How can I write to console in MS Windows with yasm? - windows

I tried this code (using interruptions), but it is for DOS:
mov eax, 42 ; write 42 to console
mov ecx, 10 ; in decimal
out_int_loop:
xor edx, edx
div ecx
push eax
add dl, '0' ; one digit
mov ah, 2 ; 2 is code for writing character
int 21h
pop eax
cmp eax, 0
jnz out_int_loop

I used WriteConsoleA function from winapi.
Also I used GetStdHandle to get stdout.
To import this functions you need these lines:
extern __imp__GetStdHandle#4
extern __imp__WriteConsoleA#20

Related

when writing 64bit reverse shell in assembly got stuck at createrprocessA api

hello i am writing windows 64bit reverse shell in assembly and after gett connected to the targetmachine ip, i want to create process to spwan a shell, fistly i try to write startinfo struct for createprocess api, but after then i pass all the parameters to the function but it doesn't work, and here is full code https://pastebin.com/6Ft2jCMX
;STARTUPINFOA+PROCESS_INFORMATION
;----------------------------------
push byte 0x12 ; We want to place (18 * 4) = 72 null bytes onto the stack
pop rcx ; Set ECX for the loop
xor r11,r11
push_loop:
push r11 ; push a null dword
loop push_loop ; keep looping untill we have pushed enough nulls
lea r12,[rsp]
mov dl,104
xor rcx,rcx
mov [r12],dword edx
mov [r12+4],rcx
mov [r12+12],rcx
mov [r12+20],rcx
mov [r12+24],rcx
xor rdx,rdx
mov dl,255
inc rdx
mov [r12+0x3c],edx
mov [r12+0x50],r14 ; HANDLE hStdInput;
mov [r12+0x58],r14 ; HANDLE hStdOutput;
mov [r12+0x60],r14 ;HANDLE hStdError;
;createprocessA_calling
sub rsp, 0x70
push 'cmdA'
mov [rsp+3],byte dl
lea rdx,[rsp]
inc rcx
mov [rsp+32],rcx
xor rcx,rcx
xor r8,r8
mov [rsp+40],r8
mov [rsp+48],r8
mov [rsp+56],r8
lea r9,[r12]
mov [rsp+64],r9
lea r9,[r12+104]
mov [rsp+72],r9
xor r9,r9
call rbx ;createprocessA
so at last when i call the createprocessA it got stuck

I'm unsure what the problem with my assembly code it works until eax is popped and replaced by a register

; Input x and y, output min of the two numbers
.586
.MODEL FLAT
INCLUDE io.h
.STACK 4096
.DATA
number DWORD ?
array DWORD 20, 15, 62, 40, 18
nbrElts DWORD 5
prompt BYTE "Enter value:", 0
string BYTE 80 DUP (?)
resultLbl BYTE "Position", 0
result BYTE 11 DUP (?), 0
.CODE
_MainProc PROC
input prompt, string, 20 ; read ASCII characters
atod string ; convert to integer
mov number, eax ; store in memory
push nbrElts ; 3rd parameter (# of elements in array)
lea eax, array ; 2nd parameter (address of array)
push eax
push number ; 1st parameter (value from user)
call searchArray ; searchArray(number, array, 5)
add esp, 12
dtoa result, eax ; convert to ASCII characters
output resultLbl, result ; output label and result
mov eax, 0 ; exit with return code 0
ret
_MainProc ENDP
; searchArray(int x, array, int y)
;
searchArray PROC
push ebp ; save base pointer
mov ebp, esp ; establish stack frame
push eax ; save registers
push ebx
push esi
push ecx
push edx
mov ebx, [ebp+8] ; x, value from user
mov esi, [ebp+12] ; address of array
mov ecx, [ebp+16] ; y, number of elements
mov edx, 1
mov ecx, 5
forLoop:
mov eax, [esi] ; a[i]
cmp eax, ebx ; eax = ebx ?
je isEqual
;cmp eax, ebx
add esi, 4
inc edx
loop forLoop
;mov eax, 0
cmp edx, 6
je notEqual
isEqual:
mov eax, edx
jmp exitCode
notEqual:
mov eax, 0
jmp exitCode
exitCode:
mov eax, edx
pop edx ; restore EBP
pop ecx ; restore EAX
pop esi
pop ebx
pop ebp
ret ; return
searchArray ENDP
END ; end of source code
The pops at the end of the function need to match the pushes at the beginning of the function. If they don't match, the stack pointer ends up in the wrong place and the ret returns to the wrong place.
In your case, you have an extra push without a corresponding pop.
The reason to push registers at the beginning and pop them at the end is to preserve their values. But you don't want to preserve the value of eax. You want to return a different value, the result of the function. So there is absolutely no reason to push eax.

