Take a look at the program shown in Fig 412 replicated below

Take a look at the program shown in Fig. 4.12 replicated below.; this program evaluates: - (x + y - 2z + 1) .586 .MODEL FLAT .STACK 4096 .DATA x DWORD 80 y DWORD 40 z DWORD 26 .CODE main PROC mov eax, X; result:= x add eax, y; result:= x + y mov ebx, z; temp:= z add ebx, ebx; temp:= 2*z sub eax, ebx; result:= x + y - 2z inc eax; result:= x + y - 2*z + 1 neg eax; result:= - (x + y - 2*z + 1) mov eax, 0; exit with return code 0 ret main ENDP END Change the code in Fig 4.12 to perform: ((x-y)/y - 3z) - 1 As the code shows, the result is given in eax. Verity your code using ProjOne in debug mode.

Solution

; this program: ((x-y)/y - 3z) - l
.586
.MODEL FLAT
.STACK 4096

.DATA
x DWORD 80
y DWORD 40
z DWORD 26

.CODE
main PROC
mov eax,x ;result = x
sub eax,y ;result = x-y
div eax,y ;division (x-y)/y
mflo ecx ; quotient of the division
mov ebx,z ;temp = z
mult ebx,3 ; result : 3*z
mflo edx ; lower 32 bits of multiplication
sub ecx,edx ; result: (((x-y)/y)-3z)
sub ecx,1 ; result : (((x-y)/y)-3z)-1
mov ecx,0; exit with return code 0
ret
main ENDP
END