Finding the number of bytes of entered string at runtime - linux

I'm new at learning assembly x86. I have written a program that asks the user to enter a number and then checks if it's even or odd and then print a message to display this information.
The code works fine but it has one problem. It only works for 1 digit numbers:
; Ask the user to enter a number from the keyboard
; Check if this number is odd or even and display a message to say this
section .text
global _start ;must be declared for linker (gcc)
_start: ;tell linker entry point
;Display 'Please enter a number'
mov eax, 4 ; sys_write
mov ebx, 1 ; file descriptor: stdout
mov ecx, msg1 ; message to be print
mov edx, len1 ; message length
int 80h ; perform system call
;Enter the number from the keyboard
mov eax, 3 ; sys_read
mov ebx, 2 ; file descriptor: stdin
mov ecx, myvariable ; destination (memory address)
mov edx, 4 ; size of the the memory location in bytes
int 80h ; perform system call
;Convert the variable to a number and check if even or odd
mov eax, [myvariable]
sub eax, '0' ;eax now has the number value
and eax, 01H
jz isEven
;Display 'The entered number is odd'
mov eax, 4 ; sys_write
mov ebx, 1 ; file descriptor: stdout
mov ecx, msg2 ; message to be print
mov edx, len2 ; message length
int 80h
jmp outProg
isEven:
;Display 'The entered number is even'
mov eax, 4 ; sys_write
mov ebx, 1 ; file descriptor: stdout
mov ecx, msg3 ; message to be print
mov edx, len3 ; message length
int 80h
outProg:
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
msg1 db "Please enter a number: ", 0xA,0xD
len1 equ $- msg1
msg2 db "The entered number is odd", 0xA,0xD
len2 equ $- msg2
msg3 db "The entered number is even", 0xA,0xD
len3 equ $- msg3
segment .bss
myvariable resb 4
It does not work properly for numbers with more than 1 digit because it only takes in account the first byte(first digit) of the entered number so it only checks that. So I would need a way to find out how many digits(bytes) there are in the entered value that the user gives so I could do something like this:
;Convert the variable to a number and check if even or odd
mov eax, [myvariable+(number_of_digits-1)]
And only check eax which contains the last digit to see if it's even or odd.
Problem is I have no ideea how could I check how many bytes are in my number after the user has entered it.
I'm sure it's something very easy yet I have not been able to figure it out, nor have I found any solutions on how to do this on google. Please help me with this. Thank you!

