I'm writing a simple subroutine in FASM to print 32-bit unsigned integers to STDOUT. This is what I came up with:
format elf
public uprint
section ".text" executable
uprint:
push ebx
push ecx
push edx
push esi
mov ebx, 10
mov ecx, buf + 11
xor esi, esi
do:
dec ecx
xor edx, edx
div ebx
add dl, 0x30
mov [ecx], dl
inc esi
test eax, 0
jnz do
mov eax, 4
mov ebx, 1
mov edx, esi
int 0x80
pop esi
pop edx
pop ecx
pop ebx
ret
section ".data" writeable
buf rb 11
Then I wrote another program to test whether the above subroutine works properly:
format elf
extrn uprint
public _start
section ".text" executable
_start:
mov eax, 1337
call uprint
mov eax, 4
mov ebx, 1
mov ecx, newline
mov edx, 1
int 0x80
mov eax, 1
xor ebx, ebx
int 0x80
section ".data"
newline db 0x0A
I compiled both these programs to their corresponding object files and linked them to create the executable.
On executing the program however it only displayed 7 instead of 1337. As it turns out only the last digit of the number is display regardless of the number itself.
This is strange because my uprint subroutine is correct. In fact if I combine both these programs into a single program then it displays 1337 correctly.
What am I doing wrong?
I gain the distinct impression that your LINK operation is building the uprint before the _start and you're in fact entering UPRINT, not at _start as you expect.
I found out my mistake. I'm using test eax, 0 which always sets the zero flag. Hence only the first digit is processed. Intead I need to use either test eax, eax or cmp eax, 0.
Related
THE PROGRAM IS USED TO ACCEPT CHARACTERS AND DISPLAY THEM IN REVERSE ORDER
The code is included here:
section .bss
num resb 1
section .text
global _start
_start:
call inputkey
call outputkey
;Output the number entered
mov eax, 1
mov ebx, 0
int 80h
inputkey:
;Read and store the user input
mov eax, 3
mov ebx, 2
mov ecx, num
mov edx, 1
int 80h
cmp ecx, 1Ch
je .sub2
push ecx
jmp inputkey
.sub2:
push ecx
ret
outputkey:
pop ecx
;Output the message
mov eax, 4
mov ebx, 1
;mov ecx, num
mov edx, 1
int 80h
cmp ecx, 1Ch
je .sub1
jmp outputkey
.sub1:
ret
The code to compile and run the program
logic.asm
is given here:
nasm -f elf logic.asm
ld -m elf_i386 -s -o logic logic.o
./logic
There are a few problems with the code. Firstly, for the sys_read syscall (eax = 3) you supplied 2 as the file descriptor, however 2 refers to stderr, but in this case you'd want stdin, which is 0 (I like to remember it as the non-zero numbers 1 and 2 being the output).
Next, an important thing to realize about the ret instruction is that it pops the value off the top of the stack and returns to it (treating it as an address). Meaning that even if you got to the .sub2 label, you'd likely get a segfault. With this in mind, the stack also tends to not be permanent storage, as in it is not preserved throughout procedures, so I'd recommend just making your buffer larger to e.g. 256 bytes and increment a value to point to an index in the buffer. (Using a fixed-size buffer will keep you from getting into the complications of memory allocation early, though if you want to go down that route you could do an external malloc call or just an mmap syscall.)
To demonstrate what I mean by an index into the reserved buffer:
section .bss
buf resb 256
; ...
inputkey:
xor esi, esi ; clear esi register, we'll use it as the index
mov eax, 3
mov ebx, 0 ; stdin file descriptor
mov edx, 1 ; read one byte
.l1: ; loop can start here instead of earlier, since the values eax, ebx and edx remain unchanged
lea ecx, [buf+esi] ; load the address of buf + esi
int 80h
cmp [buf+esi], 0x0a ; check for a \n character, meaning the user hit enter
je .e1
inc esi
jmp .l1
.e1:
ret
In this case, we also get to preserve esi up until the output, meaning that to reverse the input, we just print in descending order.
outputkey:
mov eax, 4
mov ebx, 1 ; stdout
mov edx, 1
.l2:
lea ecx, [buf+esi]
int 80h
test esi, esi ; if esi is zero it will set the ZF flag
jz .e2:
jmp .l2
.e2:
ret
Note: I haven't tested this code, so if there are any issues with it let me know.
I'm writing an assembly program that would print even numbers between 0-9 using a loop. I encountered this problem, segmentation fault while running the code. I check other answers on the site but couldn't find an answer that satisfies my issue.
I suspect that the function nwLine might be the source of the problem.
;;this program prints even numbers from 0-8 using loop function
section .text
global _start
cr db 10
_start: ;tell linker entry point
mov ecx, 5
mov eax, '0'
evenLoop:
mov [evnum], eax ;add eax to evnum
mov eax, 4
mov ebx, 1
push ecx
mov ecx, evnum
mov edx, 1
int 80h
call nwLine
mov eax, [evnum]
sub eax, '1'
inc eax
add eax, '2'
pop ecx
loop evenLoop
nwLine: ;function to move pointer to next line
mov eax,4 ; System call number(sys_write)
mov ebx,1 ; File descriptor 1 - standard output
mov ecx, cr
mov edx, 1
int 80h ; Call the kernel
ret
mov eax,1 ;system call number (sys_exit)
int 80h ;call kernel
section .bss
evnum resb 1
if anyone knows how to solve the problem with the nwLine function, please tell me.
