;   to put under interruption
         move.w    #\2,-(sp)           ;\2=control register
         move.w    #\3,-(sp)           ;\3=data register
         move.w    #0,-(sp)            ;0 =TIMER A
         move.w    #$1F,-(sp)          ;XBTIMER           
         trap      #14                 ;XBIOS
         adda.l    #12,sp              ;we reposition SP
Setting to 1 the different bit vectors for TIMER A
         or.b      #%100000,$FFFA07    ;IERA
         or.b      #%100000,$FFFA13    ;IMRA
         
         ENDM                          ;End of the macro



TIMERB MACRO $\1,$\2,$\3 ;MACRO with 3 parameters

Zeroing the different bit vectors for TIMER B
         and.b     #%11111110,$FFFA13  ;IMRB
         and.b     #%11111110,$FFFA0F  ;ISRB
         and.b     #%11111110,$FFFA0B  ;IPRB
         and.b     #%11111110,$FFFA07  ;IERA
XBTIMER and installing the routine on TIMER B
         pea       \1                  ;\1=LABEL of the start of the routine
                                       ;   to put under interruption
         move.w    #\2,-(sp)           ;\2=control register
         move.w    #\3,-(sp)           ;\3=data register
         move.w    #1,-(sp)            ;1 = TIMER B
         move.w    #$1F,-(sp)          ;XBTIMER      
         trap      #14                 ;XBIOS
         adda.l    #12,sp              ;we reposition SP
Setting to 1 the different bit vectors
         or.b      #%1,$FFFA07         ;IERB
         or.b      #%1,$FFFA13         ;IMRB
         
         ENDM                          ;END of the MACRO


2) Exercise nr°3: 

  --------------
  Two solutions exist to stop TIMERS A and B.

  The first consists of setting the first 4 bits of the TACR registers
  (For TIMER A) or TBCR (For TIMER B) to 0.

  So we would write:

    BCLR  #0,$FFFA19    ;bit 0 of the TACR register
    BCLR  #1,$FFFA19    ;bit 1
    BCLR  #2,$FFFA19    ;bit 2
    BCLR  #3,$FFFA19    ;bit 3

  For TIMER A (In SUPERVISOR MODE!)   

  Or 

    BCLR  #0,$FFFA1B    ;bit 0 of the TBCR register
    BCLR  #1,$FFFA1B    ;bit 1
    BCLR  #2,$FFFA1B    ;bit 2
    BCLR  #3,$FFFA1B    ;bit 3

  For TIMER B (In SUPERVISOR MODE!)



  There is also an XBIOS function that allows to block an
  interruption of the MFP 68901: It is the JDISINT function of code 26.

  We just need to pass the IPL of the MFP that must be stopped.
  It is with this function that we will create our MACROS:



  For TIMER A (Level 13/6 in the MFP), we will write:

STOPTIMEA MACRO 

           move.w    #13,-(sp)    ;TIMER A ( IPL 13/6)
           move.w    #26,-(sp)    ;JDISINT
           trap      #14          ;XBIOS
           addq.l    #4,sp        ;we reposition SP
           ENDM                   

  For TIMER B (Level 8/6 in the MFP), we will write:

STOPTIMEB MACRO 

           move.w    #8,-(sp)     ;TIMER B ( IPL 8/6 )     
           move.w    #26,-(sp)    ;JDISINT
           trap      #14          ;XBIOS
           addq.l    #4,sp        ;we reposition SP
           ENDM


3) Exercise nr°3: 

  --------------
  Here is the listing of the program that varies the palette of colors
  under interruption.
  There was no difficulty. 



              TEXT

              INCLUDE       "INIT_TOS.L"    ;Setblock
              INCLUDE       "MACROS.L"      ;The MACROS
              INCLUDE       "TIMERS.L"      ;The file of MACROS
                                            ;for the TIMERS

              SETBLOCK            ;initialisation

              SUPER               ;SUPERVISOR MODE

              PRINTLINE     CA    ;text
              WAIT                ;wait

              TIMERA        ROUTINE,200,%111     ;We install the routine
                                                 ;on TIMER A with the
                                                 ;data register=200
                                                 ;and control register
                                                 ;=%111

              USER                ;USER MODE for
         
              KEEP          2000  ;return to DESKTOP with the program
The routine under interruption that changes the colors

ROUTINE cmpi #30,TESTE ;Do we have 30 in 'TESTE'? 

