SPO600 Lab4 String lab Option 1: Adding Calculator

Hello, welcome to lab 4 for the SPO600(Software Portability and Optimization).This is a lab for the week 4 materials. My team chose to work on option 1 and option 4.

Introduction

This blog is for option 1, we created an adding calculator using 6502 assembly language. I will create a subroutine for the user to insert 2 numbers and add them together.

Requirements

1. Create a subroutine which enables the user to enter two numbers of up to two digits. Indicate where the cursor is, and allow the user to use the digit keys (0-9), backspace, and enter keys. Return the user's input value in the accumulator (A) register.
2. Using this subroutine, write a program which add the two numbers (each of which is in the range 0-99) and print the result.

Coding

define  SCINIT  $ff81 ; initialize/clear screen
define  CHRIN  $ffcf ; input character from keyboard
define  CHROUT  $ffd2 ; output character to screen
define  SCREEN  $ffed ; get screen size
define  PLOT    $fff0 ; get/set cursor coordinates

define RIGHT  $81
define LEFT  $83
define ENTER  $0d
define BACKSPACE $08

define NUM1    $15;
define  NUM2    $16;

jsr SCINIT

ldy #$00
jsr firstNumPrint ; ask for input for first number
jsr getNum; get the first number
jsr storeFirstNum  ; then store the first number
ldy #$00
jsr secondNumPrint  ; ask for input for second number
jsr getNum; get the second number
jsr storeSecondNum  ; store the second number
ldy #$00
jsr resultPrintString  ; print a string 'Result'
jsr printResult ; print the result
jmp mainLoop  ; go back to the first step

getNum:
     sec
     jsr PLOT
     ldx #$15
     clc
     jsr PLOT

getNumLoop:
     sec
     jsr PLOT
     jsr CHRIN


charCheck: 
     cmp #BACKSPACE ; if user enter backspace, it erase the #$15 digit
     beq move_back

     cmp #RIGHT ; if user enter right arrow, it goes to the first digit
     beq move_right

     cmp #LEFT ; if user enter left arrow, it goes to the second digit
     beq move_left

     cmp #ENTER ; if user enter enter, it goes to the next process
     beq move

printNum:
     cmp #$30
     bcc getNumLoop

     clc
     cmp #$3a
     bcs getNumLoop

     jsr CHROUT

     sec
     jsr PLOT
     cpx #$17
     bne getNumLoop
     dex
     clc
     jsr PLOT
     jmp getNumLoop

move_back:
 cpx #$15
 beq getNumLoop
 jsr CHROUT
 jmp getNumLoop

move_left: 
     cpx #$15 ; first digit
     beq getNumLoop
     jsr CHROUT
     jmp getNumLoop

move_right: 
 cpx #$16 ; second digit
     beq getNumLoop
     jsr CHROUT
     jmp getNumLoop

move:
     sec
     jsr PLOT
     ldx #$15 ; first digit
     clc
     jsr PLOT
     sec
     jsr PLOT

     clc
     sbc #$2F ; to calculate it, it should be subtracted by #$2f

     asl
     asl
     asl
     asl

     pha

     ldx #$16
     clc
     jsr PLOT
     sec
     jsr PLOT

     clc
     sbc #$2F ; to calculate it, it should be subtracted by #$2f
     pha

     ldx #$00
     iny
     clc
     jsr PLOT
     sec
     jsr PLOT

     pla
     tax
     pla

     rts

storeFirstNum:
     sta NUM1
     txa
     eor NUM1
     sta NUM1
     rts

storeSecondNum:
     sta NUM2
     txa
     eor NUM2
     sta NUM2
     rts

printResult:
     sec
     jsr PLOT
     ldx #$15
     clc
     jsr PLOT
     sec
     jsr PLOT

     sed
     lda NUM1
     adc NUM2
     cld
     pha

     bcc outputAddition
     ldx #$14
     clc
     jsr PLOT
     sec
     jsr PLOT
     lda #$31
     jsr CHROUT

outputAddition:
     lsr
     lsr
     lsr
     lsr
     clc
     adc #$30 ; as the received number does not fit for ASCII, it needs to add 
     jsr CHROUT

     pla
     and #$0F
     clc
     adc #$30 ; as the received number does not fit for ASCII, it needs to add 
     jsr CHROUT

     sec
     jsr PLOT
     ldx #$00
     iny
     clc
     jsr PLOT

     rts

firstNumPrint:
     lda firstNum,y
     beq goback_main
     jsr CHROUT
     iny
     bne firstNumPrint

secondNumPrint:
 lda secondNum,y
        beq goback_main
        jsr CHROUT
        iny
        bne secondNumPrint

resultPrintString:
 lda result,y
        beq goback_main
        jsr CHROUT
        iny
        bne resultPrintString

goback_main:
     rts

firstNum:
dcb "E","N","T","E","R",32,"F","I","R","S","T",32,"N","U","M","B","E","R",":",32,32,"0","0"
dcb 00


secondNum:
dcb "E","N","T","E","R",32,"S","E","C","O","N","D",32,"N","U","M","B","E","R",":",32,"0","0"
dcb 00

result:
dcb "R","E","S","U","L","T",":"
dcb 00

This is the results:

Thoughts & reflection

The assembly language program is getting harder than before, even harder than the third lab. In this lab, I spent a lot of time on how to get the user input, store the two numbers and the results, and display the results on the screen. As I said in each blog about the 6502 assembly language, It is a very low level programming language and it frequently accesses the memory and registers. The operation that we designed for this program is quite easy in any higher level language. But in assembler language, I'm still not familiar with the syntax. I hope more and more practices would help me to solve the problem easily.