I have the two words stored in variables just called first and second. How do I compare the characters of each string to see which one is larger?
section .data
greet: db "Type a word: "
greetL: equ $-greet
section .bss
first: resb 20
second: resb 20
section .text
global _start
_start:
nop
mov eax, 4
mov ebx, 1
mov ecx, greet
mov edx, greetL
int 80H
mov eax, 3
mov ebx, 0
mov ecx, first
mov edx, 20
int 80H
then I just do the same thing to get the second word
I dunno, cmp I suppose...
xor ecx, ecx ; index
top
mov al, [first + ecx]
cmp [second + ecx], al
jb first_is_bigger
ja second_is_bigger
cmp al, LF ; LF equ 10
je they_are_equal
inc ecx
jmp top
That's untested. I've been having a lot of brain flatulence lately - better check me. If the pesky user enters more than 20 characters, you may be in deep muddy water...
Related
I'm trying to make a very simple assembly program run, however I seem to get segfaults whatever I do.
Here is my code (should print 'a' on a linux machine)
section .data
buffer times 50 db 97
pointer db 0
section .text
global _start
_start:
mov ECX , pointer
mov EDX , [buffer + ECX]
mov EAX , 4
mov EBX , 1
mov ECX , EDX
mov EDX , 1
int 0x80
It causes a segfault on the first MOV but it seems obvious to me that it should work.
I reduced it to almost nothing and it still segfault.
section .data
msg db "hello"
section .text
global _start
_start:
mov EAX,1
I've run this succesfully:
section .text
global _start
_start:
mov ax, 0b
dec ax
sub ax, 11111111b
mov bx, 97
add ax, bx
mov [INVENTORY], ax ; put a in first inventory pos
mov eax, 4
mov ebx, 1
mov ecx, INVENTORY
mov edx, 1
int 0x80
mov ax, [INVENTORY]
add ax, 1
mov [INVENTORY + 1], ax ; put b in second inventory pos
mov [VAR], ax
mov eax, 4
mov ebx, 1
mov ecx, VAR
mov edx, 1
int 0x80
mov eax, 4
mov ebx, 1
mov ecx, '\n'
mov edx, 1
int 0x80
mov eax,1
int 0x80
_newline:
section .data
VAR DW 0
INVENTORY TIMES 8 DW 0
Is it possible that it has to do with the symbols I use for newlines or tabs? I generate the assembly from java and I use \t for tabs and \n for new lines (and spaces so it doesn't look too bad.
I'm using NASM and I'm running it here:
https://www.tutorialspoint.com/compile_assembly_online.php
Thank you!
If you are just trying to print out a set of 'a's.
section .data
buffer times 50 db 97
len.buffer equ $-buffer
pointer db 0
section .text
global _start
_start:
; ssize_t write(int fd, const void *buf, size_t count);
; i386 ebx ecx edx esi edi ebp
mov EAX , 4 ; write syscall
mov EBX , 1 ; std out
lea ecx, [buffer] ; buffer
mov edx, len.buffer ; size
int 0x80
_exit:
mov eax, 1 ; exit syscall
int 0x80
output:
./yvon_001
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\ndavid#ubuntuserver00A:~/asm$ ./yvon_001
In the end #MichealPetch was right, I needed to add an EXIT syscall at the end of the code. The sample I tried still did a SEGFAULT because I was moving pointer instead of [pointer] in a registry.
Thanks for comments and answers!
I have code here that reads in input to determine the dimensions of a matrix/2d array and then reads in numbers one by one. Then it is SUPPOSED to spit out the smallest number.
However my comparison operator doesn't seem to be working? I've tried putting both as registers, with different variables and so on but for some reason eax with whatever happened to be the first entry in the array is ALWAYS smaller than the next number, even if this is clearly not the case.
So it always skips reassignment.