You actually want movzx eax, byte [myvariable+(number_of_digits-1)] to only load 1 byte, not a dword. Or just directly test memory with test byte [...], 1. You can skip the sub because '0' is an even number; subtracting to convert from ASCII code to integer digit doesn't change the low bit.
But yes, you need least significant digit, the last (highest address) in printing / reading order.
A read system call returns the number of bytes read in EAX. (Or negative error code). This will include a newline if the user hit return, but not if the user redirected from a file that didn't end with a newline. (Or if they submitted input on a terminal using control-d after typing some digits). The most simple and robust way would be to simply loop looking for the first non-digit in the buffer.
But the "clever" / fun way would be to check if [mybuffer + eax - 1] is a digit, and if so use it. Otherwise check the previous byte. (Or just assume there's a newline and always check [mybuffer + eax - 2], the 2nd-last byte of what was read. (Or off the start of the buffer if the user just pressed return.)
(To efficiently check for an ASCII digit; sub al, '0' / cmp al, 9 / ja non_digit. See double condition checking in assembly / What is the idea behind ^= 32, that converts lowercase letters to upper and vice versa?)
Just for fun, here's a more compact version that always just checks the 2nd-last byte of the read() input. (It doesn't check for being a digit, and it reads outside the buffer for input lengths of 0 or 1, e.g. pressing control-D or return.) Also for read errors, e.g. redirect with strace ./oddeven <&- to close its stdin.
Note the interesting part:
; check if the low digit is even or odd
mov ecx, msg_even
mov edx, msg_odd ; these don't set flags and actually could be done after TEST
test byte [mybuf + eax - 2], 1 ; check the low bit of 2nd-last byte of the read input
cmovnz ecx, edx
;Display selected message
mov eax, 4 ; sys_write
mov ebx, 1 ; file descriptor: stdout
mov edx, msg_odd.len
int 80h ; write(1, digit&1 ? msg_odd : msg_even, msg_odd.len)
I used cmov, but a simple branch over a mov ecx, msg_odd would work. You don't need to duplicate the whole setup for the system call, just run it with the right pointer and length. (ECX and EDX values, and I padded the odd message with a space so I could use the same length for both.)
And this is a homebrewed static_assert(msg_odd.len == msg_even.len), using NASM's conditional directives (https://nasm.us/doc/nasmdoc4.html). It's not just a separate preprocessor like C has, it can use NASM numeric equ expressions.
%if msg_odd.len != msg_even.len
; homebrew assert with NASM preprocessor, since I chose to skip doing a 2nd cmov for the length
%warn we assume both messages have the same length
%endif
The full thing. I outside of the part shown above, I just tweaked comments to sometimes simplify when I thought it was too redundant, and used meaningful label names.
Also, I put .rodata and .bss at the top because NASM complained about referencing msg_odd.len before it was defined. (You previously had your strings in .data, but read-only data should generally go in .rodata, so the OS can share those pages between runs of the same program because they stay clean.)
Other fixes:
Linux/Unix uses 0xa line endings, \n not \n\r.
stdin is fd 0. 2 is stderr. (2 happens to work because terminal emulators normally run the shell with all 3 file descriptors referring to the same read+write open file description for the tty).
; Ask the user to enter a number from the keyboard
; Check if this number is odd or even and display a message to say this
section .rodata
msg_prompt db "Please enter a number: ", 0xA
.len equ $- msg_prompt
msg_odd db "The entered number is odd ", 0xA ; padded with a space for same length as even
.len equ $- msg_odd
msg_even db "The entered number is even", 0xA
.len equ $- msg_even
section .bss
mybuf resb 128
.len equ $ - mybuf
section .text
global _start
_start: ; ld defaults to starting at the top of the .text section, but exporting a symbol silences the warning and can make GDB work more easily.
; Display prompt
mov eax, 4 ; sys_write
mov ebx, 1 ; file descriptor: stdout
mov ecx, msg_prompt
mov edx, msg_prompt.len
int 80h ; perform system call
mov eax, 3 ; sys_read
xor ebx, ebx ; file descriptor: stdin
mov ecx, mybuf
mov edx, mybuf.len
int 80h ; read(0, mybuf, len)
; return value in EAX: negative for error, 0 for EOF, or positive byte count
; for this toy program, lets assume valid input ending with digit\n
; the newline will be at [mybuf + eax - 1]. The digit before that, at [mybuf + eax - 2].
; If the user just presses return, we'll access before the end of mybuf, and may segfault if it's at the start of a page.
; check if the low digit is even or odd
mov ecx, msg_even
mov edx, msg_odd ; these don't set flags and actually could be done after TEST
test byte [mybuf + eax - 2], 1 ; check the low bit of 2nd-last byte of the read input
cmovnz ecx, edx
;Display selected message
mov eax, 4 ; sys_write
mov ebx, 1 ; file descriptor: stdout
mov edx, msg_odd.len
int 80h ; write(1, digit&1 ? msg_odd : msg_even, msg_odd.len)
%if msg_odd.len != msg_even.len
; homebrew assert with NASM preprocessor, since I chose to skip doing a 2nd cmov for the length
%warning we assume both messages have the same length
%endif
mov eax, 1 ;system call number (sys_exit)
xor ebx, ebx
int 0x80 ; _exit(0)
assemble + link with nasm -felf32 oddeven.asm && ld -melf_i386 -o oddeven oddeven.o

