Computer science

THIS IS Lab #$7$template FOR the assignment

; Lab #$7$template $(2024$Matthew W$.$Phillips$)$

section $.$data

name db 'Student Name Here', $0$xA ; TODO: Add your name inside the single quotes

len equ $ $-$name ; $($don$\text{'}$t delete$)$

section $.$bss

x resw $1$; Used for atoi $($don$\text{'}$t delete$)$

temp resb $1$; used for atoi $($don$\text{'}$t delete$)$

s$3$resb $3$; used for itoa $($don$\text{'}$t delete$)$

; TODO: Add uninitialized data as needed

section $.$text

global $\_$start

$\_$start:

; TODO: Call print$\_$name

; TODO: Get input buffer from keyboard $(100-994)$; Call input

; TODO: Convert input buffer into integer value; Call atoi$\_3$

; TODO: Add $5$to the value in integer value; Call add$\_5$

; TODO: Convert value into ASCII string for printing; Call itoa$\_3$

; TODO: print the buffer...; Call print

; TODO: Exit program; Call exit

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; DON'T CHANGE ANYTHING BELOW THIS

LINE! ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

add$\_5$:

add eax, $5$; Add $5$to the value

ret ; return to caller

print$\_$name:

mov ecx, name ; set output buffer

mov edx, len ; set length

call print ; call subroutine

ret ; return to caller

print:

mov eax, $4$; sys$\_$write

mov ebx, $1$; stdout

int $0$x$80$; system call

ret ; return to caller

input:

mov eax, $3$; sys$\_$read

mov ebx, $0$; stdin

int $0$x$80$; system call

ret ; return to caller

atoi$\_3$:

mov ecx, eax ; transfer contents into a register we aren't otherwise using

$($ecx$)$

xor eax, eax ; clear out previous bits

mov al$,$cl ; move $1$st byte of input to al register

sub al$,\text{'}0\text{'}$; convert ASCII to integer digit

mov bx$,100$; We are going to multiply by $100$

mul bx ; ax $=$ax$($al$)*$bx

mov $[$x$],$ax ; Store first part of integer into memory

xor eax, eax ; clear out previous bits

mov al$,$ch ; move $2$nd byte of input to al register

sub al$,\text{'}0\text{'}$; convert ASCII to integer digit

mov bx$,10$; Multiply by $10$for $10\text{'}$s column

mul bx ; ax $=$ax$($al$)*$bx

mov $[$temp$],$al ; Place al into temporary storage...

xor eax, eax ; clear out previous bits

mov eax, $[$x$]$; Place old value into eax to build on$...$

add eax, $[$temp$]$; Add temporary value to this...

mov $[$x$],$eax ; Place back into $"$x$"$

xor eax, eax ; clear out previous bits

shr ecx, $16$; shift $2-$bytes to the right to reveal last two$-$bytes of

register...

mov al$,$cl ; move $3$rd byte of input to al register

sub al$,\text{'}0\text{'}$; convert ASCII to integer digit

add eax, $[$x$]$; add x to this value...

ret ; return to caller

itoa$\_3$:

xor dx$,$dx ; Clear dx register prior to div operation $($always$!)$

mov bx$,100$; move divisor into bx for divisor

div bx ; ax $=$ax $/$bx

add ax$,\text{'}0\text{'}$; convert digit to ASCII

mov $[$s$3],$byte ax ; move byte to memory

mov ax$,$dx ; add remainder to ax$...$

xor dx$,$dx ; make sure to clear out the modulo register prior to repeat

use of div!

mov bx$,10$; set divisor as $10$now$...$

div bx ; ax $=$ax $/$bx

add ax$,\text{'}0\text{'}$; convert to ASCII

mov $[$s$3+1],$byte ax ; move byte to memory

add dx$,\text{'}0\text{'}$; convert to ASCII

mov $[$s$3+2],$byte dx ; move byte to memory

xor eax, eax ; clear out the reg.

mov eax, $[$s$3]$; transfer for return value...

ret ; return to caller

exit:

mov eax,$1$; sys$\_$exit

mov ebx,$0$; exit is normal

int $0$x$80$; system call