Assembly Programming using SASM on Windows, with an example using int 0x80 (Linux system calls)

I need help. I'm trying to run the program (NASM) below in SASM.
SYS_EXIT equ 1
SYS_READ equ 3
SYS_WRITE equ 4
STDIN equ 0
STDOUT equ 1
segment .data
msg1 db "Enter a digit ", 0xA,0xD
len1 equ $- msg1
msg2 db "Please enter a second digit", 0xA,0xD
len2 equ $- msg2
msg3 db "The sum is: "
len3 equ $- msg3
segment .bss
num1 resb 2
num2 resb 2
res resb 1
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov eax, SYS_WRITE
mov ebx, STDOUT
mov ecx, msg1
mov edx, len1
int 0x80
mov eax, SYS_READ
mov ebx, STDIN
mov ecx, num1
mov edx, 2
int 0x80
mov eax, SYS_WRITE
mov ebx, STDOUT
mov ecx, msg2
mov edx, len2
int 0x80
mov eax, SYS_READ
mov ebx, STDIN
mov ecx, num2
mov edx, 2
int 0x80
mov eax, SYS_WRITE
mov ebx, STDOUT
mov ecx, msg3
mov edx, len3
int 0x80
; moving the first number to eax register and second number to ebx
; and subtracting ascii '0' to convert it into a decimal number
mov eax, [num1]
sub eax, '0'
mov ebx, [num2]
sub ebx, '0'
; add eax and ebx
add eax, ebx
; add '0' to to convert the sum from decimal to ASCII
add eax, '0'
; storing the sum in memory location res
mov [res], eax
; print the sum
mov eax, SYS_WRITE
mov ebx, STDOUT
mov ecx, res
mov edx, 1
int 0x80
exit:
mov eax, SYS_EXIT
xor ebx, ebx
int 0x80
I had this error:
[20:53:11] Warning! Errors have occurred in the build:
c:/program files (x86)/sasm/mingw/bin/../lib/gcc/mingw32/4.6.2/../../../libmingw32.a(main.o): In function 'main':
C:\MinGW\msys\1.0\src\mingwrt/../mingw/main.c:73: undefined reference to `WinMain#16'
Also, how do I limit users input up to 4 digits only?
global _start should change to global main and Linux system calls should be replaced by Windows API function calls and declared as external. Modern versions of Windows doesn't approve use of system calls due to malware or badware risks, so deprecated (permanent) system call codes. Every modern version of Windows has different system call number codes, though you can find them on internet, you shouldn't rely on them unless you want to revise your assembly code for each version of Windows thus reducing portability and increasing workload. There are significant differences between Linux/Mac and Windows in the way of handling registers, stack and function names.

Reading from STDIN and printing to STDOUT with nasm assembly?

As the title suggest I seem to be having a hard time converting the below code to do the exact same thing, which is to read from stdin and stdout. My professor wants us to stop using int 80h and switch over to using gcc. I've had no problems with reading input with the below code however, switching over to gcc is where I start getting segmentation core dump errors.
section .bss
buf resb 1 ; 1000-byte buffer (in data section)
section .text
global _start
_start:
loop1: mov edx, 1 ; max length
mov ecx, buf ; buffer
mov ebx, 0 ; stdin
mov eax, 3 ; sys_read
int 80h
cmp eax, 0 ; end loop if read <= 0
jle lpend1
mov edx, eax ; length
mov ecx, buf ; buffer
mov ebx, 1 ; stdout
mov eax, 4 ; sys_write
int 80h
jmp loop1 ; go back for more
lpend1:
mov eax, 1
mov ebx, 0
int 80h
My attempt at converting the above to perform the same task
SECTION .data
format: db "%c",0
SECTION .bss
buff: resb 1
SECTION .text
extern printf
extern scanf
global main
main:
loop1:
push buff ;buff will hold the characters in the string/file
push format ;expect character for every buff
call scanf
add esp, 8 ;clear stack
cmp eax, 0 ;if eax is equal to 0 then EOF
je lpend1 ;jump to end main func
xor eax, eax ;clear eax
mov eax, buff ;mov buff to eax register
push eax ;push eax onto the stack
mov eax, format ;mov the string format to eax
push eax ;push onto the stack
call printf ;call printf, prints to screen
add esp, 8 ;clear the stack
jmp loop1 ;jump back to top and repeat
lpend1:
ret ;end of main

How to echo memory location use NASM [duplicate]