Code: Just skip straight to cmp eax,[num]
My guess is something is causing num, perhaps how its declared? Is always 'larger' than eax, is there extra fluff I am not aware of?
segment .bss
num: resw 1 ;For storing a number, to be read of printed....
nod: resb 1 ;For storing the number of digits....
temp: resb 2
matrix1: resw 200
m: resw 1
n: resw 1
i: resw 1
j: resw 1
small: resb 4 ; temp variable
buff resb 4
segment .data
msg1: db "Enter the number of rows in the matrix : "
msg_size1: equ $-msg1
msg2: db "Enter the elements one by one(row by row) : "
msg_size2: equ $-msg2
msg3: db "Enter the number of columns in the matrix : "
msg_size3: equ $-msg3
msg4: db "The Smallest Number is... : "
msg_size4: equ $-msg4
tab: db 9 ;ASCII for vertical tab
new_line: db 10 ;ASCII for new line
segment .text
global _start
_start:
;; code for reading number of rows and columns, this works fine.
mov eax, 4
mov ebx, 1
mov ecx, msg1
mov edx, msg_size1
int 80h
;; read in rows
mov ecx, 0
call read_num
mov cx, word[num]
mov word[m], cx
mov eax, 4
mov ebx, 1
mov ecx, msg3
mov edx, msg_size3
int 80h
;; read in columns
mov ecx, 0
call read_num
mov cx, word[num]
mov word[n], cx
mov eax, 4
mov ebx, 1
mov ecx, msg2
mov edx, msg_size2
int 80h
;; Reading in each element and storing it into the array
mov esi, 0
mov ebx, matrix1
;; init loop
mov word[i], 0
mov word[j], 0
;; Outer loop
i_loop:
mov word[j], 0
;; Inner Loop
j_loop:
;; A function
call read_num
;; Result of that function is now stored int he matrix
mov dx , word[num]
mov word[ebx + 2 * esi], dx
inc esi ;Incrementing array index by one....
inc word[j]
mov cx, word[j]
cmp cx, word[n]
jb j_loop
;; End Inner Loop
inc word[i]
mov cx, word[i]
cmp cx, word[m]
jb i_loop
;; End Outer Loop
;; Now begins the code to find the smallest number
mov eax, [matrix1]
;; Moves first element of Matrix1 into eax
;;; saves eax into small
mov [small], eax
;Loop through the matrix, check each number if its smaller than the first number in the array. AT the end print said number.
;Reading each element of the matrix.(Storing the elements in row major order).......
mov esi, 0
mov edi, matrix1
;; Reinit loop cters
mov word[i], 0
mov word[j], 0
;; Loop
i_loop2:
mov word[j], 0
j_loop2:
;eax will contain the array index and each element is 2 bytes(1 word) long
mov dx, word[edi+2*esi] ;
mov word[num] , dx
cmp eax,[num] ; compares eax and ebx
jle skip ;if eax is SMALLER than ebx, we can safely skip reassignment
;as our current value is already smallest.
mov eax, [num] ; stores new smallest number if the new number was smaller.
mov [small], eax
;; reassignment code is always skipped.
skip:
inc esi
inc word[j]
mov cx, word[j]
cmp cx, word[n]
jb j_loop2
inc word[i]
mov cx, word[i]
cmp cx, word[m]
jb i_loop2
; code to output the smallest number
;; Some ui text.
mov eax, 4
mov ebx, 1
mov ecx, msg4
mov edx, msg_size4
int 80h
mov ecx, 0
;; Now the actual smallest number
mov eax, 4 ; system_write
mov ebx, 1 ; stdout
mov ecx, [small] ; move smallest element to accumulator
add ecx, 48 ; convert to ascii representation
mov [buff], ecx ; move to memory
mov ecx, buff
mov edx, 4 ; size, 4 bytes
int 80h
exit:
mov eax, 1
mov ebx, 0
int 80h
;Function to read a number from console and to store that in num
read_num:
pusha
mov word[num], 0
loop_read:
mov eax, 3
mov ebx, 0
mov ecx, temp
mov edx, 1
int 80h
cmp byte[temp], 10
je end_read
mov ax, word[num]
mov bx, 10
mul bx
mov bl, byte[temp]
sub bl, 30h
mov bh, 0
add ax, bx
mov word[num], ax
jmp loop_read
end_read:
popa
ret
Oh yes, several hours of fidgeting and seconds after this post I figured out exactly why. Turns out yes, the way I declared "num" yes indeed added in extra information. I changed it to resb 4 and it works.
Crabbaskets.
