DO NOT READ THIS ARTICLE. I wrote bullshit in English.
This lab I have finished once, so this article is just for fun.
Phase-1
0000000000400ee0 <phase_1>:
400ee0: 48 83 ec 08 sub $0x8,%rsp
400ee4: be 00 24 40 00 mov $0x402400,%esi
400ee9: e8 4a 04 00 00 callq 401338 <strings_not_equal>
400eee: 85 c0 test %eax,%eax
400ef0: 74 05 je 400ef7 <phase_1+0x17>
400ef2: e8 43 05 00 00 callq 40143a <explode_bomb>
400ef7: 48 83 c4 08 add $0x8,%rsp
400efb: c3 retq
Just scan the memory of address [0x402400] . GDB command is x /s 0x402400 .
(gdb) x /s 0x402400
0x402400: "Border relations with Canada have never been better."
We can see that the answer is Border relations with Canada have never been better. .
Phase-2
0000000000400efc <phase_2>:
400efc: 55 push %rbp
400efd: 53 push %rbx
400efe: 48 83 ec 28 sub $0x28,%rsp
400f02: 48 89 e6 mov %rsp,%rsi
400f05: e8 52 05 00 00 callq 40145c <read_six_numbers>
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp)
400f0e: 74 20 je 400f30 <phase_2+0x34>
400f10: e8 25 05 00 00 callq 40143a <explode_bomb>
400f15: eb 19 jmp 400f30 <phase_2+0x34>
400f17: 8b 43 fc mov -0x4(%rbx),%eax
400f1a: 01 c0 add %eax,%eax
400f1c: 39 03 cmp %eax,(%rbx)
400f1e: 74 05 je 400f25 <phase_2+0x29>
400f20: e8 15 05 00 00 callq 40143a <explode_bomb>
400f25: 48 83 c3 04 add $0x4,%rbx
400f29: 48 39 eb cmp %rbp,%rbx
400f2c: 75 e9 jne 400f17 <phase_2+0x1b>
400f2e: eb 0c jmp 400f3c <phase_2+0x40>
400f30: 48 8d 5c 24 04 lea 0x4(%rsp),%rbx
400f35: 48 8d 6c 24 18 lea 0x18(%rsp),%rbp
400f3a: eb db jmp 400f17 <phase_2+0x1b>
400f3c: 48 83 c4 28 add $0x28,%rsp
400f40: 5b pop %rbx
400f41: 5d pop %rbp
400f42: c3 retq
From this code read_six_number , we can see that we should input six numbers. And the we can know that there is a loop, and the most important line of code is :
400f1a: 01 c0 add %eax,%eax
Now, we can guess that it should be a geometric series, whose ratio is 2. What is the first element of the number series? We can see:
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp)
400f0e: 74 20 je 400f30 <phase_2+0x34>
400f10: e8 25 05 00 00 callq 40143a <explode_bomb>
OK, we can see that if (%rsp) is not equal to 0x1 , it will call function explode_bomb .
Thus, the answer is : 1 2 4 8 16 32 .
Phase-3
0000000000400f43 <phase_3>:
400f43: 48 83 ec 18 sub $0x18,%rsp
400f47: 48 8d 4c 24 0c lea 0xc(%rsp),%rcx
400f4c: 48 8d 54 24 08 lea 0x8(%rsp),%rdx
400f51: be cf 25 40 00 mov $0x4025cf,%esi
400f56: b8 00 00 00 00 mov $0x0,%eax
400f5b: e8 90 fc ff ff callq 400bf0 <__isoc99_sscanf@plt>
400f60: 83 f8 01 cmp $0x1,%eax
400f63: 7f 05 jg 400f6a <phase_3+0x27>
400f65: e8 d0 04 00 00 callq 40143a <explode_bomb>
400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp)
400f6f: 77 3c ja 400fad <phase_3+0x6a>
400f71: 8b 44 24 08 mov 0x8(%rsp),%eax
400f75: ff 24 c5 70 24 40 00 jmpq *0x402470(,%rax,8)
400f7c: b8 cf 00 00 00 mov $0xcf,%eax
400f81: eb 3b jmp 400fbe <phase_3+0x7b>
400f83: b8 c3 02 00 00 mov $0x2c3,%eax
400f88: eb 34 jmp 400fbe <phase_3+0x7b>
400f8a: b8 00 01 00 00 mov $0x100,%eax
400f8f: eb 2d jmp 400fbe <phase_3+0x7b>
400f91: b8 85 01 00 00 mov $0x185,%eax
400f96: eb 26 jmp 400fbe <phase_3+0x7b>
400f98: b8 ce 00 00 00 mov $0xce,%eax
400f9d: eb 1f jmp 400fbe <phase_3+0x7b>
400f9f: b8 aa 02 00 00 mov $0x2aa,%eax
400fa4: eb 18 jmp 400fbe <phase_3+0x7b>
400fa6: b8 47 01 00 00 mov $0x147,%eax
400fab: eb 11 jmp 400fbe <phase_3+0x7b>
400fad: e8 88 04 00 00 callq 40143a <explode_bomb>
400fb2: b8 00 00 00 00 mov $0x0,%eax
400fb7: eb 05 jmp 400fbe <phase_3+0x7b>
400fb9: b8 37 01 00 00 mov $0x137,%eax
400fbe: 3b 44 24 0c cmp 0xc(%rsp),%eax
400fc2: 74 05 je 400fc9 <phase_3+0x86>
400fc4: e8 71 04 00 00 callq 40143a <explode_bomb>
400fc9: 48 83 c4 18 add $0x18,%rsp
400fcd: c3 retq
Look at this line of code:
400f51: be cf 25 40 00 mov $0x4025cf,%esi
Run x /s 0x4025cf in GDB, it will print: "%d %d" . Thus, we should input 2 int in this phase.
