Given a byte string S of length l, obtain the string D of length l-1 as D(i) = S(i) * S(i+1)

Question

Given a byte string S of length l, obtain the string D of length l-1 as D(i) = S(i) * S(i+1) (each element of D is the product of two consecutive elements of S).

here is my code.

bits 32

global start        

extern exit
import exit msvcrt.dll  

segment data use32 class=data
    s db 1, -3, 4, 7 ; declaring the string of bytes s
    l equ $ - s - 1 ; compute the length of the string l - 1
    d times l dw 0 ; reserve l bytes for the destination string and initialize it
segment code use32 class=code
    start:
        mov ecx, l
        
        mov esi, 0
        
        jecxz end_prog
        repeat_this:
            mov al, [s + esi] ; in al - the element on the esi position
            mov dh, [s + esi + 1] ; in dh - the element after the esi position
            imul dh ; multiplying consecutive terms
            mov [d + esi], ax; put the result in d
            inc esi ; incrementing esi to go to the next position
        loop repeat_this  
    
        end_prog:
        
        push dword 0
        call [exit]

my question is did I make a mistake by declaring d times l dw 0 since I know that a byte * byte = word? Is everything else okay or should I pay attention to some details that I'm missing?

Answer 1

Inconsistent addressing of d

d is an array of 2-byte elements (d times l dw 0) and mov [d + esi], ax indexes into it with esi as the offset. However, esi is increased by 1 every iteration, so the resulting access pattern is that the stores keep stepping half on top of the previous store, shown by such a diagram: (storing -3, -12 and 28 in sequence)

FD FF
   F4 FF
      1C 00

This is possible to do, but probably unintentional.

Normally my recommendation would be to increase the register by 2 in the loop, but esi is also used to index s so that won't work. You can change the store to double esi though, for example:

mov [d + esi*2], ax

Or,

mov [d + esi + esi], ax

If you change d to be an array of bytes then the store should be mov [d + esi], al, with the original address but storing only 1 byte. Storing overlapping words would put the right data in d, but it's strange and stores a stray byte just after the end of d.

Calling exit

exit performs C library termination procedures, but the C library runtime environment wasn't used or even initialized (this code is in start, not main), so it's not necessary. It makes more sense to call ExitProcess (kernel32.dll), or just return (that does not work on Linux, but this is a Windows program and there it does work).

The loop instruction

loop is not great, it has some niche uses but typically it should be avoided. jecxz by contrast is OK, just unusual.

Minor things

The best way to zero esi isn't mov esi, 0, it's xor esi, esi, unless the flags must not be affected.

The two loads of the current and next element of S could be merged into a single word load, if you want. For example mov ax, [s + esi] followed by imul ah.