I am looking for a way to print an integer in assembler (the compiler I am using is NASM on Linux), however, after doing some research, I have not been able to find a truly viable solution. I was able to find a description for a basic algorithm to serve this purpose, and based on that I developed this code:
global _start
section .bss
digit: resb 16
count: resb 16
i: resb 16
section .data
section .text
_start:
mov dword[i], 108eh ; i = 4238
mov dword[count], 1
L01:
mov eax, dword[i]
cdq
mov ecx, 0Ah
div ecx
mov dword[digit], edx
add dword[digit], 30h ; add 48 to digit to make it an ASCII char
call write_digit
inc dword[count]
mov eax, dword[i]
cdq
mov ecx, 0Ah
div ecx
mov dword[i], eax
cmp dword[i], 0Ah
jg L01
add dword[i], 48 ; add 48 to i to make it an ASCII char
mov eax, 4 ; system call #4 = sys_write
mov ebx, 1 ; file descriptor 1 = stdout
mov ecx, i ; store *address* of i into ecx
mov edx, 16 ; byte size of 16
int 80h
jmp exit
exit:
mov eax, 01h ; exit()
xor ebx, ebx ; errno
int 80h
write_digit:
mov eax, 4 ; system call #4 = sys_write
mov ebx, 1 ; file descriptor 1 = stdout
mov ecx, digit ; store *address* of digit into ecx
mov edx, 16 ; byte size of 16
int 80h
ret
C# version of what I want to achieve (for clarity):
static string int2string(int i)
{
Stack<char> stack = new Stack<char>();
string s = "";
do
{
stack.Push((char)((i % 10) + 48));
i = i / 10;
} while (i > 10);
stack.Push((char)(i + 48));
foreach (char c in stack)
{
s += c;
}
return s;
}
The issue is that it outputs the characters in reverse, so for 4238, the output is 8324. At first, I thought that I could use the x86 stack to solve this problem, push the digits in, and pop them out and print them at the end, however when I tried implementing that feature, it flopped and I could no longer get an output.
As a result, I am a little bit perplexed about how I can implement a stack in to this algorithm in order to accomplish my goal, aka printing an integer. I would also be interested in a simpler/better solution if one is available (as it's one of my first assembler programs).
One approach is to use recursion. In this case you divide the number by 10 (getting a quotient and a remainder) and then call yourself with the quotient as the number to display; and then display the digit corresponding to the remainder.
An example of this would be:
;Input
; eax = number to display
section .data
const10: dd 10
section .text
printNumber:
push eax
push edx
xor edx,edx ;edx:eax = number
div dword [const10] ;eax = quotient, edx = remainder
test eax,eax ;Is quotient zero?
je .l1 ; yes, don't display it
call printNumber ;Display the quotient
.l1:
lea eax,[edx+'0']
call printCharacter ;Display the remainder
pop edx
pop eax
ret
Another approach is to avoid recursion by changing the divisor. An example of this would be:
;Input
; eax = number to display
section .data
divisorTable:
dd 1000000000
dd 100000000
dd 10000000
dd 1000000
dd 100000
dd 10000
dd 1000
dd 100
dd 10
dd 1
dd 0
section .text
printNumber:
push eax
push ebx
push edx
mov ebx,divisorTable
.nextDigit:
xor edx,edx ;edx:eax = number
div dword [ebx] ;eax = quotient, edx = remainder
add eax,'0'
call printCharacter ;Display the quotient
mov eax,edx ;eax = remainder
add ebx,4 ;ebx = address of next divisor
cmp dword [ebx],0 ;Have all divisors been done?
jne .nextDigit
pop edx
pop ebx
pop eax
ret
This example doesn't suppress leading zeros, but that would be easy to add.
I think that maybe implementing a stack is not the best way to do this (and I really think you could figure out how to do that, saying as how pop is just a mov and a decrement of sp, so you can really set up a stack anywhere you like by just allocating memory for it and setting one of your registers as your new 'stack pointer').
I think this code could be made clearer and more modular if you actually allocated memory for a c-style null delimited string, then create a function to convert the int to string, by the same algorithm you use, then pass the result to another function capable of printing those strings. It will avoid some of the spaghetti code syndrome you are suffering from, and fix your problem to boot. If you want me to demonstrate, just ask, but if you wrote the thing above, I think you can figure out how with the more split up process.
; Input
; EAX = pointer to the int to convert
; EDI = address of the result
; Output:
; None
int_to_string:
xor ebx, ebx ; clear the ebx, I will use as counter for stack pushes
.push_chars:
xor edx, edx ; clear edx
mov ecx, 10 ; ecx is divisor, devide by 10
div ecx ; devide edx by ecx, result in eax remainder in edx
add edx, 0x30 ; add 0x30 to edx convert int => ascii
push edx ; push result to stack
inc ebx ; increment my stack push counter
test eax, eax ; is eax 0?
jnz .push_chars ; if eax not 0 repeat
.pop_chars:
pop eax ; pop result from stack into eax
stosb ; store contents of eax in at the address of num which is in EDI
dec ebx ; decrement my stack push counter
cmp ebx, 0 ; check if stack push counter is 0
jg .pop_chars ; not 0 repeat
mov eax, 0x0a
stosb ; add line feed
ret ; return to main
; eax = number to stringify/output
; edi = location of buffer
intToString:
push edx
push ecx
push edi
push ebp
mov ebp, esp
mov ecx, 10
.pushDigits:
xor edx, edx ; zero-extend eax
div ecx ; divide by 10; now edx = next digit
add edx, 30h ; decimal value + 30h => ascii digit
push edx ; push the whole dword, cause that's how x86 rolls
test eax, eax ; leading zeros suck
jnz .pushDigits
.popDigits:
pop eax
stosb ; don't write the whole dword, just the low byte
cmp esp, ebp ; if esp==ebp, we've popped all the digits
jne .popDigits
xor eax, eax ; add trailing nul
stosb
mov eax, edi
pop ebp
pop edi
pop ecx
pop edx
sub eax, edi ; return number of bytes written
ret

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