I'm trying to input into my program... All it does is run through and print a '0' to the screen. I'm pretty sure that the PRINTDECI function works, I made it a while ago and it works. Do I just have to loop over the input code and only exit when I enter a certain value? I'm not sure how I would do that... Unless it's by ACSII values which might suck.... Anyways, here's my code (Yasm(nasm clone), Intel Syntax):
GLOBAL _start
SECTION .text
PRINTDECI:
LEA R9,[NUMBER + 18] ; last character of buffer
MOV R10,R9 ; copy the last character address
MOV RBX,10 ; base10 divisor
DIV_BY_10:
XOR RDX,RDX ; zero rdx for div
DIV RBX ; rax:rdx = rax / rbx
ADD RDX,0x30 ; convert binary digit to ascii
TEST RAX,RAX ; if rax == 0 exit DIV_BY_10
JZ CHECK_BUFFER
MOV byte [R9],DL ; save remainder
SUB R9,1 ; decrement the buffer address
JMP DIV_BY_10
CHECK_BUFFER:
MOV byte [R9],DL
SUB R9,1
CMP R9,R10 ; if the buffer has data print it
JNE PRINT_BUFFER
MOV byte [R9],'0' ; place the default zero into the empty buffer
SUB R9,1
PRINT_BUFFER:
ADD R9,1 ; address of last digit saved to buffer
SUB R10,R9 ; end address minus start address
ADD R10,1 ; R10 = length of number
MOV RAX,1 ; NR_write
MOV RDI,1 ; stdout
MOV RSI,R9 ; number buffer address
MOV RDX,R10 ; string length
SYSCALL
RET
_start:
MOV RCX, SCORE ;Input into Score
MOV RDX, SCORELEN
MOV RAX, 3
MOV RBX, 0
SYSCALL
MOV RAX, [SCORE]
PUSH RAX ;Print Score
CALL PRINTDECI
POP RAX
MOV RAX,60 ;Kill the Code
MOV RDI,0
SYSCALL
SECTION .bss
SCORE: RESQ 1
SCORELEN EQU $-SCORE
Thanks for any help!
- Kyle
As a side note, the pointer in RCX goes to a insanely large number according to DDD... So I'm thinking I have to get it to pause and wait for me to type, but I have no idea how to do that...
The 'setup' to call syscall 0 (READ) on x86_64 system is:
#xenon:~$ syscalls_lookup read
read:
rax = 0 (0x0)
rdi = unsigned int fd
rsi = char *buf
rdx = size_t count
So your _start code should be something like:
_start:
mov rax, 0 ; READ
mov rdi, 0 ; stdin
mov rsi, SCORE ; buffer
mov rdx, SCORELEN ; length
syscall
The register conventions and syscall numbers for x86_64 are COMPLETELY different than those for i386.
Some conceptual issues you seem to have:
READ does not do ANY interpretation on what you type, you seem to be expecting it to let you type a number (say, 57) and have it return the value 57. Nope. It'll return '5', '7', 'ENTER', 'GARBAGE'... Your SCORELEN is probably 8 (length of resq 1), so you'll read, AT MOST, 8 bytes. or Characters, if you wish to call them that. And unless you type the EOF char (^D), you'll need to type those 8 characters before the READ call will return to your code.
You have to convert the characters you receive into a value... You can do it the easy way and link with ATOI() in the C library, or write your own parser to convert the characters into a value by addition and multiplication (it's not hard, see code below).
Used below, here as a reference:
#xenon:~$ syscalls_lookup write
write:
rax = 1 (0x1)
rdi = unsigned int fd
rsi = const char *buf
rdx = size_t count
Ugh.... So many... I'll just rewrite bits:
global _start
section .text
PRINTDECI:
; input is in RAX
lea r9, [NUMBER + NUMBERLEN - 1 ] ; + space for \n
mov r10, r9 ; save end position for later
mov [r9], '\n' ; store \n at end
dec r9
mov rbx, 10 ; base10 divisor
DIV_BY_10:
xor rdx, rdx ; zero rdx for div
div rbx : rax = rdx:rax / rbx, rdx = remainder
or dl, 0x30 ; make REMAINDER a digit
mov [r9], dl
dec r9
or rax, rax
jnz DIV_BY_10
PRINT_BUFFER:
sub r10, r9 ; get length (r10 - r9)
inc r9 ; make r9 point to initial character
mov rax, 1 ; WRITE (1)
mov rdi, 1 ; stdout
mov rsi, r9 ; first character in buffer
mov rdx, r10 ; length
syscall
ret
MAKEVALUE:
; RAX points to buffer
mov r9, rax ; save pointer
xor rcx, rcx ; zero value storage
MAKELOOP:
mov al, [r9] ; get a character
or al, al ; set flags
jz MAKEDONE ; zero byte? we're done!
