This question already has answers here:
Read and print user input with x86 assembly (GNU/Linux)
(2 answers)
X86 read from stdin and write to stdout without referring the standard library
(1 answer)
Linux NASM detect EOF
(1 answer)
Closed 2 years ago.
section .bss ; you can change
user_input: resb 256 ;256 byte of memory is reserved and can be accessed through the name user_input
user_input_size equ $- user_input
section .data ;you cant change
output: db "Enter your print statement: "
output_size equ $- output
section .text ;assembly code
global _start
_start:
mov eax, 4
mov ebx, 1
mov ecx, output
mov edx, output_size
int 80
mov eax, 3
mov ebx, 0
mov ecx, user_input
mov edx, user_input_size
int 80
push ecx ; not sure about this
push edx ; not sure about this
mov eax, 4
mov ebx, 0
pop ecx
pop edx
int 80
mov eax, 1
mov ebx, 0
int 80
i want to print out "Enter your print statement: ", have the user input a statement and have that inputted statement returned to the screen. The first part is working. The others not so much. When the user enters a statement is it stored eax or ecx(user_input).
Related
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 ;
This question already has answers here:
Infinite loop after cin c++
(1 answer)
Timeout error C++
(2 answers)
Closed 4 years ago.
My assignment is to ask the user for input and the output would return their name.
Here is my code :
section .data
user: db 'Hello.What is your name ? '
user_L: equ $-user ;length of the UserNameMsg segment
hello: db ' Hello'
hello_L: equ $-hello
section .bss
user_v: resb 23
section .text
global _start
_start:
mov eax, 4
mov ebx, 1
mov ecx, user
mov edx, user_L
int 0x80
mov eax, 3
mov ebx, 0
mov ecx, user
mov edx, 23
int 0x80
mov eax, 4
mov ebx, 1
mov ecx, hello
mov edx, hello_L
int 0x80
mov eax,4
mov ebx, 1
mov ecx, user_v
mov edx, 23
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
When i ran my code in JDoodle's online assembler I get the error saying:
Timeout - Some common reasons for Timeout Your Program may have a endless loop
I'm not sure what to do to fix this error.
This question already has answers here:
What are file descriptors, explained in simple terms?
(13 answers)
Closed 4 years ago.
SEGMENT .data ; nothing here
SEGMENT .text ; sauce
global _start
_start:
pop ECX ; get ARGC value
mov EAX, 4 ; sys_write()
mov EBX, 1 ; /dev/stdout
;^^^^^^^^^^^
mov EDX, 1 ; a single byte
int 0x80
mov EAX, 1 ; sys_exit()
mov EBX, 0 ; return 0
int 0x80
SEGMENT .bss ; nothing here
why mov EBX, ""1"" ; /dev/stdout ?
where document I can find "1" ?
On unix systems, a process normally have 3 common I/O channels connected to it, called stdin/stdout/stderr. These have the corresponding file descriptor values 0, 1 and 2.
This is documented at: http://pubs.opengroup.org/onlinepubs/9699919799/functions/stdin.html (See also the wikipedia page )
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 started learning how to write programs using the NASM assembly programming language. I wrote this simple program that prompts the user to enter two numbers and then adds the two operands together. I got it to compile with no errors or warnings, but when it prompts the user for the two numbers and it begins to add the two numbers it prints out segmentation fault and program ends. I know a segmentation fault is the equivalent to an access reading / writing violation exception in the Win32 world. But, because I don't know how to debug NASM code; I can't figure out what is wrong. I suspect it has to do with an invalid pointer; but I don't know. Here is the code below:
section .data
msg1: db 'Please Enter A Number: ', 0
length1: equ $ - msg1
msg2: db 'Please Enter A Second Number: ', 0
length2: equ $ - msg2
section .bss
operand1: resb 255
operand2: resb 255
answer: resb 255
section .text
global _start
_start:
; Print first message
mov eax, 4
mov ebx, 1
mov ecx, msg1
mov edx, length1
int 80h
; Now read value
mov eax, 3
mov ebx, 1
mov ecx, operand1
mov edx, 255
int 80h
; Print second message
mov eax, 4
mov ebx, 1
mov ecx, msg2
mov edx, length2
int 80h
; Now read second value
mov eax, 3
mov ebx, 1
mov ecx, operand2
mov edx, 255
int 80h
; Now add operand1 and operand2 and print answer
mov eax, 4
mov ebx, 1
xor ecx, ecx ; Make the ecx register 0
mov ecx, operand1
add ecx, operand2
mov edx, 510
int 80h
(Aside: you should be reading from STDIN_FILENO = 0, not STDOUT_FILENO = 1. Also, you're writing a NUL character and you shouldn't.)
The problem is that operand1 and operand2 are addresses to memory locations holding characters you've read. When you add them, you get a pointer to invalid memory.
You'll have to convert them to integers, add them, and convert back to a string, before you can write it out.
Value in ecx is an address of string that is to be printed when you call int 80h. Last part does not make sense
mov eax, 4
mov ebx, 1
xor ecx, ecx ; Make the ecx register 0
mov ecx, operand1
add ecx, operand2 ; **<<< invalid memory address now in ECX !!!**
mov edx, 510
int 80h
because you are adding address of string operand1 and address of string operand2 and trying to print whatever is located ant resulting address which is most likely points to nowhere.
To debug your program with gdb you can do:
nasm -f elf64 -g -l q1.lst q1.asm
gcc -o q1 q1.o
I replaced the "_start" with "main" so that gcc won't complain, and you can skip the 64 in "-f elf64" if you are building on 32 bit platform.
gdb q1
Here is an example f gdb session:
(gdb) br main
Breakpoint 1 at 0x4004d0: file q1.asm, line 20.
(gdb) r
Starting program: /home/anonymous/Projects/asm/q1
Breakpoint 1, main () at q1.asm:20
20 mov eax, 4
(gdb) n
21 mov ebx, 1
(gdb) n
22 mov ecx, msg1
(gdb) n
23 mov edx, length1
(gdb) p msg1
$1 = 1634036816
(gdb)