The following code:
section .bss
name: resb 50
section .text
global _start
_start:
PUSH EBP
MOV EBP, ESP
MOV EDX, len
MOV ECX, msg
MOV EBX, 1
MOV EAX, 4
INT 0x80
MOV EDX, 50
MOV ECX, name
MOV EBX, 0
MOV EAX, 3
INT 0x80
MOV EBX, 1
MOV EAX, 4
INT 0x80
MOV EDX, cm
MOV ECX, ex
MOV EBX, 1
MOV EAX, 4
INT 0x80
MOV EBX, 0
MOV EAX, 1
INT 0x80
section .data
msg db 'Hello!',0xa
ex db '!',0xa
len equ $ - msg
cm equ $ - ex
I intended to make a simple I/O program that printed Hello!, asked for a char and would print %c!.
Input being | and output being :, I get the following:
:Hello!
:!
|4
:4
:!
How do I make it so that it returns the following
:Hello!
|4
:4!
As Damien_The_Unbeliever says, your equs want to come immediately after the string they're supposed to measure. After your sys_read, eax will be the number of characters read, including the linefeed that ends the reading. You probably don't want to print the linefeed (in this case - sometimes you would). So:
mov edx, eax
dec edx
Or if you want to do it in one instruction:
lea edx, [eax - 1]
As it stands, edx still holds 50, so your next sys_write will print 50 characters. It will NOT stop at a zero or any other string-terminator. ecx will still contain name, but I would reload it just for clarity.
By rights, you should check for an error return (eax would be negative) after each and every int 0x80 but an error is unlikely here.
On NASM in Arch Linux, how can I append the character zero ('0') to a 32 bit variable? My reason for wanting to do this is so that I can output the number 10 by setting a single-digit input to 1 and appending a zero. I need to figure out how to append the zero.
The desirable situation:
Please enter a number: 9
10
Using this method, I want to be able to do this:
Please enter a number: 9999999
10000000
How can I do this?
Thanks in advance,
RileyH
Well, as Bo says... but I was bored. You seem resistant to doing this the easy way (convert your input to a number, add 1, and convert it back to text) so I tried it using characters. This is what I came up with. It's horrid, but "seems to work".
; enter a number and add 1 - the hard way!
; nasm -f elf32 myprog.asm
; ld -o myprog myprog.o -melf_i386
global _start
; you may have these in an ".inc" file
sys_exit equ 1
sys_read equ 3
sys_write equ 4
stdin equ 0
stdout equ 1
stderr equ 2
LF equ 10
section .data
prompt db "Enter a number - not more than 10 digits - no nondigits.", LF
prompt_size equ $ - prompt
errmsg db "Idiot human! Follow instructions next time!", LF
errmsg_size equ $ - errmsg
section .bss
buffer resb 16
fakecarry resb 1
section .text
_start:
nop
mov eax, sys_write
mov ebx, stdout
mov ecx, prompt
mov edx, prompt_size
int 80h
mov eax, sys_read
mov ebx, stdin
mov ecx, buffer + 1 ; leave a space for an extra digit in front
mov edx, 11
int 80h
cmp byte [buffer + 1 + eax - 1], LF
jz goodinput
; pesky user has tried to overflow us!
; flush the buffer, yell at him, and kick him out!
sub esp, 4 ; temporary "buffer"
flush:
mov eax, sys_read
; ebx still okay
mov ecx, esp ; buffer is on the stack
mov edx, 1
int 80h
cmp byte [ecx], LF
jnz flush
add esp, 4 ; "free" our "buffer"
jmp errexit
goodinput:
lea esi, [buffer + eax - 1] ; end of input characters
mov byte [fakecarry], 1 ; only because we want to add 1
xor edx, edx ; count length as we go
next:
; check for valid decimal digit
mov al, [esi]
cmp al, '0'
jb errexit
cmp al, '9'
ja errexit
add al, [fakecarry] ; from previous digit, or first... to add 1
mov byte [fakecarry], 0 ; reset it for next time
cmp al, '9' ; still good digit?
jna nocarry
; fake a "carry" for next digit
mov byte [fakecarry], 1
mov al, '0'
cmp esi, buffer + 1
jnz nocarry
; if first digit entered, we're done
; save last digit and add one ('1') into the space we left
mov [esi], al
inc edx
dec esi
mov byte [esi], '1'
inc edx
dec esi
jmp done
nocarry:
mov [esi], al
inc edx
dec esi
cmp esi, buffer
jnz next
done:
inc edx
inc edx
mov ecx, esi ; should be either buffer + 1, or buffer
mov ebx, stdout
mov eax, sys_write
int 80h
xor eax, eax ; claim "no error"
exit:
mov ebx, eax
mov eax, sys_exit
int 80h
errexit:
mov edx, errmsg_size
mov ecx, errmsg
mov ebx, stderr
mov eax, sys_write
int 80h
mov ebx, -1
jmp exit
;-----------------------------
Is that what you had in mind?
