section .data
bufChar: equ 0
section .bss
bufNum: resb 1
bufMult: resb 1
.
.
.
leerNumero:
xor eax,eax
mov [bufNum],eax
add eax,1
mov [bufMult],eax
inicioLeerNumero:
mov edx,1
mov ecx,bufChar
mov ebx,0
mov eax,3
int 80h
cmp byte [ecx + edx - 1],10 ; Segfaults here.
je rLeerNumero
cmp byte [ecx + edx - 1],48
jl noNumero
cmp byte [ecx + edx - 1],57
jg noNumero
sub eax,48
mul word [bufMult]
jo overflow
add [bufNum],eax
jo overflow
mov eax,10
mul word [bufMult]
jo overflow
mov [bufMult],eax
jmp inicioLeerNumero
rLeerNumero:
mov eax,bufNum
ret
noNumero:
mov eax,errorNumero
mov ebx,lErrorNumero
call imprimir
jmp salir
overflow:
mov eax,errorOverflow
mov ebx,lErrorOverflow
call imprimir
jmp salir
This code should of work, at least in paper it does. I need to do some homework completely in assembly without linking the C Library, hence why i am re-inventing the wheel and making a method to read a number from the console into EAX.
I am having a mysterious segfault at the line marked with the comment and i fail to see how i coud be trying to access misaligned memory... any ideas on how could this be failing?
Any chance that int 80h is changing ecx or edx on you causing a bad pointer read? If you can read the registers in a debugger before and after that instruction, you could confirm that.
I had declared bufChar as .data, obviously mov'ing into a constant would segfault the thing. Sadly, i wasted a week wrapping my head around this, just realized so.
Related
I have written an assembly code to print numbers from 1 to 9 but the code only prints 1 and no other element other than 1 is printed and only one output is received.It means that the loop is also not being run. I cant figure out what is wrong with my code.
section .bss
lena equ 1024
outbuff resb lena
section .data
section .text
global _start
_start:
nop
mov cx,0
incre:
inc cx
add cx,30h
mov [outbuff],cx
cmp cx,39h
jg done
cmp cx,39h
jl print
print:
mov rax,1 ;sys_write
mov rdi,1
mov rsi,outbuff
mov rdx,lena
syscall
jmp incre
done:
mov rax,60 ;sys_exit
mov rdi,0
syscall
My OS is 64 bit linux. this code is built using nasm with the following commands : nasm -f elf64 -g -o num.o num.asm and ld -o num num.asm
Answer rewritten after some experimentation.
There two errors in your code, and a few inefficiencies.
First, you add 0x30 to the number (to turn it from the number 1 to the ASCII 1). However, you do that increment inside the loop. As a result, your first iteration cx is 0x31, second 0x62 ("b"), third 0x93 (invalid UTf-8 sequence) etc.
Just initialize cx to 0x30 and remove the add from inside the loop.
But there's another problem. RCX is clobbered during system calls. Replacing cx with r12 causes the program to work.
In addition to that, you pass the buffer's length to write, but it only has one character. The program so far:
section .bss
lena equ 1024
outbuff resb lena
section .data
section .text
global _start
_start:
nop
mov r12,30h
incre:
inc r12
mov [outbuff],r12
cmp r12,39h
jg done
cmp r12,39h
jl print
print:
mov rax,1 ;sys_write
mov rdi,1
mov rsi,outbuff
mov rdx,1
syscall
jmp incre
done:
mov rax,60 ;sys_exit
mov rdi,0
syscall
Except even now, the code is extremely inefficient. You have two compares on the same condition, one of them branches to the very next instruction.
Also, your code would be much much much faster and smaller if you moved the breaking condition to the end of the code. Also, cx is a 16 bit register. r12 is a 64 bit register. We actually only need 8 bits. Using larger registers than needed means all of our immediates waste up space in memory and the cache. We therefor switch to the 8 bit variant of r12. After these changes, we get:
section .bss
lena equ 1024
outbuff resb lena
section .data
section .text
global _start
_start:
nop
mov r12b,30h
incre:
inc r12b
mov [outbuff],r12b
mov rax,1 ;sys_write
mov rdi,1
mov rsi,outbuff
mov rdx,1
syscall
cmp r12b,39h
jl incre
mov rax,60 ;sys_exit
mov rdi,0
syscall
There's still lots more you can do. For example, you call the write system call 9 times, instead of filling the buffer and then calling it once (despite the fact that you've allocated a 1024 bytes buffer). It will probably be faster to initialize r12 with zero (xor r12, r12) and then add 0x30. (not relevant for the 8 bit version of the register).
global _start
section .data
var dq 12494F04A6344129h
msg db "The number of times 4 present in the given number"
len equ $-msg
novar db 00
section .bss
section .text
mov dl,0Ah
mov cl,10
_start :
mov rsi,var
up: mov al,byte ptr [rsi]
mov ah,00
div dl
cmp ah,04
je dn
jne dn1
dn: inc byte[novar]
dn1: inc rsi
dec cl
jne up
jmp exit
exit: mov eax,4
mov ebx,1
mov ecx,msg
mov edx,len
int 80h
mov eax,4
mov ebx,1
mov ecx,novar
mov edx,1
int 80h
mov eax,1
mov ebx,0
int 80h
Nasm doesn't use "ptr" - that won't even assemble.