And we can see that there are many annoying jmp 0x400fb9 instructions in this assembly code, so we can infer that there is a switch-case structure.
The first 2 lines of code below shows that the first number we input (stored in [%rsp + 8]) is less equal than 7, otherwise it will call explode_bomb . The last 2 lines tell us that the program will jump to an address according to our first number.
400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp)
400f6f: 77 3c ja 400fad <phase_3+0x6a>
400f71: 8b 44 24 08 mov 0x8(%rsp),%eax
400f75: ff 24 c5 70 24 40 00 jmpq *0x402470(,%rax,8)
We know that the first number (now it has been copied to %rax) can be one of number in [0,1,2,3,4,5,6,7]. So scan the address [0x402470 + %rax * 8] :
(gdb) x /x 0x402470 + 8 * 0 => 0x402470: 0x00400f7c => 0 0xcf
(gdb) x /x 0x402470 + 8 * 1 => 0x402478: 0x00400fb9 => 1 0x137
(gdb) x /x 0x402470 + 8 * 2 => 0x402480: 0x00400f83 => 2 0x2c3
(gdb) x /x 0x402470 + 8 * 3 => 0x402488: 0x00400f8a => 3 0x100
(gdb) x /x 0x402470 + 8 * 4 => 0x402490: 0x00400f91 => 4 0x185
(gdb) x /x 0x402470 + 8 * 5 => 0x402498: 0x00400f98 => 5 0xce
(gdb) x /x 0x402470 + 8 * 6 => 0x4024a0: 0x00400f9f => 6 0x2aa
(gdb) x /x 0x402470 + 8 * 7 => 0x4024a8: 0x00400fa6 => 7 0x147
When the first number we input is 0, the program will jump to 0x400f7c. And the code at 0x400f7c is :
400f7c: b8 cf 00 00 00 mov $0xcf,%eax
So the second number is 0xcf = 207. Total answer is 0 207.
