Reversing elements of an array assembly

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Hi,

I am trying to reverse the element of an array in assembly, here is my code:

C#

.data
	
	array DWORD 1, 5, 6, 8, 0Ah, 1Bh, 1Eh, 22h, 2Ah, 32h  ;array to be reversed

.code
main proc
       mov esi, 0
	mov edi, 0
	mov eax, 0
	mov ebx, 0

	mov esi, OFFSET array							;move first element address to esi
	mov edi, OFFSET array + SIZEOF array - TYPE array  ;move last element address to edi
	
	mov ecx, LENGTHOF array / 2     ;sets the counter in the reverseLoop

reverseLoop:
	mov eax, [esi]    ;move the element in esi to eax
	mov ebx, [edi]	  ;move the element in edi to ebx

	xchg eax, ebx     ;exchange the two elements

	mov [esi], eax  ;move the element in eax, to the address in esi
	mov [edi], ebx  ;move the element in ebx, to the address in edi

	add esi, TYPE array ;increase esi to take the next element in the array (from the left)
	sub edi, TYPE array ;decrease edi to take the next element in the array (from the right)

	loop reverseLoop

	invoke ExitProcess,0
main endp
end main</pre>

This code works so far with DWORD array, however, if i change the size of the array to BYTE i get a problem. In the first line in reverseLoop, "mov eax, [esi]", the eax register does not get the first element in the array as it used to when the size of the array was DWORD. Instead it seems to get some random number. Any ideas why it is doing that?

What I have tried:

I could not really figure out what the problem is.

Posted 11-Jun-16 10:50am

Member 12569032

Updated 24-Nov-17 16:24pm

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PIEBALDconsult 11-Jun-16 17:57pm

You should not have deleted the original version of this question; you should have used "Improve question".

Member 12569032 11-Jun-16 18:05pm

I did, but I accidently deleted the question and then tried to improve, but then saw that it was deleted.

3 solutions

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Patrice T · Answer 1 · 2016-06-11T13:08:00

First of all, you can simplify your code

reverseLoop:
	mov eax, [esi]    ;move the element in esi to eax
	mov ebx, [edi]	  ;move the element in edi to ebx
 
	xchg eax, ebx     ;exchange the two elements
 
	mov [esi], eax  ;move the element in eax, to the address in esi
	mov [edi], ebx  ;move the element in ebx, to the address in edi
 
	add esi, TYPE array ;increase esi to take the next element in the array (from the left)
	sub edi, TYPE array ;decrease edi to take the next element in the array (from the right)
 
	loop reverseLoop

by

reverseLoop:
	mov eax, [esi]    ;move the element in esi to eax
	mov ebx, [edi]	  ;move the element in edi to ebx
 
	xchg eax, ebx     ;exchange the two elements
 
	mov [esi], ebx  ;move the element in ebx, to the address in esi
	mov [edi], eax  ;move the element in eax, to the address in edi
 
	add esi, TYPE array ;increase esi to take the next element in the array (from the left)
	sub edi, TYPE array ;decrease edi to take the next element in the array (from the right)
 
	loop reverseLoop

Second, why don't you use byte sized registers to move bytes ?
Like AL or AH and BL or BH ?
I know, it is old fashioned, but it look the most straight forward to me.

[Update]
To handle all possibilities as fast as possible, you need some code like:

C++

if (TYPE is DWORD)
    optimized code for DWORD (mostly your sample code)
else if (TYPE is WORD)
    optimized code for WORD
else if (TYPE is BYTE)
    optimized code for BYTE
end if

To handle all possibilities with a single piece of code, you must operate at smallest possible data size. It is possible but will be slow.

array DWORD 1,2,3,4,5,6,7,8,9,10,11,12

will give

array DWORD 9,10,11,12,5,6,7,8,1,2,3,4

array WORD 1,2,3,4,5,6,7,8,9,10,11,12

will give

array WORD 11,12,9,10,7,8,5,6,3,4,1,2

array BYTE 1,2,3,4,5,6,7,8,9,10,11,12

will give

array BYTE 12,11,10,9,8,7,6,5,4,3,2,1

in your main loop, you need another loop to handle the different data sizes.

reverseLoop:
    for dx=1 to (TYPE array)
        mov ah, [esi]    ;move the element in esi to ah
        mov bh, [edi]	  ;move the element in edi to bh
 
        mov [esi], bh  ;move the element in bh, to the address in esi
        mov [edi], ah  ;move the element in ah, to the address in edi
        add esi, 1
        add esi, 1
    next
    sub edi, TYPE array
    sub edi, TYPE array ;decrease edi to take the next element in the array (from the right)
 
    loop reverseLoop

Pseudo code, I let you deal with details

OriginalGriff · Answer 2 · 2016-06-11T11:02:00

mov eax, [esi]loads eax with a DWORD value, not a byte.
And that's a real problem, because byte pointers aren't limited to DWORD addresses.
DWORDs are always on an address which is a multiple of 4, byte addresses are a multiple of one. So when you pass a byte address to a DWORD instruction, the value loaded is not a byte, and it's not four bytes starting at that address. It's four bytes starting from the byte address rounded down to the previous multiple of four.
So it looks like random data, but it isn't!
Use byte access with byte data.

Member 13540174 · Answer 3 · 2017-11-24T16:24:00

mov esi, 0
mov edi, 0
mov eax, 0
mov ebx, 0

mov esi, OFFSET array ;move first element address to esi
mov edi, OFFSET array ;move last element address to edi

mov ecx, LENGTHOF array ;sets the counter in the reverseLoop

reverseLoop:
mov eax, [esi] ;move the element in esi to eax
mov ebx, [edi] ;move the element in edi to ebx
add esi, TYPE array ;increase esi to take the next element in the array (from the left)
sub edi, TYPE array ;decrease edi to take the next element in the array (from the right)
call writehex
call crlf
loop reverseLoop

Reversing elements of an array assembly

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