This question already has answers here:
X86 NASM Assembly converting lower to upper and upper to lowercase characters
(5 answers)
X86 Assembly Converting lower-case to uppercase
(1 answer)
Closed 3 years ago.
I want to change the string to all caps, although I am having trouble getting the length of the input. What i have tried so far is moving the address of the message into a registrar then indexing through the string and also increment a counter variable. Then comparing the char in the address to a '.' (signifying the end of the message) and if its found not to be equal it will recall this block of statements. At least this is what I want my code to do. Not sure if this is even the right logic. I know there are alot of errors and its messy but I'm learning so please just focus on my main question. thank you! EDIT: the input i use is 'this is a TEST.'
;nasm 2.11.08
SYS_Write equ 4
SYS_Read equ 3
STDIN equ 0
STDOUT equ 1
section .bss
message resb 15
counter resb 2
section .data
msg1: db 'Enter input (with a period) that I will turn into all capitals!',0xa ;msg for input
len1 equ $- msg1
section .text
global _start
_start:
mov eax, SYS_Write ; The system call for write (sys_write)
mov ebx, STDOUT ; File descriptor 1 - standard output
mov ecx, msg1 ; msg to print
mov edx, len1 ; len of message
int 0x80 ; Call the kernel
mov eax, SYS_Read ;system call to read input
mov ebx, STDIN ;file descriptor
mov ecx, message ;variable for input
mov edx, 15 ;size of message
int 0x80 ;kernel call
mov [counter], byte '0'
getLen:
mov eax, message
add eax, [counter]
inc byte [counter]
cmp eax, '.'
jne getLen
mov eax, SYS_Write ; this is to print the counter to make sure it got the right len
mov ebx, STDOUT
mov ecx, counter
mov edx, 2
int 0x80
jmp end
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
mov eax, [message]
;add eax, counter
cmp eax, 90
jg toUpper
toUpper:
sub eax, 32
mov [message], eax
mov eax, SYS_Write ; The system call for write (sys_write)
mov ebx, STDOUT ; File descriptor 1 - standard output
mov ecx, message ; Put the offset of hello in ecx
mov edx, 10 ; helloLen is a constant, so we don't need to say
int 0x80 ; Call the kernel
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
end:
mov eax,1 ; The system call for exit (sys_exit)
mov ebx,0 ; Exit with return code of 0 (no error)
int 0x80 ;
Related
Task: output the number in hexadecimal form to the console. After that print some string (let it be "String after num").
The first part is successful, but the second is not.
The input number is stored in memory by the num label.
String is stored in memory by the line label.
String length - lines.
Code:
global _start
section .data
num db 01111110b
temp db 0
line db 10, "String after num", 10
lines equ $-line
section .text
_start:
call write_hex ; write num in hex format
mov eax, 4 ; write "Hello world!"
mov ebx, 1 ;
mov ecx, line ;
mov edx, lines ;
int 80H ;
mov eax, 1 ; exit
xor ebx, ebx
int 80H
write_hex:
mov eax, [num]
mov [temp], eax
shr byte [num], 4
call to_hex_digit
call write_digit
mov eax, [temp]
mov [num], eax
and byte [num], 1111b
call to_hex_digit
call write_digit
ret
to_hex_digit:
add [num], byte '0'
cmp [num], byte '9'
jle end
add [num], byte 7
end: ret
write_digit:
mov eax, 4
mov ebx, 1
mov ecx, num
mov edx, 1
int 80H
ret
Output:
Thanks for any help.
I used an invalid register to store a temporary value. I replaced the register which will work with the temp from eax to al. See Jester's answer.
This question already has answers here:
In NASM labels next to each other in memory are printing both strings instead of first one
(1 answer)
How does $ work in NASM, exactly?
(2 answers)
Closed 4 years ago.
running the code below generates a file with Welcome to jj Shashwat as content. what i didn't get is why does it writes Shashwat at the end of the file, Shashwat is in a totally different variable. Any idea why does this happen?
section .text
global _start ;must be declared for using gcc
_start:
;create the file
mov eax, 8
mov ebx, file_name
mov ecx, 0777 ;read, write and execute by all
int 0x80 ;call kernel
mov [fd_out], eax
; close the file
mov eax, 6
mov ebx, [fd_out]
;open the file for reading
mov eax, 5
mov ebx, file_name
mov ecx, 2 ;for read/write access
mov edx, 0777 ;read, write and execute by all
int 0x80
mov [fd_out], eax
; write into the file
mov edx,len ;number of bytes
mov ecx, msg ;message to write
mov ebx, [fd_out] ;file descriptor
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
; close the file
mov eax, 6
mov ebx, [fd_out]
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
file_name db 'myfile.txt', 0
msg db 'Welcome to jj', 0
mgafter db ' Shashwat', 0
lntwo equ $-mgafter
len equ $-msg
section .bss
fd_out resb 1
fd_in resb 1
info resb 26
That's because you said len equ $-msg after defining both msg and msgafter, so len is set to the length of both msg and msgafter, making your write call write both strings. This is because len equ $-msg means “set len to be the difference between the current location ($) and the location of msg.”
To fix this, move the len equ $-msg line right after the definition of msg.