and rax, 0x0f ; strip off high nybble and zero rest of RAX (we're lazy!)
add rcx, rcx ; value = value * 2
mov rdx, rcx ; save it
add rcx, rcx ; value = value * 4
add rcx, rcx ; value = value * 8
add rcx, rdx ; value = value * 8 + value * 2 (== value * 10)
add rcx, rax ; add new digit
jmp MAKELOOP ; do it again
MAKEDONE:
mov rax, rcx ; put value in RAX to return
ret
_start:
mov rax, 0 ; READ (0)
mov rdi, 0 ; stdin
mov rsi, SCORE ; buffer
mov rdx, SCORELEN ; length
syscall
; RAX contains HOW MANY CHARS we read!
; -OR-, -1 to indicate error, really
; should check for that, but that's for
; you to do later... right? (if RAX==-1,
; you'll get a segfault, just so you know!)
add rax, SCORE ; get position of last byte
movb [rax], 0 ; force a terminator at end
mov rax, SCORE ; point to beginning of buffer
call MAKEVALUE ; convert from ASCII to a value
; RAX now should have the VALUE of the string of characters
; we input above. (well, hopefully, right?)
mov [VALUE], rax ; store it, because we can!
; it's stored... pretend it's later... we need value of VALUE!
mov rax, [VALUE] ; get the VALUE
call PRINTDECI ; convert and display value
; all done!
mov rax, 60 ; EXIT (60/0x3C)
mov rdi, 0 ; exit code = 0
syscall
section .bss
SCORE: resb 11 ; 10 chars + zero terminator
SCORELEN equ $-SCORE
NUMBER: resb 19 ; 18 chars + CR terminator
NUMBERLEN equ $-NUMBER
I'm going to say that this should work first time, it's off-the-cuff for me, haven't tested it, but it should be good. We read up to 10 chars, terminate it with a zero, convert to a value, then convert to ascii and write it out.
To be more proper, you should save registers to the stack in each subroutine, well, certain ones, and really, only if you're going to interface with libraries... doing things yourself lets you have all the freedom you want to play with the registers, you just have to remember what you put where!
Yes, someone is going to say "why didn't you just multiply by 10 instead of weird adding?" ... uh... because it's easier on the registers and I don't have to set it all up in rdx:rax. Besides, it's just as readable and understandable, especially with the comments. Roll with it! This isn't a competition, it's learning!
Machine code is fun! Gotta juggle all the eggs in your head though... no help from the compiler here!
Technically, you should check return result (RAX) of the syscalls for READ and WRITE, handle errors appropriately, yadda yadda yadda.... learn to use your debugger (gdb or whatever).
Hope this helps.
Related
I am trying to write skewed Text on Terminal via assembly to get my skills going.
Example [] = spaces :
H
[]E
[][]L
[][][] L
[][][][] O
But I seem to run into a timeout, because I am trying to move a character between registers / get the character via memory address - like getting a char at an index. But it won't really work. I tried working with allocating a byte to increment in the .bss section before, but that lead to a different error which I couldn't work around (relocation truncated to fit: R_X86_64_8 against `.bss')
Maybe there is a better approach to handle this, and if you could help me get on the right path I would be so so happy. To get more context, here is the 'problematic' snippet of my latest code.