I have a program below that tries to take input from the user and repeat that same string until the user enters it again. (It's a personal learning project)
However, I am having some severe diffuculty in getting it to perform correctly. In a past thread here, you can see the input, pun intended, that other users have provided on this problem.
%include "system.inc"
section .data
greet: db 'Hello!', 0Ah, 'Please enter a word or character:', 0Ah
greetL: equ $-greet ;length of string
inform: db 'I will now repeat this until you type it back to me.', 0Ah
informL: equ $-inform
finish: db 'Good bye!', 0Ah
finishL: equ $-finish
newline: db 0Ah
newlineL: equ $-newline
section .bss
input: resb 40 ;first input buffer
check: resb 40 ;second input buffer
section .text
global _start
_start:
greeting:
mov eax, 4
mov ebx, 1
mov ecx, greet
mov edx, greetL
sys.write
getword:
mov eax, 3
mov ebx, 0
mov ecx, input
mov edx, 40
sys.read
sub eax, 1 ;remove the newline
push eax ;store length for later
instruct:
mov eax, 4
mov ebx, 1
mov ecx, inform
mov edx, informL
sys.write
pop edx ;pop length into edx
mov ecx, edx ;copy into ecx
push ecx ;store ecx again (needed multiple times)
mov eax, 4
mov ebx, 1
mov ecx, input
sys.write
mov eax, 4 ;print newline
mov ebx, 1
mov ecx, newline
mov edx, newlineL
sys.write
mov eax, 3 ;get the user's word
mov ebx, 0
mov ecx, check
mov edx, 40
sys.read
sub eax, 1
push eax
xor eax, eax
checker:
pop ecx ;length of check
pop ebx ;length of input
mov edx, ebx ;copy
cmp ebx, ecx ;see if input was the same as before
jne loop ;if not the same go to input again
mov ebx, check
mov ecx, input
secondcheck:
mov dl, [ebx]
cmp dl, [ecx]
jne loop
inc ebx
inc ecx
dec eax
jnz secondcheck
jmp done
loop:
pop edx
mov ecx, edx
push ecx
mov eax, 4
mov ebx, 1
mov ecx, check
sys.write ;repeat the word
mov eax, 4
mov ebx, 1
mov ecx, newline
mov edx, newlineL
sys.write
mov eax, 3 ;replace new input with old
mov ebx, 0
mov ecx, check
mov edx, 40
sys.read
jmp checker
done:
mov eax, 1
mov ebx, 0
sys.exit
Example output would yield:
Hello!
Please enter a word or character:
INPUT: Nick
I will now repeat this until you type it back to me.
Nick
INPUT: Nick
N
INPUT: Nick
INPUT: Nick
And that goes on forever until is ^C it to death. Any ideas on the problem?
Thanks.
instruct leaves two items on the stack, which are consumed by checker the first time round the loop. But they are not replaced for the case where you go round the loop again. This is the most fundamental problem in your code (there may be others).
You could find this by running with a debugger and watching the stack pointer esp; but it can be seen just by looking at the code -- if you take everything out except for the stack manipulation and branches, you can clearly see that the checker -> loop -> back to checker path pops three items but only pushes one:
greeting:
...
getword:
...
push eax ;store length for later
instruct:
...
pop edx ;pop length into edx
...
push ecx ;store ecx again (needed multiple times)
...
push eax
checker:
pop ecx ;length of check
pop ebx ;length of input
...
jne loop ;if not the same go to input again
...
secondcheck:
...
jne loop
...
jnz secondcheck
jmp done
loop:
pop edx
...
push ecx
...
jmp checker
done:
...
There are better ways to keep long-lived variables than trying to shuffle them around on the stack like this with push and pop.
Keep them in a data section (the .bss you already have would be suitable) instead of on the stack.
Allocate some space on the stack, and load/store them there directly. e.g. sub esp, 8 to reserve two 32-bit words, then access [esp] and [esp+4]. (The stack should be aligned to a 32-bit boundary, so always reserve a multiple of 4 bytes.) Remember to add esp, 8 when you've finished using it.
(These are essentially the equivalent of what a C compiler would do for global (or static) variables, and local variables, respectively.)