.data section annoying error with nasm [duplicate] - linux

I wrote the following simple program, but nasm refuses to compile it.
section .text
global _start
_start:
mov rax, 0x01
mov rdi, 0x01
mov rsi, str
mov rdx, 0x03
syscall
mov rax, 60
syscall
segment .data
str db 'Some string'
nasm -f elf64 main.asm
main.asm:15: error: comma, colon, decorator or end of line expected after operand
As I read in this answer this is because str is an instruction mnemonic. So I added a colon to str and now it compiles fine. But what about the line
mov rsi, str
str is an instruction mnemonic here, but it still compiles fine. Why?

As the NASM manual explains, other than macro definitions and directives, the format of a NASM source line has some combination of these four fields:
label: instruction operands ; comment
After it sees mov as the mnemonic, it's no longer considering the remaining tokens as possible instruction mnemonics. Assembly language is strictly one instruction per statement.
If you wanted the bytes that encode an str ax instruction as the immediate operand for mov-sign-extended-imm32, you'd have to do that yourself with a numeric constant. NASM syntax doesn't have a way to do that for you, so its parser doesn't need to recurse into the operands once its found a mnemonic.
Or instead of encoding str manually, use a db to emit the bytes of the mov instruction.
db 0x48, 0xc7, 0xc6 ; REX.W prefix, opcode for mov r/m64,imm32, ModR/M = rsi destination
str [rax+1] ; a disp8 makes this 4 bytes long.
;; same machine code as
mov rsi, strict dword 0x0148000f ; str [rax+1]
;; nasm optimizes it to mov esi, imm32 without strict dword.
;; I guess I should have used that 5-byte instruction for the DB version...

Related

Compact shellcode to print a 0-terminated string pointed-to by a register, given puts or printf at known absolute addresses?

