Modify the array sum.asm file to find the largest number in the array. $$ You do this by using the cmp arg$1,$ arg$2$ command. This command compares the two values by subtracting them $($but it does not actually change anything$).$ Instead, the results are stored in the flags register for use by a jmp command. You would use the jump command to store the larger number in a variable with each iteration of the loop. There is an example of how to use the jump commands to do this is below. $$ Hint: You may need more registers to do this. How to use jump commands for conditions:

cmp A$,$ B

; you will not use all these commands and you must use them in the order you choose

jg greater$\_$ ; A $>$ B

jle less$\_$ ; A $<=$ B

je equal$\_$ ; A $=$ B

jge greater$\_$eq$\_$ ; A $>=$ B

jmp label$\_$ ; unconditional

THIS IS THE SUM.ASM FILE

section $.$data

array dd $1,2,3,4,5$ ; Array of $532-$bit integers

len equ $5$ ; Array length

output db "Sum is: $",0$ ; Output prefix

section $.$bss

result resb $12$ ; Reserve space for the result string $($max $10$ digits $+$ newline $+$ null terminator$)$

section $.$text

global $\_$start ; Entry point for the program

$\_$start:

; Initialize variables

mov ecx, len ; Loop counter

mov eax, $0$ ; Sum accumulator

mov esi, $0$ ; Array index

sum$\_$loop:

; Add current element to sum

add eax, $[$array $+$ esi $*4]$ ; $4$ bytes per element

; Move to next element

inc esi

; Check loop condition

dec ecx

jnz sum$\_$loop ; Continue if not zero

; At this point, EAX contains the sum

; Convert EAX $($the sum$)$ to a string and store in 'result'

mov edi, result $+11$ ; Start filling result from the end

mov byte $[$edi$],0$ ; Null terminator

dec edi

; If the sum is zero, handle it as a special case

cmp eax, $0$

jne convert$\_$to$\_$string

mov byte $[$edi$],\text{'}0\text{'}$

jmp print$\_$result

convert$\_$to$\_$string:

; Convert integer in EAX to a decimal ASCII string

$.$convert$\_$loop:

mov edx, $0$

mov ebx, $10$

div ebx ; EAX $=$ EAX $/10,$ EDX $=$ EAX $\%10$

add dl$,\text{'}0\text{'}$ ; Convert remainder to ASCII

mov $[$edi$],$ dl ; Store ASCII character

dec edi ; Move backwards in result

test eax, eax ; Check if EAX is zero

jnz $.$convert$\_$loop ; Continue if not zero

; Set EDI to the start of the converted number

inc edi

print$\_$result:

; Print "Sum is: $"$ followed by the converted number

; First, print the "Sum is: $"$ prefix

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

mov ebx, $1$ ; file descriptor $($stdout$)$

mov ecx, output ; pointer to message

mov edx, $8$ ; message length

int $0$x$80$ ; call kernel

; Calculate length of the result number string

mov eax, result $+11$ ; End of buffer

sub eax, edi ; Length of converted number

mov edx, eax ; Move length to EDX for sys$\_$write

; Now, print the result number string

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

mov ebx, $1$ ; file descriptor $($stdout$)$

mov ecx, edi ; pointer to converted number

int $0$x$80$ ; call kernel

; Exit the program

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

xor ebx, ebx ; exit code $0$

int $0$x$80$ ; call kernel