Related

Finding number is Even/Odd in assembly

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`

Display contents of register

hi i need help displaying contents of a register.my code is below.i have been able to display values of the data register but i want to display flag states. eg 1 or 0. and it would be helpful if to display also the contents of other registers like esi,ebp.
my code is not printing the states of the flags ..what am i missing
section .text
global _start ;must be declared for using gcc
_start : ;tell linker entry point
mov eax,msg ; moves message "rubi" to eax register
mov [reg],eax ; moves message from eax to reg variable
mov edx, 8 ;message length
mov ecx, [reg];message to write
mov ebx, 1 ;file descriptor (stdout)
mov eax, 4 ;system call number (sys_write)
int 0x80 ;call kernel
mov eax, 100
mov ebx, 100
cmp ebx,eax
pushf
pop dword eax
mov [save_flags],eax
mov edx, 8 ;message length
mov ecx,[save_flags] ;message to write
mov ebx, 1 ;file descriptor (stdout)
mov eax, 4 ;system call number (sys_write)
int 0x80
mov eax, 1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
msg db "rubi",10
section .bss
reg resb 100
save_flags resw 100
I'm not going for anything fancy here since this appears to be a homework assignment (two people have asked the same question today). This code should be made as a function, and it can have its performance enhanced. Since I don't get an honorary degree or an A in the class it doesn't make sense to me to offer the best solution, but one you can work from:
BITS_TO_DISPLAY equ 32 ; Number of least significant bits to display (1-32)
section .text
global _start ; must be declared for using gcc
_start : ; tell linker entry point
mov edx, msg_len ; message length
mov ecx, msg ; message to write
mov ebx, 1 ; file descriptor (stdout)
mov eax, 4 ; system call number (sys_write)
int 0x80 ; call kernel
mov eax, 100
mov ebx, 100
cmp ebx,eax
pushf
pop dword eax
; Convert binary to string by shifting the right most bit off EAX into
; the carry flag (CF) and convert the bit into a '0' or '1' and place
; in the save_flags buffer in reverse order. Nul terminate the string
; in the event you ever wish to use printf to print it
mov ecx, BITS_TO_DISPLAY ; Number of bits of EAX register to display
mov byte [save_flags+ecx], 0 ; Nul terminate binary string in case we use printf
bin2ascii:
xor bl, bl ; BL = 0
shr eax, 1 ; Shift right most bit into carry flag
adc bl, '0' ; bl = bl + '0' + Carry Flag
mov [save_flags-1+ecx], bl ; Place '0'/'1' into string buffer in reverse order
dec ecx
jnz bin2ascii ; Loop until all bits processed
mov edx, BITS_TO_DISPLAY ; message length
mov ecx, save_flags ; address of binary string to write
mov ebx, 1 ; file descriptor (stdout)
mov eax, 4 ; system call number (sys_write)
int 0x80
mov eax, 1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
msg db "rubi",10
msg_len equ $ - msg
section .bss
save_flags resb BITS_TO_DISPLAY+1 ; Add one byte for nul terminator in case we use printf
The idea behind this code is that we continually shift the bits (using the SHR instruction) in the EAX register to the right one bit at a time. The bit that gets shifted out of the register gets placed in the carry flag (CF). We can use ADC to add the value of the carry flag (0/1) to ASCII '0' to get an ASCII value of '0` and '1'. We place these bytes into destination buffer in reverse order since we are moving from right to left through the bits.
BITS_TO_DISPLAY can be set between 1 and 32 (since this is 32-bit code). If you are interested in the bottom 8 bits of a register set it to 8. If you want to display all the bits of a 32-bit register, specify 32.
Note that you can pop directly into memory.
And if you want to binary dump register and flag data with write(2), your system call needs to pass a pointer to the buffer, not the data itself. Use a mov-immediate to get the address into the register, rather than doing a load. Or lea to use a RIP-relative addressing mode. Or pass a pointer to where it's sitting on the stack, instead of copying it to a global!
mov edx, 8 ;message length
mov ecx,[save_flags] ;message to write ;;;;;;; <<<--- problem
mov ebx, 1 ;file descriptor (stdout)
mov eax, 4 ;system call number (sys_write)
int 0x80
Passing a bad address to write(2) won't cause your program to receive a SIGSEGV, like it would if you used that address in user-space. Instead, write will return EFAULT. And you're not checking the return status from your system calls, so your code doesn't notice.
mov eax,msg ; moves message "rubi" to eax register
mov [reg],eax ; moves message from eax to reg variable
mov ecx, [reg];
This is silly. You should just mov ecx, msg to get the address of msg into ecx, rather than bouncing it through memory.
Are you building for 64bit? I see you're using 8 bytes for a message length. If so, you should be using the 64bit function call ABI (with syscall, not int 0x80). The system-call numbers are different. See the table in one of the links at x86. The 32bit ABI can only accept 32bit pointers. You will have a problem if you try to pass a pointer that has any of the high32 bits set.
You're probably also going to want to format the number into a string, unless you want to pipe your program's output into hexdump.

Why do I need to use [ ] (square brackets) when moving data from register to memory, but not when other way around?

