如何在装配NASM中打印数字？

Question

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.

Answer 1

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 .

Answer 2

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, ie add 4 bytes per parameter passed to the function.

Answer 3

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.

Answer 4

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.

Answer 5

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

Answer 6

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