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Printing floating point numbers from x86-64 seems to require %rbp to be saved
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Why does printf still work with RAX lower than the number of FP args in XMM registers?
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Why does eax contain the number of vector parameters?
(1 answer)
Closed 6 days ago.
I have been trying to call printf from x64 asm, but it prints garbage
Here is the code:
;print all fib numbers from 0 until 100
global main
extern printf
;printf notes
;mov al 0 is neccesary
;rdi: format string
;rsi, rdx, r8... args
section .text
main:
mov r8, 0 ;counter
mov r9, 0 ;nth value
mov r10, 1;n + 1 value
mov al, 0
Loop:
mov rsi, r9
mov rdx, r10
mov rdi, format_str
call printf
mov r11, r10
add r10, r9
mov r9, r11
inc r8
cmp r8, 100
jl Loop
mov rax, 60
mov rdi, 0
syscall
section .rodata
format_str: db "%3d: %30d\\n", 0
Note the mov al. If it is inside the loop, the program prints garbage. But if it is outside the loop, then the program segfaults.
I compile with gcc and I simply want it to output as it would if it were called from c
Related
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Garbage in string output function
(1 answer)
can't compare user input with number, nasm elf64
(2 answers)
Closed 3 years ago.
This question is similar to others, but it is not a duplicate. I am learning NASM (assembly).
My program has the following goal. I give as input a word (e.g "aab"). I want to compare each char in the string to a predefined value.
Here is a snippet.
;try to compare
xor rax,rax ;set iteration index to 0
lea rsi, [userpass]
back_to_loop:
cmp rsi, 'a'
je _exit
inc rsi ;increase index
jmp back_to_loop
!!!A good help would be editing the program in such a way, that the iteration would stop as soon as it finds an 'a' in the string. If I found how that is done, I could modify it to my own needs.
My entire effort so far:
section .text
global _start
_start:
;read input
xor rax, rax
mov rdi, rax
mov rsi, userpass
mov rdx, rax
add rdx, 0x64 ; 100
syscall
;try to compare
xor rax,rax ;set iteration index to 0
lea rsi, [userpass]
back_to_loop:
cmp rsi, 'a'
je _exit
inc rsi ;increase index
jmp back_to_loop
_exit:
mov rax, 1
xor rbx, rbx
int 80h
section .bss
num resb 64
section .data
userpass times 100 db 0 ;input
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Why do x86-64 Linux system calls modify RCX, and what does the value mean?
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What are the calling conventions for UNIX & Linux system calls (and user-space functions) on i386 and x86-64
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Closed 5 years ago.
I am trying to write a program that prints an integer, the integer is stored in rax register, I don't know why it only prints the first digit of the integer, in this case it's 3.
The program divides rax by 10 and stores the remainder+48 in the reserved memory to display it later. we repeat this as long as rax is different than 0.
code:
section .bss
integer resb 100
section .text
global _start
_start:
mov rax, 321
call _printRAX
mov rax, 60
mov rdi, 0
syscall
_printRAX:
mov rbx, 10
mov rcx, integer
mov [rcx], rbx
_loop1:
inc rcx
mov rdx, 0
div rbx
add rdx, 48
mov [rcx], dl
cmp rax, 0
jne _loop1
_loop2:
mov rax, 1
mov rdi, 1
mov rsi, rcx
mov rdx, 1
syscall
dec rcx
cmp rcx, integer
jge _loop2
ret
output:
3
Try It Online
This question already has answers here:
Why should EDX be 0 before using the DIV instruction?
(2 answers)
How do I print an integer in Assembly Level Programming without printf from the c library? (itoa, integer to decimal ASCII string)
(5 answers)
Closed 2 years ago.
Previous version of this question (that question originally had a different problem, even though the code there is now the same as the code in this question)
I am trying to make code to display a number on console in Linux 64 bit NASM, without the use of c/c++ functions (pure assembly). The code compiles and links fine but it will not give output...
It displays just newline for some time and then displays '7' forever. I am new to Assembly so I don't know what is wrong. Please help... Here is the code:
section .data
num: dq 102 ;my default number to get the reverse of (for now)
nl: db 0x0a
nlsize: equ $-nl
ten: dq 10
section .bss
rem: resq 1
remsize: equ $-rem
section .text
global _start
_start:
cmp qword [num], 0
jng _exit ;jump to _exit if num is not greater than 0
mov rax, [num] ;move the number to rax
mov rbx, [num] ;move the number to rbx as well so that i have original number in register to subtract and get the remainder
mov rcx, [ten] ;move 10 to rcx to be the divisor
div rcx ;divide number in rax by 10
mov [num], rax ;get the quotient to get the remaining number for quotient
mul rcx ;multiply number in rax by 10
sub rbx, rax ;subtract rbx - rax and store the value in rax (right?)