Background: I am a beginner trying to understand how to golf assembly, in particular to solve an online challenge.
EDIT: clarification: I want to print the value at the memory address of RDX. So “SUPER SECRET!”
Create some shellcode that can output the value of register RDX in <= 11 bytes. Null bytes are not allowed.
The program is compiled with the c standard library, so I have access to the puts / printf statement. It’s running on x86 amd64.
$rax : 0x0000000000010000 → 0x0000000ac343db31
$rdx : 0x0000555555559480 → "SUPER SECRET!"
gef➤ info address puts
Symbol "puts" is at 0x7ffff7e3c5a0 in a file compiled without debugging.
gef➤ info address printf
Symbol "printf" is at 0x7ffff7e19e10 in a file compiled without debugging.
Here is my attempt (intel syntax)
xor ebx, ebx ; zero the ebx register
inc ebx ; set the ebx register to 1 (STDOUT
xchg ecx, edx ; set the ECX register to RDX
mov edx, 0xff ; set the length to 255
mov eax, 0x4 ; set the syscall to print
int 0x80 ; interrupt
hexdump of my code
My attempt is 17 bytes and includes null bytes, which aren't allowed. What other ways can I lower the byte count? Is there a way to call puts / printf while still saving bytes?
FULL DETAILS:
I am not quite sure what is useful information and what isn't.
File details:
ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 3.2.0, BuildID[sha1]=5810a6deb6546900ba259a5fef69e1415501b0e6, not stripped
Source code:
void main() {
char* flag = get_flag(); // I don't get access to the function details
char* shellcode = (char*) mmap((void*) 0x1337,12, 0, MAP_PRIVATE | MAP_ANONYMOUS, -1, 0);
mprotect(shellcode, 12, PROT_READ | PROT_WRITE | PROT_EXEC);
fgets(shellcode, 12, stdin);
((void (*)(char*))shellcode)(flag);
}
Disassembly of main:
gef➤ disass main
Dump of assembler code for function main:
0x00005555555551de <+0>: push rbp
0x00005555555551df <+1>: mov rbp,rsp
=> 0x00005555555551e2 <+4>: sub rsp,0x10
0x00005555555551e6 <+8>: mov eax,0x0
0x00005555555551eb <+13>: call 0x555555555185 <get_flag>
0x00005555555551f0 <+18>: mov QWORD PTR [rbp-0x8],rax
0x00005555555551f4 <+22>: mov r9d,0x0
0x00005555555551fa <+28>: mov r8d,0xffffffff
0x0000555555555200 <+34>: mov ecx,0x22
0x0000555555555205 <+39>: mov edx,0x0
0x000055555555520a <+44>: mov esi,0xc
0x000055555555520f <+49>: mov edi,0x1337
0x0000555555555214 <+54>: call 0x555555555030 <mmap#plt>
0x0000555555555219 <+59>: mov QWORD PTR [rbp-0x10],rax
0x000055555555521d <+63>: mov rax,QWORD PTR [rbp-0x10]
0x0000555555555221 <+67>: mov edx,0x7
0x0000555555555226 <+72>: mov esi,0xc
0x000055555555522b <+77>: mov rdi,rax
0x000055555555522e <+80>: call 0x555555555060 <mprotect#plt>
0x0000555555555233 <+85>: mov rdx,QWORD PTR [rip+0x2e26] # 0x555555558060 <stdin##GLIBC_2.2.5>
0x000055555555523a <+92>: mov rax,QWORD PTR [rbp-0x10]
0x000055555555523e <+96>: mov esi,0xc
0x0000555555555243 <+101>: mov rdi,rax
0x0000555555555246 <+104>: call 0x555555555040 <fgets#plt>
0x000055555555524b <+109>: mov rax,QWORD PTR [rbp-0x10]
0x000055555555524f <+113>: mov rdx,QWORD PTR [rbp-0x8]
0x0000555555555253 <+117>: mov rdi,rdx
0x0000555555555256 <+120>: call rax
0x0000555555555258 <+122>: nop
0x0000555555555259 <+123>: leave
0x000055555555525a <+124>: ret
Register state right before shellcode is executed:
$rax : 0x0000000000010000 → "EXPLOIT\n"
$rbx : 0x0000555555555260 → <__libc_csu_init+0> push r15
$rcx : 0x000055555555a4e8 → 0x0000000000000000
$rdx : 0x0000555555559480 → "SUPER SECRET!"
$rsp : 0x00007fffffffd940 → 0x0000000000010000 → "EXPLOIT\n"
$rbp : 0x00007fffffffd950 → 0x0000000000000000
$rsi : 0x4f4c5058
$rdi : 0x00007ffff7fa34d0 → 0x0000000000000000
$rip : 0x0000555555555253 → <main+117> mov rdi, rdx
$r8 : 0x0000000000010000 → "EXPLOIT\n"
$r9 : 0x7c
$r10 : 0x000055555555448f → "mprotect"
$r11 : 0x246
$r12 : 0x00005555555550a0 → <_start+0> xor ebp, ebp
$r13 : 0x00007fffffffda40 → 0x0000000000000001
$r14 : 0x0
$r15 : 0x0
(This register state is a snapshot at the assembly line below)
●→ 0x555555555253 <main+117> mov rdi, rdx
0x555555555256 <main+120> call rax
Since I already spilled the beans and "spoiled" the answer to the online challenge in comments, I might as well write it up. 2 key tricks:
Create 0x7ffff7e3c5a0 (&puts) in a register with lea reg, [reg + disp32], using the known value of RDI which is within the +-2^31 range of a disp32. (Or use RBP as a starting point, but not RSP: that would need a SIB byte in the addressing mode).
This is a generalization of the code-golf trick of lea edi, [rax+1] trick to create small constants from other small constants (especially 0) in 3 bytes, with code that runs less slowly than push imm8 / pop reg.
The disp32 is large enough to not have any zero bytes; you have a couple registers to choose from in case one had been too close.
Copy a 64-bit register in 2 bytes with push reg / pop reg, instead of 3-byte mov rdi, rdx (REX + opcode + modrm). No savings if either push needs a REX prefix (for R8..R15), and actually costs bytes if both are "non-legacy" registers.
See other answers on Tips for golfing in x86/x64 machine code on codegolf.SE for more.
bits 64
lea rsi, [rdi - 0x166f30]
;; add rbp, imm32 ; alternative, but that would mess up a call-preserved register so we might crash on return.
push rdx
pop rdi ; copy RDX to first arg, x86-64 SysV calling convention
jmp rsi ; tailcall puts
This is exactly 11 bytes, and I don't see a way for it to be smaller. add r64, imm32 is also 7 bytes, same as LEA. (Or 6 bytes if the register is RAX, but even the xchg rax, rdi short form would cost 2 bytes to get it there, and the RAX value is still the fgets return value, which is the small mmap buffer address.)
The puts function pointer doesn't fit in 32 bits, so we need a REX prefix on any instruction that puts it into a register. Otherwise we could just mov reg, imm32 (5 bytes) with the absolute address, not deriving it from another register.
$ nasm -fbin -o exploit.bin -l /dev/stdout exploit.asm
1 bits 64
2 00000000 488DB7D090E9FF lea rsi, [rdi - 0x166f30]
3 ;; add rbp, imm32 ; we can avoid messing up any call-preserved registers
4 00000007 52 push rdx
5 00000008 5F pop rdi ; copy to first arg
6 00000009 FFE6 jmp rsi ; tailcall
$ ll exploit.bin
-rw-r--r-- 1 peter peter 11 Apr 24 04:09 exploit.bin
$ ./a.out < exploit.bin # would work if the addresses in my build matched yours
My build of your incomplete .c uses different addresses on my machine, but it does reach this code (at address 0x10000, mmap_min_addr which mmap picks after the amusing choice of 0x1337 as a hint address, which isn't even page aligned but doesn't result in EIVAL on current Linux.)
Since we only tailcall puts with correct stack alignment and don't modify any call-preserved registers, this should successfully return to main.
Note that 0 bytes (ASCII NUL, not NULL) would actually work in shellcode for this test program, if not for the requirement that forbids it.
The input is read using fgets (apparently to simulate a gets() overflow).
fgets actually can read a 0 aka '\0'; the only critical character is 0xa aka '\n' newline. See Is it possible to read null characters correctly using fgets or gets_s?
Often buffer overflows exploit a strcpy or something else that stops on a 0 byte, but fgets only stops on EOF or newline. (Or the buffer size, a feature gets is missing, hence its deprecation and removal from even the ISO C standard library! It's literally impossible to use safely unless you control the input data). So yes, it's totally normal to forbid zero bytes.
BTW, your int 0x80 attempt is not viable: What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code? - you can't use the 32-bit ABI to pass 64-bit pointers to write, and the string you want to output is not in the low 32 bits of virtual address space.
Of course, with the 64-bit syscall ABI, you're fine if you can hardcode the length.
push rdx
pop rsi
shr eax, 16 ; fun 3-byte way to turn 0x10000` into `1`, __NR_write 64-bit, instead of just push 1 / pop
mov edi, eax ; STDOUT_FD = __NR_write
lea edx, [rax + 13 - 1] ; 3 bytes. RDX = 13 = string length
; or mov dl, 0xff ; 2 bytes leaving garbage in rest of RDX
syscall
But this is 12 bytes, as well as hard-coding the length of the string (which was supposed to be part of the secret?).
mov dl, 0xff could make sure the length was at least 255, and actually much more in this case, if you don't mind getting reams of garbage after the string you want, until write hits an unmapped page and returns early. That would save a byte, making this 11.
(Fun fact, Linux write does not return an error when it's successfully written some bytes; instead it returns how many it did write. If you try again with buf + write_len, you would get a -EFAULT return value for passing a bad pointer to write.)

