If you have no clue what I am talking about, check out my previous post. I have fixed the flaws, and now it will parse the commands it is handed on the command line, and stick them into an integer. It will now print it to stdout.

This has been an awesome learning experience for me. Especially with regards to how to do recursion, how to debug a pure assembly program with gdb and whatnot.

Here comes the code, as I said before, see my previous thread if you don't know what I am talking about, as well as instructions on how to compile the program.

; File: addnumbers2.asm
; Bert JW Regeer
; 
; 2008-01-27
; 
; Function:
;	Add numbers together that are provided as arguments to the program in argv[1] and argv[2].
;
; Known limitations:
;	As of right now, the numbers that are provided may not add up to anything more than 9.
;	This will hopefully be fixed in the next revision. Floating point numbers will not work.
;	Any input that is larger than an integer will cause overflows, and thus will not work.
 
section .data

	; Define some strings that are going to be used throughout the program

	; This string is to let the user know they failed to provide the proper amount of arguments.
args	db	"Program addnumbers: ", 0xa, 0x9, "addnumbers  ", 0xa, 0x9, "Arguments 1 and 2 are required.", 0xa, 0x9, "Anything that will cause addition to overflow an int (2,147,483,647), will fail! :P", 0xa
largs	equ	$ - args

	; This string contains part of the output that we are going to send to the terminal. The last two
	; bytes will be filled automatically by the program, before it is output to stdout.
msg	db	'Answer: ',	0
lmsg	equ	$ - msg

num1	dd	0
num2	dd	0

section .bss
	; This is where I am going to store the output of my conversion from an integer to a char
answer	resb	64

section .text
global start				; Linker defined entry point. Mac OS X this is start. 
global _start				; FreeBSD and others _start.


_start:
start:
	push	ebp			; 
	mov	ebp, esp 		; Set up the stack frame

	mov	ecx, [ebp + 4]		; Get argc, we check if it set to at least 3
	mov	edx, ebp		; Put the base pointer into edx, so we can use that in 
					; our dereferences coming up
	add	edx, 8			; Add 8. We want to skip ebp and argc

	cmp	ecx, 3			; Check if we have at least 3 arguments to the program. 
					; At least two arguments are required, and the 3rd one is 
					; the name of the program
	jl	exit			; If the value in ecx is less than 3, jump to exit

	mov	esi, 1			; Set the index to 1
	
	mov	eax, [edx + esi * 4]	; Move the pointer to the character array into eax
	push	eax			; Push eax onto the stack
	push	num1			; Push the pointer to num1 onto the stack
	call	ctoi			; Call my char to int function
	add	esp, byte 8		; Put the stack pointer back to where it was.

	inc	esi			; Increase the index
	
	mov	eax, [edx + esi * 4]	; Move the pointer to the character array into eax
	push	eax			; Push eax onto the stack
	push	num2			; Push the pointer to num2 onto the stack
	call	ctoi			; Call my char to int function
	add	esp, byte 8		; Put the stack pointer back to where it was.

	mov	eax, [num1]		; Move value stored in num1 into eax
	add	eax, [num2]		; Add num2 to eax, this will now be stored in eax

	push	eax			; Push the new calculated number onto the stack
	call	itoa			; Convert the integer to a character array

	push	dword lmsg		; Push the length of the string
	push	msg			; Push the location of the string in memory
	push	dword 0x1		; Push the file descriptor to write to
	mov	eax,4			; Move the syscall number into eax
	push	eax			; Push the syscall onto the stack
	int	0x80			; Interrupt 80, go to kernel
	add	esp, byte 16		; Clean up the stack
 
	push	answer			; Push answer onto the stack
	call	len			; Get it's length

	push	edi			; Push the length onto the stack
	push	answer			; Push the pointer to the character string onto the stack
	push	dword 0x1		; Push the file descriptor to write to
	mov	eax,4			; Push the syscall number into eax
	push	eax			; Push the syscall onto the stack
	int	0x80			; Interrupt 80, go to kernel
	add	esp, byte 16		; Clean up the stack

	jmp	done			; Program is done. Jump to done

exit:
	; This label is jumped to when we want to exit the program and let the user know how
	; to run the program. Like for instance what paramaters to send the program.
					; Call sys_write
	push	dword largs		; Push the length of the string
	push	dword args		; Push the location of the string in memory
	push	dword 0x1		; Push the file descriptor to write to
	mov	eax,4			; Move the syscall number into eax
	push	eax			; Push the syscall onto the stack
	int	0x80			; Interrupt 80, go to kernel
	add	esp, byte 16		; Clean up the stack

done:
	; This is the label we jump to when we want to exit the program, we set the exit code
	; to 0.
					; Call sys_exit
	push	dword 0x0		; Push the value to return to the operating system
	mov	eax,1			; Move the syscall number into eax
	push	eax			; Push the syscall onto the stack
	int	0x80			; Interrupt 80, go to kernel

