Question

Write an assembly language program (using MC6800 instruction set) which will do the

following:

We are going to add two numbers 0x4AC0EA (addend) and 0x661B93 (augend). The three byte

addend is to be stored in locations $0100 through $0102 while the augend is to be stored in

locations $0103 through $0105. The three byte result must be stored in locations $0106

through $0108.

Answer 1

Answer :- The code has been written below-

; store 0x4AC0EA to the memory location $0100

LDA A, #$EA ; Accumulator A = 0xEA

STA A, $0100 ; keep 0xEA at memory address $0100

LDA A, #$C0 ; Accumulator A = 0xC0

STA A, $0101 ; keep 0xC0 at memory address $0101

LDA A, #$4A ; Accumulator A = 0x4A

STA A, $0102 ; keep 0x4A at memory address $0102

; store 0x661B93 to the memory location $0103

LDA A, #$93 ; Accumulator A = 0x93

STA A, $0103 ; keep 0x93 at memory address $0103

LDA A, #$1B ; Accumulator A = 0x1B

STA A, $0104 ; keep 0x1B at memory address $0104

LDA A, #$66 ; Accumulator A = 0x66

STA A, $0105 ; keep 0x66 at memory address $0105

; add the two numbers

LDA A, $0100 ; get the first addend byte in A

LDA B, $0103 ; get the first augend byte in B

ABA ; A = A + B

STA A, $0106 ; keep the added value at memory address 0x0106

LDA A, $0101 ; get the 2nd addend byte in A

LDA B, $0104 ; get the 2nd augend byte in B

ADC A, #0 ; Add the prevoius carry first (if there was any carry)

ABA ; A = A + B

STA A, $0107 ; keep the added value to location 0x0107

LDA A, $0102 ; get the 3rd addend byte in A

LDA B, $0105 ; get the 3rd augend byte in B

ADC A, #0 ; Add the prevoius carry first (if there was any carry)

ABA ; A = A + B

STA A, $0108 ; keep the added value to location 0x0107