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 and argv. ; ; Known limitations: ; 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 <number 1> <number 2>", 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