inc rcx ; increase pointer of rcx
space_loop:
push rdi ;save destination register
cmp byte [rcx], 0 ;end of string found
je eos ; jump to exit subroutine
mov rax, 4; SYS Write
mov rbx, 0 ; STDOUT
cmp rsp,[rbp] ; compare spaceIndex to String Index
je print_char
mov rdi, space ;print space
mov rdx,1 ; length = 1
syscall
inc byte [rsp] ; increase spaceIndex
jnz space_loop
print_char:
;print out char
mov rax, 4
mov rbx, 0
mov rcx,[rcx] ; <- this line breaks it dont know how to access index here
mov rdx, 1
syscall
call new_Line ; make line break
inc rcx ; increase rcx counter
inc byte [rbp] ; increase index counter
mov byte [rsp], 0 ;reset space counter
jmp space_loop; next Line
I'm trying to learn assembly. I wanted to write a simple program that counted to 20 and printed out the numbers. I know you have to subtract ascii '0' from a ascii representation of a number to turn it into it's digit, but my implementation just refuses to work. I still get 123456789:;<=>?#ABCD
Here is my code.
section .bss
num resb 1
section .text
global _start
_start:
mov eax, '1'
mov ebx, 1 ; Filehandler 1 = stdout
mov ecx, 20 ; The number we're counting to
mov edx, 1 ; Size of a number in bytes
l1:
mov [num], eax ; Put eax into the value of num
mov eax, 4 ; Put 4 into eax (write)
push ecx ; Save ecx on the stack
mov ecx, num ; print num
int 0x80 ; Do the print
pop ecx ; Bring ecx back from the stack
mov eax, [num] ; Put the value of num into eax
sub eax, '0' ; Convert to digit
inc eax ; Increment eax
add eax, '0' ; Convert back to ascii
loop l1
mov eax,1 ; System call number (sys_exit)
int 0x80 ; Call kernel
Can anyone see what the problem is? I'm totally hitting a brick wall. I'm using nasm to compile and ld to link.
I am doing a proj. in 64-bit NASM. I have to convert decimal to binary and binary to decimal.
I keep getting segmentation fault after debugging when i call printf.
extern printf
section .bss
decsave: resd 2 ; stores dec->bin conversion
binsave: resd 1
section .data ; preset constants, writeable
dec1: db '1','2','4','.','3','7','5',0
bin1: dq 01010110110101B ; 10101101.10101 note where binary point should be
ten: dq 10
debug: db "debug 124 is %ld", 10, 0
section .text ; instructions, code segment
global main ; for gcc standard linking
main: ; label
push rbp ; save rbp
;parse and convert integer portion of dec->bin
mov rax,0 ; accumulate value here
mov al,[dec1] ; get first ASCII digit
sub al,48 ; convert ASCII digit to binary
mov rbx,0 ; clear register (upper part)
mov bl,[dec1+1] ; get next ASCII digit
sub rbx,48 ; convert ASCII digit to binary
imul rax,10 ; ignore rdx
add rax,rbx ; increment accumulator
mov rbx,0
mov bl,[dec1+2]
sub rbx,48
imul rax,10
add rax,rbx
mov [decsave],rax ; save decimal portion
mov rdi, debug
mov rsi, [decsave]
mov rax,0
call printf
; return using c-style pops to return stack to correct position
; and registers to correct content
pop rbp
mov rax,0
ret ; return
; print the bits in decsave:
section .bss
abits: resb 17 ; 16 characters & zero terminator
section .data
fmts: db "%s",0
section .text
; shift decimal portion into abits as ascii
mov rax,[decsave] ; restore rax to dec. portion
mov rcx,8 ; for printing 1st 8 bits
loop3: mov rdx,0 ; clear rdx ready for a bit
shld rdx,rax,1 ; top bit of rax into rdx
add rdx,48 ; make it ASCII
mov [abits+rcx-1],dl ; store character
ror rax,1 ; next bit into top of rax
loop loop3 ; decrement rcx, jump non zero
mov byte [abits+7],'.' ; end of dec. portion string
mov byte [abits+8],0 ; end of "C" string
push qword abits ; string to print
push qword fmts ; "%s"
call printf
add rsp,8
mov rax,[decsave+16] ; increment to fractional portion
mov rcx,16 ; for printing 3 bits as required in the directions
loop4: mov rdx,0 ; clear rdx ready for a bit
shld rdx,rax,1 ; top bit of rax into rdx
add rdx,48 ; make it ASCII
mov [abits+rcx-1],dl ; store character
ror rax,1 ; next bit into top of rax
loop loop4 ; decrement rcx, jump non zero
mov byte [abits+3],10 ; end of "C" string at 3 places
mov byte [abits+4],0 ; end of "C" string
push qword abits ; string to print
push qword fmts ; "%s"
call printf
add rsp,8
Is there a any other way to get around it?