	; We never return to this function, so no need to clean the stack. :P

ctoi:
	; char to i. We actually convert entire character array's to integers.
	;
	; We get two paramaters on the stack. The first one we grab is the pointer to the place to store
	; the number. The second is the pointer to the character array.

	push	ebp			; Push the old base pointer onto the stack
	mov	ebp, esp		; Create a new base pointer
	push	esi			; Store all the original registers
	push	eax
	push	ebx
	push	ecx	
	push	edx			; Push edx, so that we can overwrite it
	sub	esp, 4			; We get another storage space on the stack
	mov	[esp], dword 10		; This is the number we are going to multiply by

	mov	eax, [ebp + 12]		; Move the pointer to the character array into eax
	push	eax			; Push the pointer to the character array onto the stack
	call	len			; Call the string length versoin
	add	esp, byte 4		; Reclaim the space we lost when we pushed eax onto the stack
	
	mov	ebx, [ebp + 8]		; This is where we are going to store the numbers
	mov	esi, [ebp + 12]		; This is the pointer to the character array
	movzx	ecx, di			; move with extended zero edi.
	mov	edi, 0			; Clean up edi

	ctoi_loop:
	mov	eax, [ebx]		; Move the value stored in ebx into eax
	mul	dword [esp]		; Move it over a 10s place.
	mov	[ebx], eax		; Move the new number back into ebx

	movzx	eax, byte [esi + edi]	; Move the character into eax
	movsx	eax, al			; We just want the lower part of the character

	sub	eax, 0x30		; Subtract 0x30, ASCII 0 so that it is an actual number
	add	[ebx], eax		; Add the new number to the old number that has been multiplied by 10
	inc	edi			; Increase the counter
	loop ctoi_loop			; Loop into cx is 0

	add	esp, byte 4
	pop	edx			; Restore all the registers
	pop	ecx
	pop	ebx
	pop	eax
	pop	esi
	mov	esp, ebp		; Make esp the original base pointer again
	pop	ebp			; Pop the original base pointer into the register
	ret				; Return caller

itoa:
	; Recursive function. This is going to convert the integer to the character.
	push	ebp			; Setup a new stack frame
	mov	ebp, esp
	push	eax			; Save the registers
	push	ebx
	push	ecx
	push	edx

	mov	eax, [ebp + 8]		; eax is going to contain the integer
	mov	ebx, dword 10		; This is our "stop" value as well as our value to divide with
	mov	ecx, answer		; Put a pointer to answer into ecx
	push	ebx			; Push ebx on the field for our "stop" value

	itoa_loop:
	cmp	eax, ebx		; Compare eax, and ebx
	jl	itoa_unroll		; Jump if eax is less than ebx (which is 10)
	xor	edx, edx		; Clear edx
	div	ebx			; Divide by ebx (10)
	push	edx			; Push the remainder onto the stack
	jmp	itoa_loop		; Jump back to the top of the loop
	itoa_unroll:			
	add	al, 0x30		; Add 0x30 to the bottom part of eax to make it an ASCII char
	mov	[ecx], byte al		; Move the ASCII char into the memory references by ecx
	inc	ecx			; Increment ecx
	pop	eax			; Pop the next variable from the stack
	cmp	eax, ebx		; Compare if eax is ebx
	jne	itoa_unroll		; If they are not equal, we jump back to the unroll loop
					; else we are done, and we execute the next few commands
	mov	[ecx], byte 0xa		; Add a newline character to the end of the character array
	inc	ecx			; Increment ecx
	mov	[ecx], byte 0		; Add a null byte to ecx, so that when we pass it to our
					; len function it will properly give us a length

	pop	edx			; Restore registers
	pop	ecx
	pop	ebx
	pop	eax
	mov	esp, ebp		
	pop	ebp
	ret

len:
	; Returns the length of a string. The string has to be null terminated. Otherwise this function
	; will fail miserably. 
	; Upon return. edi will contain the length of the string.

	push	ebp			; Save the previous stack pointer. We restore it on return
	mov	ebp, esp		; We setup a new stack frame
	push	eax			; Save registers we are going to use. edi returns the length of the string
	push 	ecx
	
	mov	ecx,  [ebp + 8]		; Move the pointer to eax; we want an offset of one, to jump over the return address

	mov	edi, 0			; Set the counter to 0. We are going to increment this each loop

	len_loop:			; Just a quick label to jump to
	movzx	eax, byte [ecx + edi]	; Move the character to eax.
	movsx	eax, al			; Move al to eax. al is part of eax.
	inc	di			; Increase di.
	cmp	eax, 0			; Compare eax to 0.
	jnz 	len_loop		; If it is not zero, we jump back to len_loop and repeat.

	dec	di			; Remove one from the count

	pop	ecx			; Restore registers
	pop	eax
	mov	esp, ebp		; Set esp back to what ebp used to be.
	pop	ebp			; Restore the stack frame
	ret				; Return to caller