We've learned only a touch (by touch I mean nothing in class, only theoretical talk of how registers work) of assembly in school and my professor wants us to do something in assembly a little harder than adding two integers. I've been doing a bunch of research and have come up with some code. I am using Microsoft Visual Code on a Linux VM. From my understanding, the code puts input1 into ECX, then puts input2 into ECX, then this essentially appends IF the length of input1 is at least 10 bytes. Not entirely sure how this is working. If I use less than 10 bytes for the first input I get the next input printed on a new line.
I've been trying to understand how [variable] and variable, value vs. address if I understand correctly, fit into this and have played around with reserved byte variables. I'm not asking for my project to be done, but I feel like I'm blindly trying this as opposed to without direction. Thanks in advance guys!
section .text
global main ;must be declared for using gcc
extern printf ;C library used to print
extern exit ;C library used to exit
extern scanf ;C library used to input user entry
main: ;tell linker entry point, called main due to C library usage
;Reads prompt message
mov eax, 4 ;system call number (sys_write)
mov ebx, 1 ;file descriptor (stdout)
mov ecx, prompt1 ;message to write called prompt
mov edx, len1 ;message length
int 0x80 ;call kernel
;Accept user input
mov eax, 3 ;system call number (sys_read)
mov ebx, 0 ;file descriptor (stdin)
mov ecx, input1 ;variable to read into, i.e. input
mov edx, 25 ;input length
int 0x80 ;call kernel
;Reads prompt message
mov eax, 4 ;system call number (sys_write)
mov ebx, 1 ;file descriptor (stdout)
mov ecx, prompt2 ;message to write called prompt
mov edx, len2 ;message length
int 0x80 ;call kernel
;Accept user input
mov eax, 3 ;system call number (sys_read)
mov ebx, 0 ;file descriptor (stdin)
mov ecx, input2 ;variable to read into, i.e. input
mov edx, 25 ;input length
int 0x80 ;call kernel
;cld
mov al, 0
mov ecx, 50
mov edi, input3
repne scasb
dec edi
mov ecx, 2
mov esi, input4
rep movsb
int 0x80
;Display user input
mov eax, 4 ;system call number (sys_write)
mov ebx, 1 ;file descriptor (stdout)
mov ecx, input1 ;message to write
mov edx, 50 ;message length
int 0x80 ;call kernel
call exit
section .data
prompt1 db 'Please enter your first string: ', 0xa ;string to print for user input
len1 equ $ - prompt1 ;length of string prompt
prompt2 db 'Please enter your second string: ',0xa ;string to print for user input
len2 equ $ - prompt2 ;length of string prompt
input3: times 10 db 0 ;variable input is of 10 bytes, null string terminated
input4: times 1 db 0
section .bss
input1 resb 50
input2 resb 25
Related
This question already has answers here:
X86 NASM Assembly converting lower to upper and upper to lowercase characters
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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 ;
I'm trying to find whether a given number (Input by user) is even or odd.
I'm simply applying AND operation on binary digits of a no. with 1, If the number is odd then operation will result 0 and we will Output Number is odd, otherwise we will output Number is even.
Although logic seems simple, But it's not working in the below code. I'm not getting where is the problem in the code. Can anybody tell me where is the problem
section .data
userMsg db 'Please enter a number'
lenuserMsg equ $ - userMsg
even_msg db 'Even Number!'
len1 equ $ - even_msg
odd_msg db 'Odd Number!'
len2 equ $ - odd_msg
section .bss
num resb 5 ;Reserved 5 Bytes for Input
section .text
global _start ;must be declared for linker (gcc)
_start:
;User Prompt
mov ebx, 1 ;file descriptor (stdout)
mov ecx, userMsg ;message to write 'Please enter a number'
mov edx, lenuserMsg ;message length
mov eax, 4 ;system call number (sys_write)
int 0x80 ;call kernel
;Taking user input
mov ebx, 0 ;(stdin)
mov ecx, num
mov edx, 5 ;i/p length
mov eax, 3 ;system call number (sys_read)
int 0x80 ;call kernel
mov ax, [num]
and ax, 1
jz evnn ;Jump on Even
;Printing No. is Odd
mov ebx, 1 ;file descriptor (stdout)
mov ecx, odd_msg ;message to write 'Odd Number!'
mov edx, len2 ;message length
mov eax, 4 ;system call number (sys_write)
int 0x80 ;call kernel
jmp outprog ;Jump to exit
;Printing No. is Even
evnn:
mov ebx, 1 ;file descriptor (stdout)
mov ecx, even_msg ;message to write 'Even Number!'
mov edx, len1 ;message length
mov eax, 4 ;system call number (sys_write)
int 0x80 ;call kernel
;Exit
outprog:
mov eax, 1 ;system call number (sys_exit)
int 0x80 ;call kernel
Just focus on the real problem at hand, shall we? If say an ASCII char is put in AL register, just turn it into a digit and the rest should just be natural. In computing (binary numbers and systems), integers oddness or evenness is determined by the bit 0. If it is 1, it is an odd number. If it is 0, it is an even number. (I am surprised that nobody has specifically put enough emphasis on this thus far).