Thank you.
As Jester pointed out, if the vararg function is not using sse, then al must be zero. There is a bigger issue here:
With the x86-64 calling convention, parameters are not passed on the stack as they are for 32bit, but instead passed through registers. Which registers all depend on what OS your program is written for.
x86 calling conventions
This is my first attempt in 64-bit assembly under Linux. I am using FASM.
I am converting a 64-bit register hex value to string. It is working fine until it reaches the final digit. I can't figure out exactly what's wrong with my code. Maybe there is something about 64-programming that I don't know or with the syscall (I am a linux noob as well)
format ELF64 executable 3
entry start
segment readable executable
start:
mov rax,3c5677h ;final '7' is not displayed
push rax
call REG
;call line
xor edi,edi ;exit
mov eax,60
syscall
;----------------------------------
REG:push rbp ;stack frame setup
mov rbp,rsp
sub rsp,8 ;space for local char
mov rax,[rbp+16];arg
lea r9,[rsp-8] ;local char
mov rcx,16 ;divisor
mov rsi,16 ;16 hex digits for register
.begin: ;get the digit
xor rdx,rdx ;by division
div rcx ;of 16
push rdx ;from back to front
dec rsi
test rsi,rsi
jz .disp
jmp .begin
.disp: ;convert and display digit
inc rsi
pop rax ;In reverse order
add rax,30h ;convert digit to string
cmp rax,39h ;if alpha
jbe .normal
add rax,7 ;add 7
.normal:
mov [r9],rax ;copy the value
push rsi ;save RSI for syscall
mov rsi,r9 ;address of char
mov edx,1 ;size
mov edi,1 ;stdout
mov eax,1 ;sys_write
syscall
pop rsi ;restore RSI for index
cmp rsi,16
je .done
jmp .disp
.done:
add rsp,8 ;stack balancing
pop rbp
ret
Thanks in advance for your help.
I believe the problem printing the last digit comes from how you load r9. If on entry, rsp was 100. You subtract 8 (rsp = 92), then load r9 with rsp - 8 (r9 = 84). Presumably you meant r9 to 100, so try changing that to:
lea r9, [rsp+8]
For a more efficient solution, how about something more like this (assumes value in rbx):
mov r9, 16 ; How many digits to print
mov rsi, rsp ; memory to write digits to
sub rsp, 8 ; protect our stack
mov edx, 1 ; size is always 1
mov edi, 1 ; stdout is always 1
.disp:
rol rbx, 4 ; Get the next nibble
mov cl, bl ; copy it to scratch
and cl, 15 ; mask out extra bits
add cl, 0x30 ; Convert to char
cmp cl, 0x39 ; if alpha
jbe .normal
add cl, 7 ; Adjust for letters
.normal:
mov [rsi], cl ; copy the value
mov eax, 1 ; sys_write
syscall ; overwrites rcx, rax, r11
dec r9 ; Finished a digit
jnz .disp ; Are we done?
add rsp, 8 ; Done with the memory
Suppose that I have an integer number in a register, how can I print it? Can you show a simple example code?
I already know how to print a string such as "hello, world".
I'm developing on Linux.
If you're already on Linux, there's no need to do the conversion yourself. Just use printf instead:
;
; assemble and link with:
; nasm -f elf printf-test.asm && gcc -m32 -o printf-test printf-test.o
;
section .text
global main
extern printf
main:
mov eax, 0xDEADBEEF
push eax
push message
call printf
add esp, 8
ret
message db "Register = %08X", 10, 0
Note that printf uses the cdecl calling convention so we need to restore the stack pointer afterwards, i.e. add 4 bytes per parameter passed to the function.
