Question

List the SHORTEST POSSIBLE CODE (counting number of instructions) for making x, y, and z, defined as follows, get the value 1.
for an 80*86 machine

x:   dw   0xff00
y:   resb 1
z:   resw 1

edit:
I think the answer should be somethink like that:

MOV  DWORD [x+1], 0x01010001  ; 

;check:

mov eax , 0
mov al , byte[y]
print_d eax ; print 0

mov eax , 0
mov ax , word[x]
print_d eax ; print 256

mov eax , 0
mov ax , word[z]
print_d eax ; print 257 

but. it is not good…sholud print 1

Answer 1

Here’s the memory where your x, y and z are, listed as bytes (from lower addresses (x) to higher (z)):

xx XX yy zz ZZ

where xx is the least significant byte of x (0), XX is the most significant byte of x (0xFF) and likewise for y and z.

If I understand it correctly, y and z aren’t initialized (res* hints NASM syntax for memory reservation keywords).

So you want to transform this:

00 FF yy zz ZZ

into this:

01 00 01 01 00

Right?

MOV DWORD [x+1], 0x01010001 will transform it into:

00 01 00 01 01

So, it’s not correct. And you need more than 1 instruction to change 5 bytes because 32-bit instructions write at most 4 bytes at a time.

I’d say the shortest in terms of the number of instructions is 2 MOVs (NASM syntax):

mov dword [x], 0x01010001
mov byte [x+4], 0