... ;OS puts a char in AL.
sub al,30h ;turn an ASCII char to one integer digit
shr al,1 ;Lets see how the flags responds below
jc .odd ;CF is set if the first bit (right-most, bit 0) is 1.
;do Even things
;skip pass .odd
.odd:
;do Odd things
Your code does not work because when you ask the user for a number, you read in an ASCII encoded string. You will need to call atoi (ASCII to INT) first to convert the string to a "real" number as computers see it. atoi is included in glibc.
extern atoi
push eax ; pointer to your string to be converted, eg '123'
call atoi
; now eax contains your number, 123
You can also do a bit test on the least significant bit (bit 0) to find out if it is even or odd:
mov al, 01000_1101b
bt al, 0 ; copies the bit to the Carry Flag
jc its_odd ; jump if CF==1
; else - it's even (CF==0)
What BT does, it copies the bit to CF and you can do conditional jumps based on that.
mov ax, [num] loads the first 2 digits of the user's input string, and you're testing the first one. So you're actually testing whether the first character's ASCII code is even.
2 is a factor of 10, so you only need to test the low bit of the last decimal digit to determine if a base-10 number is even or odd.
And since the ASCII code for '0' is 0x30, you can just test the low bit of the last ASCII character of the string.
You don't need to call atoi() unless you need to test n % 3 or some other modulus that isn't a factor of 10. (i.e. you can test n % 2, n % 5, and n % 10 by looking at only the last digit). Note that you can't just test the low 2 bit of the low decimal digit to check for a multiple of 4, because 10 is not a multiple of 4. e.g. 100%4 = 0, but 30%4 = 2.
So, given a pointer + length, you can use TEST byte [last_char], 1 / jnz odd. e.g. after your sys_read, you have a pointer to the string in ECX, and the return value (byte count) in EAX.
;Taking user input
mov ebx, 0 ;(stdin)
mov ecx, num
mov edx, 5 ;i/p length
mov eax, 3 ;system call number (sys_read)
int 0x80 ;call kernel
; Now we make the unsafe assumption that input ended with a newline
; so the last decimal digit is at num+eax-1.
; now do anything that is common to both the odd and even branches,
; instead of duplicating that in each branch.
Then comes the actual test for odd/even: Just one test&branch on the last ASCII digit:
; We still have num in ECX, because int 0x80 doesn't clobber any regs (except for eax with the return value).
test byte [ecx + eax - 1], 1
jnz odd
`section .bss
num resb 1
section .data
msg1 db'enter a number',0xa
len1 equ $-msg1
msg2 db' is even',0xa
len2 equ $-msg2
msg3 db'is odd',0xa
len3 equ $-msg3
section .text
global _start
_start:
mov edx,len1
mov ecx,msg1
mov ebx,1
mov eax,4
int 80h
mov ecx,num
mov ebx,0
mov eax,3
int 80h
mov al,[num]
add al,30h
and al,1
jz iseven
jmp isodd
isodd:
mov edx,len3
mov ecx,msg3
mov ebx,1
mov eax,4
int 80h
jmp exit
iseven:
mov edx,len2
mov ecx,msg2
mov ebx,1
mov eax,4
int 80h
jmp exit
exit:
mov eax,1
int 80h`
I wrote this code, it reads the data from user but did not display the output. It is written in Assembly language. I am new to Assembly language. Can somebody please help me in solving this. I shall be very thankful. Thanks in advance. Here is the code:
section .data ;Data segment
userMsg db 'Please enter a number: ' ;Ask the user to enter a number
lenUserMsg equ $-userMsg ;The length of the message
dispMsg db 'You have entered: '
lenDispMsg equ $-dispMsg
section .bss ;Uninitialized data
num resb 5
section .text ;Code Segment
global _start
_start:
;User prompt
mov eax, 4
mov ebx, 1
mov ecx, userMsg
mov edx, lenUserMsg
int 80h
;Read and store the user input
mov eax, 3
mov ebx, 2
mov ecx, num
mov edx, 5 ;5 bytes (numeric, 1 for sign) of that information
int 80h
;Output the message 'The entered number is: '
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Output the number entered
mov eax, 4
mov ebx, 1
mov ecx, num
mov edx, 5
int 80h
; Exit code
mov eax, 1
mov ebx, 0
int 80h
In typical environments, file descripter 0 stands for standard input, 1 for standard output, and 2 for standard error output.
Reading from standard error output makes no sense for me.