You have to convert it in a string; if you're talking about hex numbers it's pretty easy. Any number can be represented this way:
0xa31f = 0xf * 16^0 + 0x1 * 16^1 + 3 * 16^2 + 0xa * 16^3
So when you have this number you have to split it like I've shown then convert every "section" to its ASCII equivalent.
Getting the four parts is easily done with some bit magic, in particular with a right shift to move the part we're interested in in the first four bits then AND the result with 0xf to isolate it from the rest. Here's what I mean (soppose we want to take the 3):
0xa31f -> shift right by 8 = 0x00a3 -> AND with 0xf = 0x0003
Now that we have a single number we have to convert it into its ASCII value. If the number is smaller or equal than 9 we can just add 0's ASCII value (0x30), if it's greater than 9 we have to use a's ASCII value (0x61).
Here it is, now we just have to code it:
mov si, ??? ; si points to the target buffer
mov ax, 0a31fh ; ax contains the number we want to convert
mov bx, ax ; store a copy in bx
xor dx, dx ; dx will contain the result
mov cx, 3 ; cx's our counter
convert_loop:
mov ax, bx ; load the number into ax
and ax, 0fh ; we want the first 4 bits
cmp ax, 9h ; check what we should add
ja greater_than_9
add ax, 30h ; 0x30 ('0')
jmp converted
greater_than_9:
add ax, 61h ; or 0x61 ('a')
converted:
xchg al, ah ; put a null terminator after it
mov [si], ax ; (will be overwritten unless this
inc si ; is the last one)
shr bx, 4 ; get the next part
dec cx ; one less to do
jnz convert_loop
sub di, 4 ; di still points to the target buffer
PS: I know this is 16 bit code but I still use the old TASM :P
PPS: this is Intel syntax, converting to AT&T syntax isn't difficult though, look here.
Linux x86-64 with printf
main.asm
default rel ; make [rel format] the default, you always want this.
extern printf, exit ; NASM requires declarations of external symbols, unlike GAS
section .rodata
format db "%#x", 10, 0 ; C 0-terminated string: "%#x\n"
section .text
global main
main:
sub rsp, 8 ; re-align the stack to 16 before calling another function
; Call printf.
mov esi, 0x12345678 ; "%x" takes a 32-bit unsigned int
lea rdi, [rel format]
xor eax, eax ; AL=0 no FP args in XMM regs
call printf
; Return from main.
xor eax, eax
add rsp, 8
ret
GitHub upstream.
Then:
nasm -f elf64 -o main.o main.asm
gcc -no-pie -o main.out main.o
./main.out
Output:
0x12345678
Notes:
sub rsp, 8: How to write assembly language hello world program for 64 bit Mac OS X using printf?
xor eax, eax: Why is %eax zeroed before a call to printf?
-no-pie: plain call printf doesn't work in a PIE executable (-pie), the linker only automatically generates a PLT stub for old-style executables. Your options are:
call printf wrt ..plt to call through the PLT like traditional call printf
call [rel printf wrt ..got] to not use a PLT at all, like gcc -fno-plt.
Like GAS syntax call *printf#GOTPCREL(%rip).
Either of these are fine in a non-PIE executable as well, and don't cause any inefficiency unless you're statically linking libc. In which case call printf can resolve to a call rel32 directly to libc, because the offset from your code to the libc function would be known at static linking time.
See also: Can't call C standard library function on 64-bit Linux from assembly (yasm) code
If you want hex without the C library: Printing Hexadecimal Digits with Assembly
Tested on Ubuntu 18.10, NASM 2.13.03.
It depends on the architecture/environment you are using.
For instance, if I want to display a number on linux, the ASM code will be different from the one I would use on windows.
Edit:
You can refer to THIS for an example of conversion.
I'm relatively new to assembly, and this obviously is not the best solution,
but it's working. The main function is _iprint, it first checks whether the
number in eax is negative, and prints a minus sign if so, than it proceeds
by printing the individual numbers by calling the function _dprint for
every digit. The idea is the following, if we have 512 than it is equal to: 512 = (5 * 10 + 1) * 10 + 2 = Q * 10 + R, so we can found the last digit of a number by dividing it by 10, and
getting the reminder R, but if we do it in a loop than digits will be in a
reverse order, so we use the stack for pushing them, and after that when
writing them to stdout they are popped out in right order.