mov [rem], rbx ;get the remainder from rax. this must be done right after div (WHY??????????)
call _disprem ;call _disprem to display the remainder... call returns the flow back to the caller right?
jmp _start ;get to the loop again
_exit:
mov rax, 60
mov rdi, 0
syscall
_newl:
mov rax, 1
mov rdi, 1
mov rsi, nl
mov rdx, nlsize
syscall
ret
_disprem:
mov rax, 1
mov rdi, 1
add qword [rem], 0x0000000000000030 ;since the rem variable is quadword (64 bit)
mov rsi, rem ;for getting ascii value (48 is ascii 0 in decimal) to convert the rem to character
mov rdx, remsize
syscall
sub qword[rem], 0x0000000000000030 ;get me my original number back plz thanks
call _newl
ret
This question already has answers here:
Displaying a number in x86-64 assembly with only Linux system calls [duplicate]
How do I print an integer in Assembly Level Programming without printf from the c library? (itoa, integer to decimal ASCII string)
(5 answers)
Closed 2 years ago.
I am learning assembly on Linux (NASM) x64 machine (I don't have access to 32 or 16 bit machine), and I am trying to display number on screen (reverse of number according to code but that's a start).
Number is predefined in section .data -> num.
I am quite a newbie at assembly programming and due to the lack of material on x64 assembly (really, cant find much, and all I was able to find was quite confusing) I am unable to resolve the issue.
The issue is that the code compiles an links with no errors/warnings, but it just displays some spaces (not even newline). If I remove the call _newl code from _disprem, those spaces are also gone. There is not even segment fault or something.
By the way, algorithm to get the remainder (to get the digits in a number) is num - (num / 10) * 10
section .data
num: dq 102 ;my default number to get the reverse of (for now)
nl: db 0x0a
nlsize: equ $-nl
ten: dq 10
section .bss
rem: resq 1
remsize: equ $-rem
section .text
global _start
_start:
cmp qword [num], 0
jng _exit ;jump to _exit if num is not greater than 0
mov rax, [num] ;move the number to rax
mov rbx, [num] ;move the number to rbx as well so that i have original number in register to subtract and get the remainder
mov rcx, [ten] ;move 10 to rcx to be the divisor
div rcx ;divide number in rax by 10
mov [num], rax ;get the quotient to get the remaining number for quotient
mul rcx ;multiply number in rax by 10
sub rbx, rax ;subtract rbx - rax and store the value in rax (right?)
mov [rem], rbx ;get the remainder from rax. this must be done right after div (WHY??????????)
call _disprem ;call _disprem to display the remainder... call returns the flow back to the caller right?
jmp _start ;get to the loop again
_exit:
mov rax, 60
mov rdi, 0
syscall
_newl:
mov rax, 1
mov rdi, 1
mov rsi, nl
mov rdx, nlsize
syscall
ret
_disprem:
mov rax, 1
mov rdi, 1
add qword [rem], 0x0000000000000030 ;since the rem variable is quadword (64 bit)
mov rsi, rem
mov rdx, remsize
syscall
sub qword [rem], 0x0000000000000030 ;get me my original number back plz thanks
call _newl
ret
This question already has an answer here:
What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code?
(1 answer)
Closed 4 years ago.
; NASM
push 30 ; '0'
mov rax, 4 ; write
mov rbx, 1 ; stdout
mov rcx, rsp ; ptr to character on stack
mov rdx, 1 ; length of string = 1
int 80h
The code above does not print anything to stdout. It works when i give it a ptr to a character in section .data. What am i doing wrong?
amd64 uses a different method for system calls than int 0x80, although that might still work with 32-bit libraries installed, etc. Whereas on x86 one would do:
mov eax, SYSCALL_NUMBER
mov ebx, param1
mov ecx, param2
mov edx, param3
int 0x80
on amd64 one would instead do this:
mov rax, SYSCALL_NUMBER_64 ; different from the x86 equivalent, usually
mov rdi, param1
mov rsi, param2
mov rdx, param3
syscall
For what you want to do, consider the following example:
bits 64
global _start
section .text
_start:
push 0x0a424242
mov rdx, 04h
lea rsi, [rsp]
call write
call exit
exit:
mov rax, 60 ; exit()
xor rdi, rdi ; errno
syscall
write:
mov rax, 1 ; write()
mov rdi, 1 ; stdout
syscall
ret
30 decimal is the code of the ASCII "record separator". Whatever that is, it's probably not a printable character.
30 hexadecimal (30h or 0x30 in NASM parlance), on the other hand, is the code of the ASCII "0".
Also, you need to use the 64-bit ABI.