Phase-4
; first call is func4(edx=a, esi=b, edi=x)
0000000000400fce <func4>:
400fce: 48 83 ec 08 sub $0x8,%rsp
400fd2: 89 d0 mov %edx,%eax ; eax = a
400fd4: 29 f0 sub %esi,%eax ; eax = eax - esi = a - b = t1
400fd6: 89 c1 mov %eax,%ecx ; ecx = eax = a - b
400fd8: c1 e9 1f shr $0x1f,%ecx ; ecx = ecx >> 31 = t1 >> 31 = t2
400fdb: 01 c8 add %ecx,%eax ; eax = eax + ecx = t1 + t2
400fdd: d1 f8 sar %eax ; eax = eax >> 1 = (t1+t2)>>1, t1=(t1+t2)>>1
400fdf: 8d 0c 30 lea (%rax,%rsi,1),%ecx ; ecx = eax + esi = t1 + b, t2=t1+b
400fe2: 39 f9 cmp %edi,%ecx ; ecx <= x ? () : (2 * func4())
400fe4: 7e 0c jle 400ff2 <func4+0x24> ;
400fe6: 8d 51 ff lea -0x1(%rcx),%edx ; edx = ecx - 1 = t2 - 1
400fe9: e8 e0 ff ff ff callq 400fce <func4> ; func4(edx=t2-1, b, x)
400fee: 01 c0 add %eax,%eax ; return 2 * func4()
400ff0: eb 15 jmp 401007 <func4+0x39>
400ff2: b8 00 00 00 00 mov $0x0,%eax ; eax = 0
400ff7: 39 f9 cmp %edi,%ecx ; ecx >= edi ? (return 0) : ()
400ff9: 7d 0c jge 401007 <func4+0x39>
400ffb: 8d 71 01 lea 0x1(%rcx),%esi ; esi = ecx + 1
400ffe: e8 cb ff ff ff callq 400fce <func4> ; func4()
401003: 8d 44 00 01 lea 0x1(%rax,%rax,1),%eax ; return 2 * func4() + 1
401007: 48 83 c4 08 add $0x8,%rsp
40100b: c3 retq
000000000040100c <phase_4>:
40100c: 48 83 ec 18 sub $0x18,%rsp
401010: 48 8d 4c 24 0c lea 0xc(%rsp),%rcx
401015: 48 8d 54 24 08 lea 0x8(%rsp),%rdx
40101a: be cf 25 40 00 mov $0x4025cf,%esi
; scan the address 0x4025cf
40101f: b8 00 00 00 00 mov $0x0,%eax
401024: e8 c7 fb ff ff callq 400bf0 <__isoc99_sscanf@plt>
401029: 83 f8 02 cmp $0x2,%eax
40102c: 75 07 jne 401035 <phase_4+0x29>
40102e: 83 7c 24 08 0e cmpl $0xe,0x8(%rsp)
; the 1st number x should be x <= 15, otherwise explode
401033: 76 05 jbe 40103a <phase_4+0x2e>
401035: e8 00 04 00 00 callq 40143a <explode_bomb>
40103a: ba 0e 00 00 00 mov $0xe,%edx
40103f: be 00 00 00 00 mov $0x0,%esi
401044: 8b 7c 24 08 mov 0x8(%rsp),%edi
401048: e8 81 ff ff ff callq 400fce <func4>
; func4(edx=0xe, esi=0x0, edi=x)
40104d: 85 c0 test %eax,%eax
; the returned val of func4 shouble 0
40104f: 75 07 jne 401058 <phase_4+0x4c>
401051: 83 7c 24 0c 00 cmpl $0x0,0xc(%rsp)
; the 2nd number y should be 0, otherwise explode
; so the total condition is func(0xe,0x0,x) == 0 and y = 0
401056: 74 05 je 40105d <phase_4+0x51>
401058: e8 dd 03 00 00 callq 40143a <explode_bomb>
40105d: 48 83 c4 18 add $0x18,%rsp
401061: c3 retq
phase_4 call a function func4, which is a recursion. Scan the address 0x4025cf, GDB will print "%d %d", which means we should input 2 int numbers. Assume that our input is x y. The x will be stored in %rsp + 0x8 and y in %rsp + 0xc. Then phase_4 would call func4(edx=0xe, esi=0x0, edi=x), and this function call passes parameters by register. Read the comment above.
The conditions for passing phase_4 is input x y to let func(0xe, 0x0, x) return 0 and x should be less equal than 15 and y must be 0.
Read the code of func4, we can change it into C code:
int func4(int a, int b, int x)
{
x; // edi
int t1 = a - b; // eax
int t2 = ((unsigned int)t1) >> 31; // ecx
t1 = (t1 + t2) >> 1;
t2 = t1 + b;
if (t2 <= x)
{
t1 = 0;
if (t2 < x)
return 2 * func4(a, t2 + 1, x) + 1;
return t1;
}
else
return 2 * func4(t2 - 1, b, x);
}
What we need to pay attention to is this instruction: 400fd8: shr $0x1f,%ecx. The shr implies that it will operation on a unsigned type (read the C code above) .