This is the code I have and it works fine:
section .bss
bufflen equ 1024
buff: resb bufflen
whatread: resb 4
section .data
section .text
global main
main:
nop
read:
mov eax,3 ; Specify sys_read
mov ebx,0 ; Specify standard input
mov ecx,buff ; Where to read to...
mov edx,bufflen ; How long to read
int 80h ; Tell linux to do its magic
; Eax currently has the return value from linux system call..
add eax, 30h ; Convert number to ASCII digit
mov [whatread],eax ; Store how many bytes has been read to memory at loc **whatread**
mov eax,4 ; Specify sys_write
mov ebx,1 ; Specify standart output
mov ecx,whatread ; Get the address of whatread to ecx
mov edx,4 ; number of bytes to be written
int 80h ; Tell linux to do its work
mov eax, 1;
mov ebx, 0;
int 80h
Here is a simple run and output:
koray#koray-VirtualBox:~/asm/buffasm$ nasm -f elf -g -F dwarf buff.asm
koray#koray-VirtualBox:~/asm/buffasm$ gcc -o buff buff.o
koray#koray-VirtualBox:~/asm/buffasm$ ./buff
p
2koray#koray-VirtualBox:~/asm/buffasm$ ./buff
ppp
4koray#koray-VirtualBox:~/asm/buffasm$
My question is: What is with these 2 instructions:
mov [whatread],eax ; Store how many byte reads info to memory at loc whatread
mov ecx,whatread ; Get the address of whatread in ecx
Why the first one works with [] but the other one without?
When I try replacing the second line above with:
mov ecx,[whatread] ; Get the address of whatread in ecx
the executable will not run properly, it will not shown anything in the console.
Using brackets and not using brackets are basically two different things:
A bracket means that the value in the memory at the given address is meant.
An expression without a bracket means that the address (or value) itself is meant.
Examples:
mov ecx, 1234
Means: Write the value 1234 to the register ecx
mov ecx, [1234]
Means: Write the value that is stored in memory at address 1234 to the register ecx
mov [1234], ecx
Means: Write the value stored in ecx to the memory at address 1234
mov 1234, ecx
... makes no sense (in this syntax) because 1234 is a constant number which cannot be changed.
Linux "write" syscall (INT 80h, EAX=4) requires the address of the value to be written, not the value itself!
This is why you do not use brackets at this position!

Assembly code automatically running a terminal command

Recently, I wrote a bit of assembly code that asks for the password and if the user enters the correct password as stored internally, it prints out "Correct!". Else, it prints out "Incorrect!".
Here is the code:
section .text
global _start
_start:
mov edx, len_whatis
mov ecx, whatis
mov ebx, 1
mov eax, 4
int 80h ; outputs: "What is the password?"
mov edx, 5 ; expect 5 bytes of input(so 4 numbers)
mov ecx, pass
mov ebx, 0
mov eax, 3
int 80h ; accepts intput and stores in pass
mov eax, [pass] ; move the pass variable into eax
sub eax, '0' ; change the ascii number in eax to a numerical number
mov ebx, [thepass] ; move the thepass variable into ebx
sub ebx, '0' ; change the ascii number in ebx to a numerical number
cmp eax, ebx ; compare the 2 numbers
je correct ; if they are equal, jump to correct
jmp incorrect ; if not, jump to incorrect
correct:
mov edx, len_corr
mov ecx, corr
mov ebx, 1
mov eax, 4
int 80h ; outputs: "Correct!"
mov ebx, 0
mov eax, 1
int 80h ; exits with status 0
incorrect:
mov edx, len_incor
mov ecx, incor
mov ebx, 1
mov eax, 4
int 80h ; outputs: "Incorrect!"
mov eax, 1
int 80h ; exits with status: 1
section .data
whatis db "What is the password?", 0xA
len_whatis equ $ - whatis
thepass db "12345"
corr db "Correct!", 0xA
len_corr equ $ - corr
incor db "Incorrect!", 0xA
len_incor equ $ - incor
section .bss
pass resb 5
Assemble:nasm -f elf password.s
Link:ld -m elf_i386 -s -o password password.o
(If you did try to assemble link and run this, you may notice that it checks the password incorrectly - ignore this. It is "off topic")
Then, I ran a test:
I ran the code with ./password
When I was prompted for the password, I typed in 123456, one more byte than the code expects
After I hit enter and the code exits, the terminal immediately tries to run a command 6
What is causing this behavior? Is it something to do with the assembler, or how my computer is reading the code?
EDIT:
And, when I run the code with 12345, the terminal prompts for a command twice when the program closes, as if someone just hit the enter button without entering a command.
You're only reading five bytes from standard input, so when you type 123456↵, your application ends up reading 12345 and leaving 6↵ in the buffer. That gets passed on to the shell.
If you want to read the whole line, use a larger buffer.

How should I work with dynamically-sized input in NASM Assembly?