Joining strings from registers and printing them (CPUID)

Starting to learn NASM assembly, I was looking at some assembly questions here in Stack Overflow and found this one here:
Concatenating strings from registers and printing them
I believe that this question is not duplicated because I am trying to
replicate the code in NASM and also things were not very clear in the
other question.
I decided to replicate this code in NASM, but I did not quite understand the MASM code in question.
I learned about CPUID and did some testing programs.
In order, I'd like to know how we can concatenate registers and then print them on the screen USING NASM.
I want to print 'ebx' + 'edx' + 'ecx' because this is how the CPUID output is organized by what I see in GDB.
I called CPUID with eax=1
"String" is not a very precise term. The Vendor Identification String of CPUID/EAX=0 contains only 12 ASCII characters, packed into 3 DWORD registers. There is no termination character like in C nor a length information like in PASCAL. But it's always the same registers and it's always 3*4=12 bytes. This is ideal for the write-syscall:
section .bss
buff resb 12
section .text
global _start
_start:
mov eax, 0
cpuid
mov dword [buff+0], ebx ; Fill the first four bytes
mov dword [buff+4], edx ; Fill the second four bytes
mov dword [buff+8], ecx ; Fill the third four bytes
mov eax, 4 ; SYSCALL write
mov ebx, 1 ; File descriptor = STDOUT
mov ecx, buff ; Pointer to ASCII string
mov edx, 12 ; Count of bytes to send
int 0x80 ; Call Linux kernel
mov eax, 1 ; SYSCALL exit
mov ebx, 0 ; Exit Code
int 80h ; Call Linux kernel