The first two lines - above the _start: label - are never executed, so those registers are never initialized. That's probably what causes the exception. dl is probably zero!
len is fine - it's an equate, not a variable.
You probably want to add the character '0' to novar before printing it.
None of this looks useful to me. Are you sure this is the question you're trying to answer, #Shubham Satpute?
STORY(IM A NEWBIE):
I started reading a pdf tutorial about programming in assembly(x86 intel) using the famous nasm assembler and i have a problem executing a very basic assembly code(inspired by a code about loops from the tutorial).
THE PROBLEM(JE FAILS):
This assembly code should read a digit(a character(that means '0'+digit)) from stdin and then write to the screen digit times "Hello world\n".Really easy loop :decrease digit and if digit equals zero('0' not the integer the character) jump(je) to the exit(mov eax,1\nint 0x80).
Sounds really easy but when i try to execute the output is weird.(really weird and BIG)
It runs many times throught the loop and stops when digit equals '0'(weird because until the program stops the condition digit == '0' been tested many times and it should be true)
Actually my problem is that the code fails to jump when digit == '0'
THE CODE(IS BIG):
segment .text
global _start
_start:
;Print 'Input a digit:'.
mov eax,4
mov ebx,1
mov ecx,msg1
mov edx,len1
int 0x80
;Input the digit.
mov eax,3
mov ebx,0
mov ecx,dig
mov edx,2
int 0x80
;Mov the first byte(the digit) in the ecx register.
;mov ecx,0
mov ecx,[dig]
;Use ecx to loop dig[0]-'0' times.
loop:
mov [dig],ecx
mov eax,4
mov ebx,1
mov ecx,dig
mov edx,1
int 0x80
mov eax,4
mov ebx,1
mov ecx,Hello
mov edx,Hellolen
int 0x80
;For some debuging (make the loop stop until return pressed)
;mov eax,3
;mov ebx,0
;mov ecx,some
;mov edx,2
;int 0x80
;Just move dig[0](some like character '4' or '7') to ecx register and compare ecx with character '0'.
mov ecx,[dig]
dec ecx
cmp ecx,'0'
;If comparison says ecx and '0' are equal jump to exit(to end the loop)
je exit
;If not jump back to loop
jmp loop
;Other stuff ...(like an exit procedure and a data(data,bss) segment)
exit:
mov eax,1
int 0x80
segment .data
msg1 db "Input a digit:"
len1 equ $-msg1
Hello db ":Hello world",0xa
Hellolen equ $-Hello
segment .bss
dig resb 2
some resb 2
THE OUTPUT:
Input a digit:4
4:Hello world
3:Hello world
2:Hello world
1:Hello world
0:Hello world
...
...(many loops later)
...
5:Hello world
4:Hello world
3:Hello world
2:Hello world
1:Hello world
$
That is my question:What is wrong with this code?
Could you explain that ?
AND i dont need alternative codes that will magically(without explanation) run cause i try to learn(im a newbie)
That is my problem(and my first question in Stackoverflow.com )
ECX is 32 bit, a character is just 8 bit. Use a 8 bit register, such as CL instead of ECX.
As jester mentioned, ecx comes in as a character so you probably should use cl
loop:
mov [dig],cl
...
mov cl,[dig]
dec cl
cmp cl,'0'
jne loop
You can also load ecx with movzx which clears the top bits of the register (i.e. a zero-extedning load):
...
movzx ecx, byte [dig]
loop:
mov [dig], cl ; store just the low byte, if you want to store
...
movzx ecx, byte [dig]
dec ecx
cmp ecx, '0'
jne loop
Note that it is often suggested that you do not use the al, bl, cl, dl registers as their use is not fully optimized. Whether this is still true, I do not know.
Excuse me again. I am trying understand learn assembly languaje. However I have many problems. I am trying working with strings in NASM. I have copy a string constant to string variable. The maximum size is 50. So I want verify this bound. However this program throw a segmentation fault. I use a example in MASM, so perhaps exist a use error with NASM syntax.