I'm trying to learn assembly with NASM on 64 bit Linux.
I managed to make a program that reads two numbers and adds them. The first thing I realized was that the program will only work with one-digit numbers (and results):
; Calculator
SECTION .data
msg1 db "Enter the first number: "
msg1len equ $-msg1
msg2 db "Enter the second number: "
msg2len equ $-msg2
msg3 db "The result is: "
msg3len equ $-msg3
SECTION .bss
num1 resb 1
num2 resb 1
result resb 1
SECTION .text
global main
main:
; Ask for the first number
mov EAX,4
mov EBX,1
mov ECX,msg1
mov EDX,msg1len
int 0x80
; Read the first number
mov EAX,3
mov EBX,1
mov ECX,num1
mov EDX,2
int 0x80
; Ask for the second number
mov EAX,4
mov EBX,1
mov ECX,msg2
mov EDX,msg2len
int 0x80
; Read the second number
mov EAX,3
mov EBX,1
mov ECX,num2
mov EDX,2
int 0x80
; Prepare to announce the result
mov EAX,4
mov EBX,1
mov ECX,msg3
mov EDX,msg3len
int 0x80
; Do the sum
; Store read values to EAX and EBX
mov EAX,[num1]
mov EBX,[num2]
; From ASCII to decimal
sub EAX,'0'
sub EBX,'0'
; Add
add EAX,EBX
; Convert back to EAX
add EAX,'0'
; Save the result back to the variable
mov [result],EAX
; Print result
mov EAX,4
mov EBX,1
mov ECX,result
mov EDX,1
int 0x80
As you can see, I reserve one byte for the first number, another for the second, and one more for the result. This isn't very flexible. I would like to make additions with numbers of any size.
How should I approach this?
First of all you are generating a 32-bit program, not a 64-bit program. This is no problem as Linux 64-bit can run 32-bit programs if they are either statically linked (this is the case for you) or the 32-bit shared libraries are installed.
Your program contains a real bug: You are reading and writing the "EAX" register from a 1-byte field in RAM:
mov EAX, [num1]
This will normally work on little-endian computers (x86). However if the byte you want to read is at the end of the last memory page of your program you'll get a bus error.
Even more critical is the write command:
mov [result], EAX
This command will overwrite 3 bytes of memory following the "result" variable. If you extend your program by additional bytes:
num1 resb 1
num2 resb 1
result resb 1
newVariable1 resb 1
You'll overwrite these variables! To correct your program you must use the AL (and BL) register instead of the complete EAX register:
mov AL, [num1]
mov BL, [num2]
...
mov [result], AL
Another finding in your program is: You are reading from file handle #1. This is the standard output. Your program should read from file handle #0 (standard input):
mov EAX, 3 ; read
mov EBX, 0 ; standard input
...
int 0x80
But now the answer to the actual question:
The C library functions (e.g. fgets()) use buffered input. Doing it like this would be a bit to complicated for the beginning so reading one byte at a time could be a possibility.
Thinking the way "how would I solve this problem using a high-level language like C". If you don't use libraries in your assembler program you can only use system calls (section 2 man pages) as functions (e.g. you cannot use "fgets()" but only "read()").
In your case a C program reading a number from standard input could look like this:
int num1;
char c;
...
num1 = 0;
while(1)
{
if(read(0,&c,1)!=1) break;
if(c=='\r' || c=='\n') break;
num1 = 10*num1 + c - '0';
}
Now you may think about the assembler code (I typically use GNU assembler, which has another syntax, so maybe this code contains some bugs):
c resb 1
num1 resb 4
...
; Set "num1" to 0
mov EAX, 0
mov [num1], EAX
; Here our while-loop starts
next_digit:
; Read one character
mov EAX, 3
mov EBX, 0
mov ECX, c
mov EDX, 1
int 0x80
; Check for the end-of-input
cmp EAX, 1
jnz end_of_loop
; This will cause EBX to be 0.
; When modifying the BL register the
; low 8 bits of EBX are modified.
; The high 24 bits remain 0.
; So clearing the EBX register before
; reading an 8-bit number into BL is
; a method for converting an 8-bit
; number to a 32-bit number!
xor EBX, EBX
; Load the character read into BL
; Check for "\r" or "\n" as input
mov BL, [c]
cmp BL, 10
jz end_of_loop
cmp BL, 13
jz end_of_loop
; read "num1" into EAX
mov EAX, [num1]
; Multiply "num1" with 10
mov ECX, 10
mul ECX
; Add one digit
sub EBX, '0'
add EAX, EBX
; write "num1" back
mov [num1], EAX
; Do the while loop again
jmp next_digit
; The end of the loop...
end_of_loop:
; Done
Writing decimal numbers with more digits is more difficult!

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