I'm getting a segmentation fault in my assembly program [duplicate]

The tutorial I am following is for x86 and was written using 32-bit assembly, I'm trying to follow along while learning x64 assembly in the process. This has been going very well up until this lesson where I have the following simple program which simply tries to modify a single character in a string; it compiles fine but segfaults when ran.
section .text
global _start ; Declare global entry oint for ld
_start:
jmp short message ; Jump to where or message is at so we can do a call to push the address onto the stack
code:
xor rax, rax ; Clean up the registers
xor rbx, rbx
xor rcx, rcx
xor rdx, rdx
; Try to change the N to a space
pop rsi ; Get address from stack
mov al, 0x20 ; Load 0x20 into RAX
mov [rsi], al; Why segfault?
xor rax, rax; Clear again
; write(rdi, rsi, rdx) = write(file_descriptor, buffer, length)
mov al, 0x01 ; write the command for 64bit Syscall Write (0x01) into the lower 8 bits of RAX
mov rdi, rax ; First Paramter, RDI = 0x01 which is STDOUT, we move rax to ensure the upper 56 bits of RDI are zero
;pop rsi ; Second Parameter, RSI = Popped address of message from stack
mov dl, 25 ; Third Parameter, RDX = Length of message
syscall ; Call Write
; exit(rdi) = exit(return value)
xor rax, rax ; write returns # of bytes written in rax, need to clean it up again
add rax, 0x3C ; 64bit syscall exit is 0x3C
xor rdi, rdi ; Return value is in rdi (First parameter), zero it to return 0
syscall ; Call Exit
message:
call code ; Pushes the address of the string onto the stack
db 'AAAABBBNAAAAAAAABBBBBBBB',0x0A
This culprit is this line:
mov [rsi], al; Why segfault?
If I comment it out, then the program runs fine, outputting the message 'AAAABBBNAAAAAAAABBBBBBBB', why can't I modify the string?
The authors code is the following:
global _start
_start:
jmp short ender
starter:
pop ebx ;get the address of the string
xor eax, eax
mov al, 0x20
mov [ebx+7], al ;put a NULL where the N is in the string
mov al, 4 ;syscall write
mov bl, 1 ;stdout is 1
pop ecx ;get the address of the string from the stack
mov dl, 25 ;length of the string
int 0x80
xor eax, eax
mov al, 1 ;exit the shellcode
xor ebx,ebx
int 0x80
ender:
call starter
db 'AAAABBBNAAAAAAAABBBBBBBB'0x0A
And I've compiled that using:
nasm -f elf <infile> -o <outfile>
ld -m elf_i386 <infile> -o <outfile>
But even that causes a segfault, images on the page show it working properly and changing the N into a space, however I seem to be stuck in segfault land :( Google isn't really being helpful in this case, and so I turn to you stackoverflow, any pointers (no pun intended!) would be appreciated
I would assume it's because you're trying to access data that is in the .text section. Usually you're not allowed to write to code segment for security. Modifiable data should be in the .data section. (Or .bss if zero-initialized.)
For actual shellcode, where you don't want to use a separate section, see Segfault when writing to string allocated by db [assembly] for alternate workarounds.
Also I would never suggest using the side effects of call pushing the address after it to the stack to get a pointer to data following it, except for shellcode.
This is a common trick in shellcode (which must be position-independent); 32-bit mode needs a call to get EIP somehow. The call must have a backwards displacement to avoid 00 bytes in the machine code, so putting the call somewhere that creates a "return" address you specifically want saves an add or lea.
Even in 64-bit code where RIP-relative addressing is possible, jmp / call / pop is about as compact as jumping over the string for a RIP-relative LEA with a negative displacement.
Outside of the shellcode / constrained-machine-code use case, it's a terrible idea and you should just lea reg, [rel buf] like a normal person with the data in .data and the code in .text. (Or read-only data in .rodata.) This way you're not trying execute code next to data, or put data next to code.
(Code-injection vulnerabilities that allow shellcode already imply the existence of a page with write and exec permission, but normal processes from modern toolchains don't have any W+X pages unless you do something to make that happen. W^X is a good security feature for this reason, so normal toolchain security features / defaults must be defeated to test shellcode.)

