Unfinished Bitness

Now that I’ve been successful in printing strings in whole and character by character, I thought the next logical step would be to get a string from the user.  This post deals specifically with that.  I use the CIO routine Get Record ($05) to accomplish it.  I also introduce a new instruction.  Let’s get to it.

Instructions

I only introduce one new instruction this time.  It is:

• DEY – DEcrement the Y register.  This reduces the value in the Y register by 1.

Source Code

10     *= $3600
20   PRSPTR = $CB
21   INSPTR = $CD
39 ; Defines
70   CIOV = $E456
71   ICHID = $0340 ;IOCB 0 S:
72   ICCOM = $0342 ;IOCB Command
73   ICBAL = $0344 ;Xfer Buffer Adr
74   ICBAH = $0345
75   ICBLL = $0348 ;Buffer Len
76   ICBLH = $0349
77   ROWCRS = 84
78   COLCRS = 85
80   EOL = $9B
81   EOS = $00
82   CLS = $7D
300 ; Test printing
302 TEST
310   LDA #SCLR/256
312   LDX #SCLR&255
314   JSR PRTCH
320   LDX #0
322   LDY #0
324   JSR CURXY
326   LDA #SRULE/256
328   LDX #SRULE&255
330   JSR PRTCH
340   LDX #0
350   LDY #2
360   JSR CURXY
370   LDA #SP1/256
380   LDX #SP1&255
390   JSR PRTCH
430   LDA #VSNAME/256
440   LDX #VSNAME&255
450   JSR INPSTR
460   LDX #0
470   LDY #4
480   JSR CURXY
490   LDA #SP2/256
500   LDX #SP2&255
510   JSR PRTCH
520   LDA #VSNAME/256
530   LDX #VSNAME&255
540   JSR PRTCH
550   LDA #SWELC/256
560   LDX #SWELC&255
570   JSR PRTCH
580   RTS 
5000 ; Print Subroutine
5005 PRTCH
5010 ; Store Starting String Address
5015   STX PRSPTR
5020   STA PRSPTR+1
5030   LDA #$00
5040   STA PRINDX
5050 ; Set IOCB Command
5052   LDX #$00
5055   LDA #$0B
5060   STA ICCOM,X
5070 ; Set Max Buffer Length
5072   LDA #$00
5080   STA ICBLL,X
5090   STA ICBLH,X
5100 ; Call CIO to Print
5105 PRCHLP LDY PRINDX
5108   LDA (PRSPTR),Y
5115   BEQ PRCHDN
5116   JSR CIOV
5120   INC PRINDX
5130   BNE PRCHLP
5140 PRCHDN RTS 
6000 ; Position Subroutine
6005 CURXY
6010   STX COLCRS
6020   STY ROWCRS
6030   RTS 
7000 ; Input String
7005 INPSTR
7010 ; Store Starting String Address
7015   STX ICBAL
7020   STA ICBAH
7050 ; Set IOCB Command
7052   LDX #$00
7055   LDA #$05 ; Get Record
7060   STA ICCOM,X
7070 ; Set Max Buffer Length
7072   LDA #30
7080   STA ICBLL,X
7085   LDA #0
7090   STA ICBLH,X
7100 ; Call CIO to Input
7110   JSR CIOV
7160   RTS 
10000 ; String Data
10001 SCLR .BYTE CLS
10002 SRULE .BYTE "012345678901234567890123456789012345",EOS
10005 SP1 .BYTE "Name? ",EOS
10010 SP2 .BYTE "Hello ",EOS
10020 SWELC .BYTE ", Welcome!",EOS
10050 VSNAME *= *+30
10060 PRINDX *= *+1
15000   .END

Breakdown

As usual I am only going to document what is different.

• Lines 310 to 390: Clear the screen, print the ruler, print the input prompt “Name? “.
• Lines 430 to 450: Load the Accumulator and X register with the high and low bytes of the input string address, then call the input subroutine identified by label INPSTR.
• Lines 460 to 570: Print the “Hello ” message, followed by the input string, followed by “, Welcome!”.
• Line 580: Return (end of program)
• Lines 5000 to 5140: The print string routine that prints character by character rather than the whole string as previously written about.
• Lines 6000 to 6030: The cursor positioning routine as previously written about.
• Line 7005: Start of the input string sub routine, labelled with INPSTR.
• Line 7015 to 7020: Store the string address high and lowe bytes into ICBAL and ICBAH (the IOCB string address).
• Line 7052: Load the X register with 0 to reflect IOCB 0.
• Line 7055: Load the Accumulator with 5, the IOCB command for get record.
• Line 7060: Store the Accumulator in the IOCB’s command address.
• Line 7072: Load the Accumulator with 30.  This is the maximum number of characters that will be accepted.
• Line 7080: Store the Accumulator in the IOCB’s buffer length low byte address (ICBLL).
• Line 7085: Load the Accumulator with 0.  This only needs to be defined if the maximum string length is greater than 255.
• Line 7090: Store the Accumulator in the IOCB’s buffer length high byte address (ICBLH).
• Line 7110: Call the CIO routine to get the input.
• Line 7160: Return from the subroutine.
• Line 10050: This is the input string address.  Reserving 30 bytes.
• Line 10060:  This is the print character index.  Reserving 1 byte.  I moved it from the beginning of the code to the end.