Therefore, we can find x by enumeration.
int main()
{
int x = 0;
for (x = 0; x <= 15; x++)
{
int t = func4(0xe, 0x0, x);
if (t == 0)
printf("%d ", x);
}
}
The result is:
(x,y) = (0,0), (1,0), (3,0), (7,0)
Phase-5
0000000000401062 <phase_5>:
401062: 53 push %rbx
401063: 48 83 ec 20 sub $0x20,%rsp
; 32 bytes stack for temporary variable
401067: 48 89 fb mov %rdi,%rbx
; rdi is the address of strings we input
40106a: 64 48 8b 04 25 28 00 mov %fs:0x28,%rax
401071: 00 00
; I don't understand here
401073: 48 89 44 24 18 mov %rax,0x18(%rsp)
401078: 31 c0 xor %eax,%eax
40107a: e8 9c 02 00 00 callq 40131b <string_length>
40107f: 83 f8 06 cmp $0x6,%eax
; we should input a string, whose len is 6
; do while loop begin, eax=0 is the counter
401082: 74 4e je 4010d2 <phase_5+0x70>
401084: e8 b1 03 00 00 callq 40143a <explode_bomb>
401089: eb 47 jmp 4010d2 <phase_5+0x70>
40108b: 0f b6 0c 03 movzbl (%rbx,%rax,1),%ecx
; t = str[i]
40108f: 88 0c 24 mov %cl,(%rsp)
401092: 48 8b 14 24 mov (%rsp),%rdx
401096: 83 e2 0f and $0xf,%edx
; t = str[i] & 0xf
401099: 0f b6 92 b0 24 40 00 movzbl 0x4024b0(%rdx),%edx
4010a0: 88 54 04 10 mov %dl,0x10(%rsp,%rax,1)
; array[rsp+0x10+rax] = str2[t]
4010a4: 48 83 c0 01 add $0x1,%rax
4010a8: 48 83 f8 06 cmp $0x6,%rax
; eax++ until out of str.length
4010ac: 75 dd jne 40108b <phase_5+0x29>
4010ae: c6 44 24 16 00 movb $0x0,0x16(%rsp)
4010b3: be 5e 24 40 00 mov $0x40245e,%esi
; "flyers"
4010b8: 48 8d 7c 24 10 lea 0x10(%rsp),%rdi
4010bd: e8 76 02 00 00 callq 401338 <strings_not_equal>
4010c2: 85 c0 test %eax,%eax
4010c4: 74 13 je 4010d9 <phase_5+0x77>
4010c6: e8 6f 03 00 00 callq 40143a <explode_bomb>
4010cb: 0f 1f 44 00 00 nopl 0x0(%rax,%rax,1)
4010d0: eb 07 jmp 4010d9 <phase_5+0x77>
4010d2: b8 00 00 00 00 mov $0x0,%eax
4010d7: eb b2 jmp 40108b <phase_5+0x29>
4010d9: 48 8b 44 24 18 mov 0x18(%rsp),%rax
4010de: 64 48 33 04 25 28 00 xor %fs:0x28,%rax
4010e5: 00 00
4010e7: 74 05 je 4010ee <phase_5+0x8c>
4010e9: e8 42 fa ff ff callq 400b30 <__stack_chk_fail@plt>
4010ee: 48 83 c4 20 add $0x20,%rsp
4010f2: 5b pop %rbx
4010f3: c3 retq
This phase_5 want us to input a string str (its length is 6), which is a simple phase.
There is a hash table in the program:
hex index: 0 1 2 3 4 5 6 7 8 9 A B C D E F
hash str : m a d u i e r s n f o t v b y l
phase_5 construct a hash function is: f(x) = hash_str[x & 0xf].
We aim to let f(str) = "flyers".
flyers is equal to character 9, F, E, 4, 5, 6 in the hash table, thus, str[i] & 0xf (i=0,...,5) should be 9, F, E, 4, 5, 6. We can check the ASCII table to get characters:
0x39 0x3F 0x3E 0x34 0x35 0x36
'9' '?' '>' '4' '5' '6'
The string "9?>456" is the key to pass this phase.
If you still do not understand, see the C code:
void phase5(const char *input)
{
const char *hash_str = "maduiersnfotvbyl";
char array[7];
int i = 0;
for (i = 0; i < 6; i++)
array[i] = hash_str[input[i] & 0xf];
if (sting_not_equal(array, "flyers"))
explode();
}
Evaluation
All the answers in a text files sin.ans:
unix > cat sin.ans
Border relations with Canada have never been better.
1 2 4 8 16 32
0 207
7 0 DrEvil
9?>456
4 3 2 1 6 5
20
Execute the bomb binary file:
unix > ./bomb sin.ans
Welcome to my fiendish little bomb. You have 6 phases with
which to blow yourself up. Have a nice day!
Phase 1 defused. How about the next one?
That's number 2. Keep going!
Halfway there!
So you got that one. Try this one.
Good work! On to the next...
Curses, you've found the secret phase!
But finding it and solving it are quite different...
Wow! You've defused the secret stage!
Congratulations! You've defused the bomb!
Summary
好无聊，希望早点开学🤢，👴要🤮了。