; Build : nasm -f elf -o baz.o baz.asm
; ld -m elf_i386 -o baz baz.o
section .bss
c: resb 1 ; character buffer
section .data
section .text
; writes an ascii character from eax to stdout
_cprint:
pushad ; push registers
mov [c], eax ; store ascii value at c
mov eax, 0x04 ; sys_write
mov ebx, 1 ; stdout
mov ecx, c ; copy c to ecx
mov edx, 1 ; one character
int 0x80 ; syscall
popad ; pop registers
ret ; bye
; writes a digit stored in eax to stdout
_dprint:
pushad ; push registers
add eax, '0' ; get digit's ascii code
mov [c], eax ; store it at c
mov eax, 0x04 ; sys_write
mov ebx, 1 ; stdout
mov ecx, c ; pass the address of c to ecx
mov edx, 1 ; one character
int 0x80 ; syscall
popad ; pop registers
ret ; bye
; now lets try to write a function which will write an integer
; number stored in eax in decimal at stdout
_iprint:
pushad ; push registers
cmp eax, 0 ; check if eax is negative
jge Pos ; if not proceed in the usual manner
push eax ; store eax
mov eax, '-' ; print minus sign
call _cprint ; call character printing function
pop eax ; restore eax
neg eax ; make eax positive
Pos:
mov ebx, 10 ; base
mov ecx, 1 ; number of digits counter
Cycle1:
mov edx, 0 ; set edx to zero before dividing otherwise the
; program gives an error: SIGFPE arithmetic exception
div ebx ; divide eax with ebx now eax holds the
; quotent and edx the reminder
push edx ; digits we have to write are in reverse order
cmp eax, 0 ; exit loop condition
jz EndLoop1 ; we are done
inc ecx ; increment number of digits counter
jmp Cycle1 ; loop back
EndLoop1:
; write the integer digits by poping them out from the stack
Cycle2:
pop eax ; pop up the digits we have stored
call _dprint ; and print them to stdout
dec ecx ; decrement number of digits counter
jz EndLoop2 ; if it's zero we are done
jmp Cycle2 ; loop back
EndLoop2:
popad ; pop registers
ret ; bye
global _start
_start:
nop ; gdb break point
mov eax, -345 ;
call _iprint ;
mov eax, 0x01 ; sys_exit
mov ebx, 0 ; error code
int 0x80 ; край
Because you didn't say about number representation I wrote the following code for unsigned number with any base(of course not too big), so you could use it:
BITS 32
global _start
section .text
_start:
mov eax, 762002099 ; unsigned number to print
mov ebx, 36 ; base to represent the number, do not set it too big
call print
;exit
mov eax, 1
xor ebx, ebx
int 0x80
print:
mov ecx, esp
sub esp, 36 ; reserve space for the number string, for base-2 it takes 33 bytes with new line, aligned by 4 bytes it takes 36 bytes.
mov edi, 1
dec ecx
mov [ecx], byte 10
print_loop:
xor edx, edx
div ebx
cmp dl, 9 ; if reminder>9 go to use_letter
jg use_letter
add dl, '0'
jmp after_use_letter
use_letter:
add dl, 'W' ; letters from 'a' to ... in ascii code
after_use_letter:
dec ecx
inc edi
mov [ecx],dl
test eax, eax
jnz print_loop
; system call to print, ecx is a pointer on the string
mov eax, 4 ; system call number (sys_write)
mov ebx, 1 ; file descriptor (stdout)
mov edx, edi ; length of the string
int 0x80
add esp, 36 ; release space for the number string
ret
It's not optimised for numbers with base of power of two and doesn't use printf from libc.
The function print outputs the number with a new line. The number string is formed on stack. Compile by nasm.
Output:
clockz
https://github.com/tigertv/stackoverflow-answers/tree/master/8194141-how-to-print-a-number-in-assembly-nasm