I'm trying to print Hi 10 times. This is my code.
section .data
msg db "Hi"
section .text
global _start
_start:
mov cx, 10
L1:
mov eax, 4
mov ebx, 1
mov ecx, msg
mov edx, 3
int 0x80
dec cx
jnz L1
mov eax, 1
mov ebx, 0
int 0x80
gdb reports that mov edx, 3 overwrites the cx register to some crazy value and so the loop keeps going forever.
What am i doing wrong? Is it because they are the same register?
How does one program in assembly with so few registers?
Compiling on centos with nasm and ld
Thanks
You're looking at the wrong line. The problem is "mov ecx, msg". ECX is the extended register of which CX is the lower part, so you're writing over it.
It's best to save your loop counter on the stack, because who knows that the 'int' call might change. Add 'push cx' (or ecx) after 'L1:'. and 'pop cx' after the 'int' call to preserve the contents of the register.
This code fixes it:
section .data
msg db "Hi"
counter dw 10
section .text
global _start
_start:
L1:
mov eax, 4
mov ebx, 1
mov ecx, msg
mov edx, 2
int 0x80
mov cx, [counter]
dec cx
mov [counter], cx
jnz L1
mov eax, 1
mov ebx, 0
int 0x80
move the value into a variable and then dec it then put it back
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.)
I started assembly (nasm) programming not too long ago. Now I made a C function with assembly implementation which prints an integer. I got it working using the extended registers, but when I want to write it with the x64 registers (rax, rbx, ..) my implementation fails. Does any of you see what I missed?
main.c:
#include <stdio.h>
extern void printnum(int i);
int main(void)
{
printnum(8);
printnum(256);
return 0;
}
32 bit version:
; main.c: http://pastebin.com/f6wEvwTq
; nasm -f elf32 -o printnum.o printnum.asm
; gcc -o printnum printnum.o main.c -m32
section .data
_nl db 0x0A
nlLen equ $ - _nl
section .text
global printnum
printnum:
enter 0,0
mov eax, [ebp+8]
xor ebx, ebx
xor ecx, ecx
xor edx, edx
push ebx
mov ebx, 10
startLoop:
idiv ebx
add edx, 0x30
push dx ; With an odd number of digits this will screw up the stack, but that's ok
; because we'll reset the stack at the end of this function anyway.
; Needs fixing though.
inc ecx
xor edx, edx
cmp eax, 0
jne startLoop
push ecx
imul ecx, 2
mov edx, ecx
mov eax, 4 ; Prints the string (from stack) to screen
mov ebx, 1
mov ecx, esp
add ecx, 4
int 80h
mov eax, 4 ; Prints a new line
mov ebx, 1
mov ecx, _nl
mov edx, nlLen
int 80h
pop eax ; returns the ammount of used characters
leave
ret
x64 version:
; main.c : http://pastebin.com/f6wEvwTq
; nasm -f elf64 -o object/printnum.o printnum.asm
; gcc -o bin/printnum object/printnum.o main.c -m64
section .data
_nl db 0x0A
nlLen equ $ - _nl
section .text
global printnum
printnum:
enter 0, 0
mov rax, [rbp + 8] ; Get the function args from the stac
xor rbx, rbx
xor rcx, rcx
xor rdx, rdx
push rbx ; The 0 byte of the string
mov rbx, 10 ; Dividor
startLoop:
idiv rbx ; modulo is in rdx
add rdx, 0x30
push dx
inc rcx ; increase the loop variable
xor rdx, rdx ; resetting the modulo
cmp rax, 0
jne startLoop
push rcx ; push the counter on the stack
imul rcx, 2
mov rdx, rcx ; string length
mov rax, 4
mov rbx, 1
mov rcx, rsp ; the string
add rcx, 4
int 0x80
mov rax, 4
mov rbx, 1
mov rcx, _nl
mov rdx, nlLen
int 0x80
pop rax
leave
ret ; return to the C routine
Thanks in advance!
I think your problem is that you're trying to use the 32-bit calling conventions in 64-bit mode. That won't fly, not if you're calling these assembly routines from C. The 64-bit calling convention is documented here: http://www.x86-64.org/documentation/abi.pdf
Also, don't open-code system calls. Call the wrappers in the C library. That way errno gets set properly, you take advantage of sysenter/syscall, you don't have to deal with the differences between the normal calling convention and the system-call argument convention, and you're insulated from certain low-level ABI issues. (Another of your problems is that write is system call number 1, not 4, for Linux/x86-64.)
Editorial aside: There are two, and only two, reasons to write anything in assembly nowadays:
You are writing one of the very few remaining bits of deep magic that cannot be written in C alone (a good example is the guts of libffi)
You are hand-optimizing an inner-loop subroutine that has been measured to be performance-critical and the C compiler doesn't do a good enough job on.
Otherwise just write whatever it is in C. Your successors will thank you.
EDIT: checked system call numbers.
I'm not sure if this answer is related to the problem you're seeing (since you didn't specify anything about what the failure is), but 64-bit code has a different calling convention than 32-bit code does. Both of the major 64-bit Intel ABIs (Windows & Linux/BSD/Mac OS) pass function parameters in registers and not on the stack. Your program appears to still be expecting them on the stack, which isn't the normal way to go about it.
Edit: Now that I see there is a C main() routine that calls your functions, my answer is exactly about the problem you're having.