Thread: Binary Bomb help

  1. #1
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    Binary Bomb help

    I am trying to figure our how to solve the binary bomb. I used "strings" to solve the first phase. But, now I don't know how to trace through the code. What do I need to look at? I did an objdump -d and -t, i have done a disas on main(), phase_2, read_six_number(), and sscanf().

    08048d28 <phase_2>:
    8048d28: 55 push %ebp
    8048d29: 89 e5 mov %esp,%ebp
    8048d2b: 56 push %esi
    8048d2c: 53 push %ebx
    8048d2d: 83 ec 30 sub $0x30,%esp
    8048d30: 8d 45 e0 lea 0xffffffe0(%ebp),%eax
    8048d33: 89 44 24 04 mov %eax,0x4(%esp)
    8048d37: 8b 45 08 mov 0x8(%ebp),%eax
    8048d3a: 89 04 24 mov %eax,(%esp)
    8048d3d: e8 15 04 00 00 call 8049157 <read_six_numbers>
    8048d42: 83 7d e0 00 cmpl $0x0,0xffffffe0(%ebp)
    8048d46: 79 05 jns 8048d4d <phase_2+0x25>
    8048d48: e8 c8 03 00 00 call 8049115 <explode_bomb>
    8048d4d: bb 01 00 00 00 mov $0x1,%ebx
    8048d52: 8d 75 e0 lea 0xffffffe0(%ebp),%esi
    8048d55: 89 d8 mov %ebx,%eax
    8048d57: 03 44 9e fc add 0xfffffffc(%esi,%ebx,4),%eax
    8048d5b: 39 04 9e cmp %eax,(%esi,%ebx,4)
    8048d5e: 74 05 je 8048d65 <phase_2+0x3d>
    8048d60: e8 b0 03 00 00 call 8049115 <explode_bomb>
    8048d65: 83 c3 01 add $0x1,%ebx
    8048d68: 83 fb 06 cmp $0x6,%ebx
    8048d6b: 75 e8 jne 8048d55 <phase_2+0x2d>
    8048d6d: 83 c4 30 add $0x30,%esp
    8048d70: 5b pop %ebx
    8048d71: 5e pop %esi
    8048d72: 5d pop %ebp
    8048d73: c3 ret


    (gdb) disas phase_2
    Dump of assembler code for function phase_2:
    0x08048d28 <phase_2+0>: push %ebp
    0x08048d29 <phase_2+1>: mov %esp,%ebp
    0x08048d2b <phase_2+3>: push %esi
    0x08048d2c <phase_2+4>: push %ebx
    0x08048d2d <phase_2+5>: sub $0x30,%esp
    0x08048d30 <phase_2+8>: lea -0x20(%ebp),%eax
    0x08048d33 <phase_2+11>: mov %eax,0x4(%esp)
    0x08048d37 <phase_2+15>: mov 0x8(%ebp),%eax
    0x08048d3a <phase_2+18>: mov %eax,(%esp)
    0x08048d3d <phase_2+21>: call 0x8049157 <read_six_numbers>
    0x08048d42 <phase_2+26>: cmpl $0x0,-0x20(%ebp)
    0x08048d46 <phase_2+30>: jns 0x8048d4d <phase_2+37>
    0x08048d48 <phase_2+32>: call 0x8049115 <explode_bomb>
    0x08048d4d <phase_2+37>: mov $0x1,%ebx // mov 1 to ebx
    0x08048d52 <phase_2+42>: lea -0x20(%ebp),%esi
    0x08048d55 <phase_2+45>: mov %ebx,%eax
    0x08048d57 <phase_2+47>: add -0x4(%esi,%ebx,4),%eax
    0x08048d5b <phase_2+51>: cmp %eax,(%esi,%ebx,4)
    0x08048d5e <phase_2+54>: je 0x8048d65 <phase_2+61>
    0x08048d60 <phase_2+56>: call 0x8049115<explode_bomb>
    0x08048d65 <phase_2+61>: add $0x1,%ebx
    0x08048d68 <phase_2+64>: cmp $0x6,%ebx
    0x08048d6b <phase_2+67>: jne 0x8048d55 <phase_2+45>
    0x08048d6d <phase_2+69>: add $0x30,%esp
    0x08048d70 <phase_2+72>: pop %ebx
    0x08048d71 <phase_2+73>: pop %esi
    0x08048d72 <phase_2+74>: pop %ebp
    0x08048d73 <phase_2+75>: ret
    End of assembler dump.