I am trying to make a program in x86 NASM that reads two integers from stdin and then does something with them (greatest common divisor, but thats not the question here). The question I have, is now that I have the two input integers, how do I go about turning them into actual integers to use as such? Here is the code I use to grab the strings of digits
SECTION .data
prompt: db "Enter a positive integer: "
plen: equ $-prompt
SECTION .bss
inbuf: resb 20
inbuf2: resb 20
SECTION .text
global _start
_start:
; Here I am grabbing the two integers to use in the function
nop
mov eax, 4 ; write
mov ebx, 1 ; to standard output
mov ecx, prompt ; the prompt string
mov edx, plen ; of length plen
int 80H ; interrupt with the syscall
mov eax, 3 ; read
mov ebx, 0 ; from standard input
mov ecx, inbuf ; into the input buffer
mov edx, 30 ; upto 30 bytes
int 80H ; interrupt with the syscall
mov eax, 4 ; write
mov ebx, 1 ; to standard output
mov ecx, prompt ; the prompt string
mov edx, plen ; of length plen
int 80H ; interrupt with the syscall
mov eax, 3 ; read
mov ebx, 0 ; from standard input
mov ecx, inbuf2 ; into the input buffer
mov edx, 30 ; upto 30 bytes
int 80H ; interrupt with the syscall
section .bss
num1 resb 4
result resb 4
section .data
SYS_EXIT equ 1
SYS_READ equ 3
SYS_WRITE equ 4
STDIN equ 0
STDOUT equ 1
INCREMENT equ 1
msg1 db 'Please enter an integer here:',0xA
len1 EQU $- msg1
msg2 db 'Your integer after being incremented is:',0xA
len2 EQU $- msg2
section .text
global _start:
_start:
mov eax, SYS_WRITE ; Prompting user to enter a number.
mov ebx, STDOUT
mov ecx, msg1
mov edx, len1
int 0x80
mov eax, SYS_READ ; Reading users number.
mov ebx, STDOUT
mov ecx, num1
mov edx, 4
int 0x80
mov eax, SYS_WRITE ; Writing second message.
mov ebx, STDOUT
mov ecx, msg2
mov edx, len2
int 0x80
mov eax, [num1] ; incrementing the user's number.
mov ebx, INCREMENT
add eax, ebx
mov [result], eax
mov eax, SYS_WRITE ; Printing out incremented number.
mov ebx, STDOUT
mov ecx, result
mov edx, 4
int 0x80
Just started learning basic assembly code today and I'm trying to increment user input numbers by 1. I am getting weird results when the number is any bigger than a single digit. What's the difference?
The problem is that your treat your input as numbers and not as strings. When you receive your input (string), you get an (ASCII) string as your return value. Increasing the first char(digit) by add eax, INCREMENT (compacted) does work, because the first ASCII char value of [num] contained in AL(lowest byte of EAX) is increased by one. This happens because the ASCII value of 0 is 48 and the ASCII value of 9 is 57.
If you think you increase these 'digits', you would in fact increase their ASCII values, which works well until you reach the one-digit-limit(0-9). Increasing 9 by one would result in : and not 10, as your may have expected because the ASCII value of 9 is 57 and adding one to it results in 58, which is the ASCII value of :.
So you would have to 'normalize' your 'numbers' before you operate on them with integer arithmetic. Transform them from ASCII strings to integer values.
Luckily for you, others have done this before you and optimized these approaches to the practical maximum. Look and search for atoi(ASCII to integer) and itoa(integer to ASCII). Then surround your arithmetical operations
mov eax, [num1] ; incrementing the user's number.
mov ebx, INCREMENT
add eax, ebx
mov [result], eax
with these functions.
I am trying to print a single digit integer in nasm assembly on linux. What I currently have compiles fine, but nothing is being written to the screen. Can anyone explain to me what I am doing wrong here?
section .text
global _start
_start:
mov ecx, 1 ; stores 1 in rcx
add edx, ecx ; stores ecx in edx
add edx, 30h ; gets the ascii value in edx
mov ecx, edx ; ascii value is now in ecx
jmp write ; jumps to write
write:
mov eax, ecx ; moves ecx to eax for writing
mov eax, 4 ; sys call for write
mov ebx, 1 ; stdout
int 80h ; call kernel
mov eax,1 ; system exit
mov ebx,0 ; exit 0
int 80h ; call the kernel again
This is adding, not storing:
add edx, ecx ; stores ecx in edx
This copies ecx to eax and then overwrites it with 4:
mov eax, ecx ; moves ecx to eax for writing
mov eax, 4 ; sys call for write
EDIT:
For a 'write' system call:
eax = 4
ebx = file descriptor (1 = screen)
ecx = address of string
edx = length of string
After reviewing the other two answers this is what I finally came up with.
sys_exit equ 1
sys_write equ 4
stdout equ 1
section .bss
outputBuffer resb 4
section .text
global _start
_start:
mov ecx, 1 ; Number 1
add ecx, 0x30 ; Add 30 hex for ascii
mov [outputBuffer], ecx ; Save number in buffer
mov ecx, outputBuffer ; Store address of outputBuffer in ecx
mov eax, sys_write ; sys_write
mov ebx, stdout ; to STDOUT
mov edx, 1 ; length = one byte
int 0x80 ; Call the kernel
mov eax, sys_exit ; system exit
mov ebx, 0 ; exit 0
int 0x80 ; call the kernel again
From man 2 write
ssize_t write(int fd, const void *buf, size_t count);
In addition to the other errors that have been pointed out, write() takes a pointer to the data and a length, not an actual byte itself in a register as you are trying to provide.
So you will have to store your data from a register to memory and use that address (or if it's constant as it currently is, don't load the data into a register but load its address instead).