Assembly

After it is assembled and run, the program asks for your name.  Enter a name and it is echo’d back with a welcome message:

Output and EOL Problem

As you can see from the output below there is a slight problem.  When the user terminates the input with the Return key, EOL (155 / $9B) is added to the end of the string.  So when it prints, an unexpected result is the cursor is moved to left most column on next row causing the next thing printed to be in the wrong location:

Since this is not desirable I had to revise the input routine slightly.  After the input call, I load the Y register with the value in ICBLL, which is the number of characters actually input.  Because the count includes the Return key, I decrement the value, then store and EOS (0) at the string address offset by the Y register value.  This is a pointer operation so I needed another 2 bytes in page zero to store the address.  I chose $CD.

Source Code – Revised

In the above listing, the highlighted lines reflect updates to the first source code listing.

10   *= $3600
20 PRSPTR = $CB
21 INSPTR = $CD
39 ; Defines
70 CIOV = $E456
71 ICHID = $0340 ;IOCB 0 S:
72 ICCOM = $0342 ;IOCB Command
73 ICBAL = $0344 ;Xfer Buffer Adr
74 ICBAH = $0345
75 ICBLL = $0348 ;Buffer Len
76 ICBLH = $0349
77 ROWCRS = 84
78 COLCRS = 85
80 EOL = $9B
81 EOS = $00
82 CLS = $7D
300 ; Test printing
302 TEST
310   LDA #SCLR/256
312   LDX #SCLR&255
314   JSR PRTCH
320   LDX #0
322   LDY #0
324   JSR CURXY
326   LDA #SRULE/256
328   LDX #SRULE&255
330   JSR PRTCH
340   LDX #0
350   LDY #2
360   JSR CURXY
370   LDA #SP1/256
380   LDX #SP1&255
390   JSR PRTCH
430   LDA #VSNAME/256
440   LDX #VSNAME&255
450   JSR INPSTR
460   LDX #0
470   LDY #4
480   JSR CURXY
490   LDA #SP2/256
500   LDX #SP2&255
510   JSR PRTCH
520   LDA #VSNAME/256
530   LDX #VSNAME&255
540   JSR PRTCH
550   LDA #SWELC/256
560   LDX #SWELC&255
570   JSR PRTCH
580   RTS 
5000 ; Print Subroutine
5005 PRTCH
5010 ; Store Starting String Address
5015   STX PRSPTR
5020   STA PRSPTR+1
5030   LDA #$00
5040   STA PRINDX
5050 ; Set IOCB Command
5052   LDX #$00
5055   LDA #$0B
5060   STA ICCOM,X
5070 ; Set Max Buffer Length
5072   LDA #$00
5080   STA ICBLL,X
5090   STA ICBLH,X
5100 ; Call CIO to Print
5105 PRCHLP LDY PRINDX
5108   LDA (PRSPTR),Y
5115   BEQ PRCHDN
5116   JSR CIOV
5120   INC PRINDX
5130   BNE PRCHLP
5140 PRCHDN RTS 
6000 ; Position Subroutine
6005 CURXY
6010   STX COLCRS
6020   STY ROWCRS
6030   RTS 
7000 ; Input String
7005 INPSTR
7010 ; Store Starting String Address
7015   STX ICBAL
7020   STA ICBAH
7030 ; Store Str Address in P0 Ptr
7035   STX INSPTR
7040   STA INSPTR+1
7050 ; Set IOCB Command
7052   LDX #$00
7055   LDA #$05 ; Get Record
7060   STA ICCOM,X
7070 ; Set Max Buffer Length
7072   LDA #30
7080   STA ICBLL,X
7085   LDA #0
7090   STA ICBLH,X
7100 ; Call CIO to Input
7110   JSR CIOV
7115 ; Get # chars input
7120   LDY ICBLL
7130 ; Reduce char count by 1
7135   DEY 
7140 ; Store EOS at EOL position
7145   LDA #EOS
7150   STA (INSPTR),Y
7160   RTS 
10000 ; String Data
10001 SCLR .BYTE CLS
10002 SRULE .BYTE "012345678901234567890123456789012345",EOS
10005 SP1 .BYTE "Name? ",EOS
10010 SP2 .BYTE "Hello ",EOS
10020 SWELC .BYTE ", Welcome!",EOS
10050 VSNAME *= *+30
10060 PRINDX *= *+1
15000   .END

Breakdown

• Line 21: Sets the location for the INput String PoinTeR labelled as INSPTR to page zero address $CD.
• Line 7035 to 7040: Store the string address in the pointer location INSPTR.
• Line 7120: Load the Y register with the value in ICBLL.  ICBLL holds the actual number of characters input after the CIO call completes.
• Line 7135: Decrement the Y register.  The number of characters input includes the return key which I don’t want.
• Line 7145: Load the Accumulator with EOS (0).
• Line 7150: Store the value in the Accumulator in the input string address offset by the Y register.  This will be the location of the Return (EOL / $9B).  This basically turns the EOL into EOS.  It uses indirect indexed addressing to get the actual string address stored in the location referenced (a pointer).

Final Assembly

After assembling the program with the changes above, running it produces the correct results.

First input your name, then it prints its back with a welcome message – correctly:

Another success!  Hrm, what to do next…