Try changing the program for reading
;Read and store the user input
mov eax, 3
mov ebx, 2
mov ecx, num
mov edx, 5 ;5 bytes (numeric, 1 for sign) of that information
int 80h
to
;Read and store the user input
mov eax, 3
mov ebx, 0
mov ecx, num
mov edx, 5 ;5 bytes (numeric, 1 for sign) of that information
int 80h
in order to have the system read some data from standard input.
section .data
out1: db 'Enter the number:'
out1l: equ $-out1
out2: db 'The number you entered was:'
out2l: equ $-out2
section .bss
input: resb 4
section .text
global _start
_start:
;for displaying the message
mov eax,4
mov ebx,1
mov ecx,out1
mov edx,out1l
int 80h
;for taking the input from the user
mov eax,3
mov ebx,0
mov ecx,input
mov edx,4
int 80h
;for displaying the message
mov eax,4
mov ebx,1
mov ecx,out2
mov edx,out2l
int 80h
;for displaying the input
mov eax,4
mov ebx,1
mov ecx,input
mov edx,4
int 80h
mov eax,1
mov ebx,100
int 80h
Going from high to low lever language I got to assembly. Now at the very beginning, I wrote a simple age program (I'm not clear how to get system time yet so I just used another input). I get Segmentation Fault (core dumped) after I enter the final input. Here is my code:
section .text
global _start
_start:
mov edx, lenask
mov ecx, ask
mov ebx, 1
mov eax, 4
int 0x80
mov edx, 5
mov ecx, input
mov ebx, 2
mov eax, 3
int 0x80
mov edx, lenask2
mov ecx, ask2
mov ebx, 1
mov eax, 4
int 0x80
mov edx, 5
mov ecx, input2
mov ebx, 2
mov eax, 3
int 0x80
mov eax, input2
mov ebx, input
sub eax, ebx
push eax
mov edx, lenanswer
mov ecx, answer
mov ebx, 1
mov eax, 4
int 0x80
pop eax
mov edx, 7
mov ecx, eax
mov ebx, 1
mov eax, 4
int 0x80
section .data
ask db "What is your age?"
lenask equ $-ask
ask2 db "What is today's year?"
lenask2 equ $-ask2
answer db "The age you were born was: "
lenanswer equ $-answer
section .bss
input resb 5
input2 resb 5
An example of what happens:
What is your age?45
What is today's year?2015
The age you were born was: Segmentation fault
It should have done:
What is your age?45
What is today's year?2015
The age you were born was: 1970
The problem is that int 0x80 with eax set to 4 calls the kernel's sys_write function (i.e. a write system call) which expects a pointer to a string. By providing a integer to the function call the kernel will think that the integer is a pointer to a memory location. Because 1970 is not a valid pointer it will raise a -EFAULT. To bypass this you need to code a ToString function to convert the number to a string and then pass the pointer to the converted string.
The Segmentation Fault itself is caused by not having a sys_exit call. The reason for this is that the program will continue past the end of your code (usally into a bunch of 00 00)
I want to add two-digit numbers in NASM(Linux). To add two simple numbers, I use the following code:
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov eax,'3'
sub eax, '0'
mov ebx, '4'
sub ebx, '0'
add eax, ebx
add eax, '0'
mov [sum], eax
mov ecx,msg
mov edx, len
mov ebx,1 ;file descriptor (stdout)
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
mov ecx,sum
mov edx, 1
mov ebx,1 ;file descriptor (stdout)
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
msg db "The sum is:", 0xA,0xD
len equ $ - msg
segment .bss
sum resb 1
The result of the code is 7.But when I carry number 17 in register eax forexample the result is not correct.In this case 5.Tell me please what is the problem? Thank you!
Here's your example with a little bit of cleaning up to help make it easier to read.
Suggestion: this kind of consistency will greatly improve your public image.
But hey; nice commenting, I could read your code and understand it (which is why I decided to answer you)
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov eax, '3'
sub eax, '0'
mov ebx, '4'
sub ebx, '0'
add eax, ebx
add eax, '0'
mov [sum], eax
mov ecx, msg
mov edx, len
mov ebx, 1 ;file descriptor (stdout)
mov eax, 4 ;system call number (sys_write)
int 0x80 ;call kernel
mov ecx, sum
mov edx, 1
mov ebx, 1 ;file descriptor (stdout)
mov eax, 4 ;system call number (sys_write)
int 0x80 ;call kernel
mov eax, 1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
msg db "The sum is:", 0xA,0xD
len equ $ - msg
segment .bss
sum resb 1
Okay now, as for your comment, "...But when I carry number 17 in register eax forexample the result is not correct."
I can imagine !
Question, when you "...carry number 17 in register eax..." are you doing it like this ?...
Mov Eax,"17"
If so, slow down and take a look at your code one step at a time via debug.
I believe that what you'll see is that you are actually doing this...
Mov Eax, 3137h
Although it might be
Mov Eax, 3731h
Interesting concept. I've never done anything like that. Whatever.
What's more, if you are using this place to store that same number...
sum resb 1
You only have one byte.
Best I can tell, your example code is limited to single digit numbers.
Now then, since your label sum has reserved only one byte; 8 bits, you can see the problem as you are storing 32 bits there. (Well, you're trying to; it won't work.) No clue what happens when you do that. You probably want to rethink that structure.
As for why 17 becomes 5, no clue here.
Let us know if any of this helps you. Assembly is great stuff. As you are personally experiencing, the initial thought adjustment can be strange for the brain, can't it !