    (gdb) disas read_six_numbers
    Dump of assembler code for function read_six_numbers:
    0x08049157 <read_six_numbers+0>: push %ebp
    0x08049158 <read_six_numbers+1>: mov %esp,%ebp
    0x0804915a <read_six_numbers+3>: sub $0x28,%esp
    0x0804915d <read_six_numbers+6>: mov 0xc(%ebp),%edx
    0x08049160 <read_six_numbers+9>: lea 0x14(%edx),%eax
    0x08049163 <read_six_numbers+12>: mov %eax,0x1c(%esp)
    0x08049167 <read_six_numbers+16>: lea 0x10(%edx),%eax
    0x0804916a <read_six_numbers+19>: mov %eax,0x18(%esp)
    0x0804916e <read_six_numbers+23>: lea 0xc(%edx),%eax
    0x08049171 <read_six_numbers+26>: mov %eax,0x14(%esp)
    0x08049175 <read_six_numbers+30>: lea 0x8(%edx),%eax
    0x08049178 <read_six_numbers+33>: mov %eax,0x10(%esp)
    0x0804917c <read_six_numbers+37>: lea 0x4(%edx),%eax
    0x0804917f <read_six_numbers+40>: mov %eax,0xc(%esp)
    0x08049183 <read_six_numbers+44>: mov %edx,0x8(%esp)
    0x08049187 <read_six_numbers+48>: movl $0x804a2d6,0x4(%esp)
    0x0804918f <read_six_numbers+56>: mov 0x8(%ebp),%eax
    0x08049192 <read_six_numbers+59>: mov %eax,(%esp)
    0x08049195 <read_six_numbers+62>: call 0x8048884 <sscanf@plt>
    0x0804919a <read_six_numbers+67>: cmp $0x5,%eax
    0x0804919d <read_six_numbers+70>: jg 0x80491a4 <read_six_numbers+77>
    0x0804919f <read_six_numbers+72>: call 0x8049115 <explode_bomb>
    0x080491a4 <read_six_numbers+77>: leave
    0x080491a5 <read_six_numbers+78>: ret
    End of assembler dump.


    (gdb) disas sscanf
    Dump of assembler code for function sscanf:
    0x00236bc0 <sscanf+0>: push %ebp
    0x00236bc1 <sscanf+1>: mov %esp,%ebp
    0x00236bc3 <sscanf+3>: push %ebx
    0x00236bc4 <sscanf+4>: sub $0x10,%esp
    0x00236bc7 <sscanf+7>: lea 0x10(%ebp),%eax
    0x00236bca <sscanf+10>: mov %eax,-0x8(%ebp)
    0x00236bcd <sscanf+13>: mov %eax,0x8(%esp)
    0x00236bd1 <sscanf+17>: mov 0xc(%ebp),%eax
    0x00236bd4 <sscanf+20>: call 0x1f6ce0 <__i686.get_pc_thunk.bx>
    0x00236bd9 <sscanf+25>: add $0xff41b,%ebx
    0x00236bdf <sscanf+31>: mov %eax,0x4(%esp)
    0x00236be3 <sscanf+35>: mov 0x8(%ebp),%eax
    0x00236be6 <sscanf+38>: mov %eax,(%esp)
    0x00236be9 <sscanf+41>: call 0x23ba00 <vsscanf>
    0x00236bee <sscanf+46>: add $0x10,%esp
    0x00236bf1 <sscanf+49>: pop %ebx
    0x00236bf2 <sscanf+50>: pop %ebp
    0x00236bf3 <sscanf+51>: ret
    End of assembler dump.

  2. #2
    and the hat of int overfl Salem's Avatar
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    Variations of this appear as either an online challenge or as homework -> CS:APP2e, Bryant and O'Hallaron

    I suppose one thing you could do is theorise what the "C" code looks like, then use "gcc -S prog.c" to generate prog.s and see how close you get.

    And yes, there are some sites which just blurt out the answer, if you're not really interested in learning anything.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

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