Mismatch in operand sizes

[bitRAKE]

(Using NASM on Ubuntu 7.10)

I'm messing with a simple linear transposition encryption program. It added 5 to each character in the string excluding the linefeed character. I thought I would jumble it a little more by make it add 1 the the first character 2 to the second and 3 to the third, up to 6 then start at one again.

Here is what I tried.

section .data
	msg:	db 'Message to be encrypted',0xa
	meglen:	equ $-msg

section .text
	global _start

_resetesi:

	mov	esi,1
	ret

_start:

	mov	eax,4	;Write origional String
	mov	ebx,1
	mov	ecx,msg
	mov	edx,meglen
	int	0x80
	
	mov	ebx,meglen	;Put the message in eax and the length in ebx
	dec	ebx;
	mov	eax,msg
	mov	esi,1
	

_eachchar:

	add	byte[eax], esi
	inc	eax
	inc	esi
	dec	ebx

	cmp	esi,6
	je	_resetesi
	cmp	ebx,0
	jne	_eachchar

	mov	eax,4	;Write encrypted string
	mov	ebx,1
	mov	ecx,msg
	mov	edx,meglen
	int	0x80
	
_exit:

	mov	eax,1
	mov	ebx,0
	int	0x80

When I try to compile I get the error:
encrypt.asm:29: error: mismatch in operand sizes

I tried using si instead of esi, but I still got the same error. I apologize if this question could be easily google'd, but I couldn't seem to find an explanation.

Thanks!

[Ancient Dragon]

line 29: byte is a reserved word that means one byte and you are attepting to add a 16-bit register value to it. You will probably have to use a different general register to do it

mov edx,esi
add byte[eax], dl

But when you do that you might as well replace esi with edx everywhere in that loop to avoid repeated copying.

[bitRAKE]

That got rid of the error. I just replaced esi with dl. I'm curious why you said to move esi to edx? Also could you explain a little about dl?

Thanks again

[Ancient Dragon]

>>I'm curious why you said to move esi to edx?
esi is a 32-bit register and can only be copied to another 32-bit register.

>>Also could you explain a little about dl?
Its just an 8-bit register just like al. Only eax, ebx, ecx, edx contain 32-bit, 16-bit and 8-bit registers. And you have to use an 8-bit register to do anything with just one byte of memory.

[bitRAKE]

>>Only eax, ebx, ecx, edx contain 32-bit, 16-bit and 8-bit registers.

OK, so would eax, ax and al all be the same register/memory. Can all three of those contain seperate values?

Thanks for working with me. :)

[Duoas]

> OK, so would eax, ax and al all be the same register/memory.
Yes and no. Read on.

> Can all three of those contain seperate values?
Yes and no. Read on.

AL is the low-order byte of AX. AH is the high-order byte of AX. Thus, AX is a 16-bit value (composed of two eight-bit values).

Likewise, AX is the low-order word of EAX. You can swap the low- and high-order words of EAX using the ROL opcode, say: ROL EAX, 16 .

So, if you set AL to 07h and AH to 8Ah, then it is the same as setting AX to 8A07h.

Hope this helps.

[bitRAKE]

Helped a lot!

So then, with the rotate code, you could do this:
ROL EAX, 8Ah and ROR EAX, 07h

Would EAX contain 07h8Ah?

Or does it not work like that?

[Tight_Coder_Ex]

Assuming EAX is already null

or ah, 7
or al, 8a

would give you that result

ROL SAL SHL and instructions to do bitwise shifts either left in this case or right using ROR RAR SHR.

Knowing the instruction set goes a long way to becoming an effective assembly programmer

[Duoas]

Back when people had to count nanoseconds on older processors it actually made a difference how you used certain registers.

Nowdays, just say MOV AL, 7 or MOV AX, 78Ah --whichever is needed.

LikeTight_Coder_Ex said, make sure you get a good reference. ROL, RCL, etc. are woefully underused opcodes, but very powerful. Look them up.

Follow along:

mov ax, 78Ah  ; AH == 07h, AL == 8Ah; AX == 078Ah
rol ax, 4     ; AH == 78h, AL == A0h; AX == 78A0h
rol ax, 4     ; AH == 8Ah, AL == 07h; AX == 8A07h

Hope this helps.

[bitRAKE]

Thanks a lot for the help guys! Still very confusing. I have only glanced at the Asm References thread, but if you guys have any suggestions for books or Web sites, let me know. I'm really struggling with finding references.

David