Emu8086 copying values from an array to another

Question

For my coursework, I have to write a program that copies all the values lower than 1000h from an array of signed 16-bit numbers (called inputArray) to another array (called outputArray). In this exercise, I'm required to use the string instructions lods and stos.

And the only problem I have as an example is the following:

org 100h
jmp init
value equ 8000h
numbers dw 8933h, 1240h, 0328h, 99A0h, 0F422h, 0101h
init: lea SI, numbers
mov CX, (init-numbers)/2
mov DX, 0h
cld
compare: lodsw
cmp AX, value
ja insert
nextElem: loop compare
lea DI, numbers
mov CX, DX
extract: pop AX
stosw
loop extract
end: int 20h
insert: push AX
inc DX
jmp nextElem

Answer 1

For my coursework, I have to write a program that copies all the values lower than 1000h from an array of signed 16-bit numbers (called inputArray) to another array (called outputArray). In this exercise, I'm required to use the string instructions lods and stos.

Let's look at all the key elements of your task in relation to the program code that you received/found (I think).

.lower and signed
Because the numbers involved are signed numbers, you need to use the appropriate conditional instructions. These include jl (JumpIfLess) and jg (JumpIfGreater). The example program used ja (JumpIfAbove) which suits working with unsigned numbers!

.1000h
Simply change the line value equ 8000h.

.another array
The example program stores the results on top of the original numbers. For your task you need to define a 2nd array. This additional array must be able to accomodate at most the same number of elements as the input array did:

inputArray    dw 8933h, 1240h, 0328h, 99A0h, 0F422h, 0101h
outputArray   dw 0, 0, 0, 0, 0, 0

Rather than writing it like above you can use the dup operator:

inputArray    dw 8933h, 1240h, 0328h, 99A0h, 0F422h, 0101h
outputArray   dw 6 dup (0)

.lods and stos
The program you got already uses these instructions.

---

Let's write a better example program (keeping that silly detour via the stack).

.If you put the data section below the code section, you don't need that jmp init anymore.
.Make the labels stand out by indenting the code.
.Have the operands aligned as well.
.Avoid jumping around. Don't ja insert but rather jna nextElem and put the insertion instructions close by.
.Avoid the slow loop instruction. dec cx jnz ... replace it nicely. You can even use another counting register altogether (*).

org 100h
value equ 8000h

  lea  SI, numbers
  mov  CX, (init-numbers)/2
  mov  DX, 0h
  cld
compare:
  lodsw
  cmp  AX, value
  jna  nextElem
  push AX         ;insert
  inc  DX         ;insert
nextElem:
  dec  cx
  jnz  compare
  lea  DI, numbers
                           ;No need to setup CX, just use DX (*)
extract:
  pop  AX
  stosw
  dec  dx                  ;(*)
  jnz  extract
end:
  int 20h

numbers dw 8933h, 1240h, 0328h, 99A0h, 0F422h, 0101h

---

How this should be written.

org 100h

  mov  si, offset inputArray
  mov  di, offset outputArray
  cld
Again:
  lodsw
  cmp  ax, 1000h
  jnl  IsNotLess
  stosw
IsNotLess:
  cmp  si, offset outputArray  ;Arrays must be adjacent for this to work!
  jb   Again

  int  20h

inputArray    dw 8933h, 1240h, 0328h, 99A0h, 0F422h, 0101h
outputArray   dw 6 dup (0)

The output array will have 5 elements:

1. The negative number 8933h
2. The positive but smaller than 1000h number 0328h
3. The negative number 99A0h
4. The negative number 0F422h
5. The positive but smaller than 1000h number 0101h