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!

lost in assembly NASM ELF64 world

So as part of my Computer Architecture class I need to get comfortable with Assembly, or at least comfortable enough, I'm trying to read the input to the user and then reprint it (for the time being), this is my how I tried to laid this out in pseudo code:
Declare msg variable (this will be printed on screen)
Declare length variable (to be used by the sys_write function) with long enough value
Pop the stack once to get the program name
Pop the stack again to get the first argument
Move the current value of the stack into the msg variable
Move msg to ECX (sys_write argument)
Mov length to EDX (sys_write argument)
Call sys_write using standard output
Kernel call
Call sys_exit and leave
This is my code so far
section .data
msg: db 'placeholder text',0xa;
length: dw 0x123;
section .text
global _start
_start:
pop rbx;
pop rbx;
; this is not working when I leave it in I get this error:
; invalid combination of opcode and operands
;mov msg, rbx;
mov ecx, msg;
mov edx, length;
mov eax, 4;
mov ebx, 1;
int 0x80;
mov ebx, 0;
mov eax, 1;
int 0x80;
When I leave it out (not moving the argument into msg), I get this output
placeholder text
#.shstrtab.text.data
�#�$�`��
We really just begun with NASM so ANY help will be greatly appreciated, I've been looking at this http://www.cin.ufpe.br/~if817/arquivos/asmtut/index.html#stack and http://syscalls.kernelgrok.com/ adapting the examples adapting the registry names to the best of my understanding to match http://www.nasm.us/doc/nasmdo11.html
I'm running Ubuntu 12.04, 64bit compiling (not even sure if this is the right word) NASM under ELF64, I'm sorry to ask such a silly question but I have been unable to find an easy enough tutorial for NASM that uses 64bits.
When the program is called the stack should looks like this:
+----------------+
| ... | <--- rsp + 24
+----------------+
| argument 2 | <--- rsp + 16
+----------------+
| argument 1 | <--- rsp + 8
+----------------+
| argument count | <--- rsp
+----------------+
The first argument is the name of your program and the second is the user input (if the user typed anything as an argument). So the count of the arguments is at least 1.
The arguments for system calls in 64-mode are stored in the following registers:
rax (system call number)
rdi (1st argument)
rsi (2nd argument)
rdx (3rd argument)
rcx (4th argument)
r8 (5th argument)
r9 (6th argument)
And the system call is called with syscall. The numbers of all the system calls can be found here here (yes they are different from the numbers in 32 bit mode).
This is the program which should do your stuff:
section .data
msg: db 'Requesting 1 argument!', 10 ; message + newline
section .text
global _start
_start:
cmp qword [rsp], 2 ; check if argument count is 2
jne fail ; if not jump to the fail lable
mov rax, 1 ; sys_write
mov rdi, 1 ; stdout
mov rsi, [rsp+16] ; get the address of the argument
mov rdx, 1 ; one character (length 1)
loop:
cmp byte [rsi], 0 ; check if current character is 0
je exit ; if 0 then jump to the exit lable
syscall
inc rsi ; jump to the next character
jmp loop ; repeat
fail:
mov rax, 1 ; sys_write
mov rdi, 1 ; stdout
lea rsi, [rel msg] ; move the address of the lable msg in rsi
mov rdx, 23 ; length = 23
syscall
exit:
mov rax, 60 ; sys_exit
mov rdi, 0 ; with code 0
syscall
Since the code isn't prefect in many ways you may want to modify it.
You've followed the instructions quite literally -- and this is expected output.
The stack variable that you write to the message, is just some binary value -- to be exact, it's a pointer to an array of strings containing the command line arguments.
To make sense of that, either you would have to print those strings, or convert the pointer to ascii string eg. "0x12313132".
My OS is Ubuntu 64-bit. Compiling your code produced the error:
nasm print3.asm
print3.asm:12: error: instruction not supported in 16-bit mode
print3.asm:13: error: instruction not supported in 16-bit mode
Exactly where the "pop rbx" is located.
Adding "BITS 64" to the top of the asm file solved the problem:
BITS 64
section .data
msg: db 'placeholder text',0xa;
length: dw 0x123;
...

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