              bgt           DEUX       ;If it's greater, go to 'DEUX'
              move.l        #PAL1,$45A ;Address of the new palette in
                                       ;'colorptr'
              add           #1,TESTE   ;Otherwise add 1 to 'TESTE' 
              bclr          #5,$FFFA0F ;Clear ISRA (bit 5=TIMER A)
              RTE                      ;RETURN FROM EXCEPTION

DEUX move.l #PAL2,$45A ;Address of the new palette in 

                                       ;'colorptr'
              add           #1,TESTE   ;Add 1 to 'TESTE'
              cmpi          #60,TESTE  ;Do we have 60 in 'TESTE' ?
              bne           NON        ;If NO, go to 'NON'
              move          #0,TESTE   ;Set 0 in 'TESTE'

NON bclr #5,$FFFA0F ;Clear ISRA (bit 5=TIMER A) 

              RTE                      ;RETURN FROM EXCEPTION

              DATA
Palette nr° 1

PAL1 DC.W $777,$700,$070,$000,$000,$777,$777,$777 

              DC.W          $777,$777,$777,$777,$777,$777,$777,$777
Palette nr° 2

PAL2 DC.W $777,$700,$007,$000,$000,$777,$777,$777 

              DC.W          $777,$777,$777,$777,$777,$777,$777,$777

CA DC.B 27,'E','Routine under interruption (TIMER A )' 

              DC.B          ' that changes the',13,10,'color nr°3 of the'
              DC.B          ' palette (System Variable $45A):',0

              BSS

              DS.B          500

PILE DS.B 1 ;For SETBLOCK SAUV_SP DS.L 1 ;For SUPER and USER TESTE DS.W 1 ;For the routine under interruption 

              END


4) Exercise nr°4: 

  --------------
  Here too, no major difficulty...



              TEXT

              INCLUDE       "INIT_TOS.L"    ;SETBLOCK
              INCLUDE       "MACROS.L"      ;The MACROS
              INCLUDE       "TIMERS.L"      ;The TIMERS MACROS
              INCLUDE       "MACROS_2.L"    ;The MACRO HEXA

              SETBLOCK                 ;initialisation                   

              SUPER                    ;SUPERVISOR MODE

              PRINTLINE     LA         ;text
              CCONOUT       #13        ;13 +
              CCONOUT       #10        ;10 = return to the line at
                                       ;column 1
              WAIT                     ;wait

              TIMERA        PRG,50,%111     ;We install our routine
                                            ;on TIMER A with the da-
                                            ;ta register = 50 and the
                                            ;control register=%111
The loop that displays the values of the variable

BOUCLE HEXA VAL ;Displays the L-M of 'VAL' in HEXA 

              CCONOUT       #13     ;13 +
              CCONOUT       #10     ;10 = return to the line at column 1
              WAIT                  ;wait
              cmpi.b        #'Q',d0 ;Key = Q?   
              beq           RETOUR  ;If YES, go to 'RETOUR'   
              cmpi.b        #'q',d0 ;Key = q?   
              beq           RETOUR  ;If YES, go to 'RETOUR'   
              jmp           BOUCLE  and we start again with 'BOUCLE'


The routine under interruption that just increments the L-M
in 'VAL'

PRG add.l #1,VAL ;Add 1 to the L-M in 'VAL' 

              bclr          #5,$FFFA0F      ;Clear ISRA (bit 5=TIMER A)
              RTE                           ;RETURN FROM EXCEPTION

RETOUR USER ;USER MODE 

              STOPTIMEA      ;we stop TIMER A
              TERM           ;and END!

              DATA

LA DC.B 27,'E','Routine under interruption (TIMER A )' 

              DC.B          ' that increments the',13,10,'content of the di'
              DC.B          'splayed variable ([Q] to QUIT) ...',0

              BSS

              DS.B          500

PILE DS.B 1 ;For SETBLOCK SAUV_SP DS.L 1 ;For SUPER and USER VAL DS.L 1 ;The variable to display 

              END



                          --------------------
    
 Well, the listings and the executable programs of these exercises 
 can be found in the files:

 1)  TIMERS .L    for the MACROS    TIMERA,TIMERB,STOPTIMEA,STOPTIMEB

 2)  PALETTE.L    for the program of changing the palette under
                  interruption.
 and  PALETTE.PRG
   
 3)  AFFICHE.L    for the program that displays the L-M incremented under
                  interruption.

 and  AFFICHE.PRG




 PIECHOCKI Laurent
 8, Impasse Bellevue            Continued in the file:ANNEXE.DOC
 57980  TENTELING                                     ----------
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