My program is the following:
section .data
MAXTEXTSIZE equ 50
_cte_hola db "Hola", 0
_cte_mundo db "Mundo", 0
section .bss
MAIN_d resb MAXTEXTSIZE+1
section .text
global _start
strlen:
mov bx, 0
strl01:
cmp WORD [SI+BX],0 t
je strend
inc bx
jmp strl01
strend:
ret
strcpy:
call strlen
cmp bx, MAXTEXTSIZE
jle copiarsizeok
mov bx, MAXTEXTSIZE
copiarsizeok:mov cx, bx
cld
rep movsb
mov al,0
mov BYTE [DI], al
ret
_start:
mov ds, ax
mov es, ax
mov si, [MAIN_d]
mov di, [_cte_hola]
call strcpy
mov eax, 1
mov ebx, 0
int 80h
Thanks in advance and excuse me. My question are stupid for a assembly programmer.
I believe you are trying to make 32bit program in Linux, but your examples are 16bit.
In Linux, all pointers are 32bit. So, use extended registers: esi, edi, ebx etc. You still can use 8 and 16bit registers for arithmetics and data processing but not as memory pointers.
In strlen you have to compare byte [esi+ebx], 0 not word.
Don't set the segment registers in Linux. They will be set by the OS and you can't touch them. In Linux all memory is one flat area and you don't have to use segment registers anymore.
Here's a more concrete example of how you could write your strlen function (which is the first of your problems)
section .data
MAXTEXTSIZE equ 50
_cte_hola db "Hola", 0xa, 0
_cte_mundo db "Mundo", 0
section .bss
MAIN_d resb MAXTEXTSIZE+1
section .text
global _start
strlen:
mov ebx, 0
strlen_loop:
cmp BYTE [esi+ebx], 0
je strlen_end
inc ebx
jmp strlen_loop
strlen_end:
mov eax, ebx
ret
_start:
mov esi, _cte_hola
call strlen ; Get the length of _cte_hola
mov edx, eax ; The length was stored in eax by strlen
mov ecx, _cte_hola
mov ebx,1
mov eax, 4
int 0x80 ; Write to stdout
mov eax, 1
int 0x80 ; Exit
There are definitely better ways of implementing this (I'd use repne to implement strlen, for example) but I wanted to keep it close to your implementation.
Hope this helps!
I made my own implementation of strlen in assembly, but it doesn't return the correct value. It returns the string length + 4. Consequently. I don't see why.. and I hope any of you do...
Assembly source:
section .text
[GLOBAL stringlen:] ; C function
stringlen:
push ebp
mov ebp, esp ; setup the stack frame
mov ecx, [ebp+8]
xor eax, eax ; loop counter
startLoop:
xor edx, edx
mov edx, [ecx+eax]
inc eax
cmp edx, 0x0 ; null byte
jne startLoop
end:
pop ebp
ret
And the main routine:
#include <stdio.h>
extern int stringlen(char *);
int main(void)
{
printf("%d", stringlen("h"));
return 0;
}
Thanks
You are not accessing bytes (characters), but doublewords. So your code is not looking for a single terminating zero, it is looking for 4 consecutive zeroes. Note that won't always return correct value +4, it depends on what the memory after your string contains.
To fix, you should use byte accesses, for example by changing edx to dl.
Thanks for your answers. Under here working code for anyone who has the same problem as me.
section .text
[GLOBAL stringlen:]
stringlen:
push ebp
mov ebp, esp
mov edx, [ebp+8] ; the string
xor eax, eax ; loop counter
jmp if
then:
inc eax
if:
mov cl, [edx+eax]
cmp cl, 0x0
jne then
end:
pop ebp
ret
Not sure about the four, but it seems obvious it will always return the proper length + 1, since eax is always increased, even if the first byte read from the string is zero.
Change the line
mov edx, [ecx+eax]
to
mov dl, byte [ecx+eax]
and
cmp edx, 0x0 ; null byte
to
cmp dl, 0x0 ; null byte
Because you have to compare only byte at a time.
Following is the code. Your original code got off-by-one error. For "h" it will return two h + null character.
section .text
[GLOBAL stringlen:] ; C function
stringlen:
push ebp
mov ebp, esp ; setup the stack frame
mov ecx, [ebp+8]
xor eax, eax ; loop counter
startLoop:
xor dx, dx
mov dl, byte [ecx+eax]
inc eax
cmp dl, 0x0 ; null byte
jne startLoop
end:
pop ebp
ret
More easy way here(ASCII zero terminated string only):
REPE SCAS m8
http://pdos.csail.mit.edu/6.828/2006/readings/i386/REP.htm
I think your inc should be after the jne. I'm not familiar with this assembly, so I don't really know.