Bomb lab phase 4 func4 - doubly-recursive function that calls itself with n-1 and n-2

Question

I'm having trouble with the password for phase_4 of my binary bomb.

So far, I understand... the inputs must be 2 integers (line 29)

and the second integer should <= than or equal to 2 when 2 is subtracted from it (lines 38-44). Which means it should be a number 2-4

Then, it calls func4, and compares the output of func4 to rsp (which I've done some testing and rsp is always 0) so i need to find a number that returns 0 when inputed into func4.

Whats confusing me is: if the result of func4 needs to be 0, that means it's input should be 0. Taking into account line 38 in phase_4 where it subtracts 2 from my input, that should mean my second # should be 2...? However I have tried that and it makes it explode in line 46 as it doesn't satisfy the first comparison.

I have tried numbers like (0 2) (2 0) (0 1) (0 3) and they all failed. Any suggestions to point me in the right direction? Thanks!

phase_4:
   0x000000000040101e <+0>: sub    $0x18,%rsp
   0x0000000000401022 <+4>: lea    0xc(%rsp),%rcx
   0x0000000000401027 <+9>: lea    0x8(%rsp),%rdx
   0x000000000040102c <+14>:    mov    $0x4027cd,%esi
   0x0000000000401031 <+19>:    mov    $0x0,%eax
   0x0000000000401036 <+24>:    callq  0x400c30 <__isoc99_sscanf@plt>
   0x000000000040103b <+29>:    cmp    $0x2,%eax        //check if 2 inputs
   0x000000000040103e <+32>:    jne    0x40104c <phase_4+46>
   0x0000000000401040 <+34>:    mov    0xc(%rsp),%eax   
=> 0x0000000000401044 <+38>:    sub    $0x2,%eax        
   0x0000000000401047 <+41>:    cmp    $0x2,%eax
   0x000000000040104a <+44>:    jbe    0x401051 <phase_4+51>//if unsigned eax <= 2
   0x000000000040104c <+46>:    callq  0x401554 <explode_bomb>
   0x0000000000401051 <+51>:    mov    0xc(%rsp),%esi  
   0x0000000000401055 <+55>:    mov    $0x7,%edi      
   0x000000000040105a <+60>:    callq  0x400fe6 <func4>
   0x000000000040105f <+65>:    cmp    0x8(%rsp),%eax  //comparing eax to 0
   0x0000000000401063 <+69>:    je     0x40106a <phase_4+76> 
   0x0000000000401065 <+71>:    callq  0x401554 <explode_bomb> //explode if output != 0
   0x000000000040106a <+76>:    add    $0x18,%rsp
   0x000000000040106e <+80>:    retq

Func4
   0x0000000000400fe6 <+0>: push   %r12
   0x0000000000400fe8 <+2>: push   %rbp
   0x0000000000400fe9 <+3>: push   %rbx
   0x0000000000400fea <+4>: mov    %edi,%ebx
   0x0000000000400fec <+6>: test   %edi,%edi
   0x0000000000400fee <+8>: jle    0x401014 <func4+46> //if input <= 0
   0x0000000000400ff0 <+10>:    mov    %esi,%ebp
   0x0000000000400ff2 <+12>:    mov    %esi,%eax
   0x0000000000400ff4 <+14>:    cmp    $0x1,%edi
   0x0000000000400ff7 <+17>:    je     0x401019 <func4+51>
   0x0000000000400ff9 <+19>:    lea    -0x1(%rdi),%edi
   0x0000000000400ffc <+22>:    callq  0x400fe6 <func4>
   0x0000000000401001 <+27>:    lea    (%rax,%rbp,1),%r12d
   0x0000000000401005 <+31>:    lea    -0x2(%rbx),%edi
   0x0000000000401008 <+34>:    mov    %ebp,%esi
   0x000000000040100a <+36>:    callq  0x400fe6 <func4>
   0x000000000040100f <+41>:    add    %r12d,%eax
   0x0000000000401012 <+44>:    jmp    0x401019 <func4+51>
   0x0000000000401014 <+46>:    mov    $0x0,%eax  //make return val 0
   0x0000000000401019 <+51>:    pop    %rbx
   0x000000000040101a <+52>:    pop    %rbp
   0x000000000040101b <+53>:    pop    %r12
   0x000000000040101d <+55>:    retq

Might want to explain what a binary bomb is and explain a bit more what your approach has been for solving it and why you think it's not working. — Qix - MONICA WAS MISTREATED
– Qix - MONICA WAS MISTREATED, Commented Oct 22, 2016 at 22:26
@BoPersson phase 4 is similar, but func 4 is different. I believe in my case func4 is a recursive Fibonacci function. And i know my second input must be >= 2 — helloworld
– helloworld, Commented Oct 22, 2016 at 22:52

score 2 · Accepted Answer · 2019-11-16 13:42:39Z

2

Then, it calls func4, and compares the output of func4 to rsp (which I've done some testing and rsp is always 0) so i need to find a number that returns 0 when inputed into func4.

This is incorrect. The output of func4 is compared with [rsp + 8], in which the first number was stored.

If we write the desired input as (a, b), then we have a = func4 (7, b) and 2 <= b <= 4.

To understand what func4 (x, y) does I recommend that you convert it to C. See my answer to this question for an illustration.

edited Nov 16, 2019 at 13:42

answered Oct 23, 2016 at 16:35

user1157391

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1 Comment

helloworld Over a year ago

thanks so much, yeah i figured that out yesterday. i was doing "i r rsp" thinking it automatically did the addition, but it didn't. Turns out it was just comparing the output of func4 to the first integer that i inputed. Thanks for your help!

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