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464 lines (417 loc) · 23 KB

Phase1

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   

string_not_equal 比较两个字符串,作为参数放在%esi中传入的0x402400地址处的字符串就是答案。

(gdb) x/s 0x402400
0x402400:       "Border relations with Canada have never been better."
(gdb)

Phase2

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   

一个名为read_six_numbers的函数自解释了输入是6个整数,第11-18行是一个循环,要求前一个数是后一个数的两倍。第7行指出第一个数是1,所以答案为1 2 4 8 16 32.

Phase3

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   

第5行的内存地址处有一个字符串%d %d,表明输入是两个整数。第11行要求第一个输入小于7,第14行的0x402470处是一个跳转表,根据第一个输入值决定执行哪个分支。所有的分支都会走到0x400fbe处,只是赋给%eax的值不同,这个值会与第二个输入比较。这个阶段有6组解,随便选取跳转表的第一项,发现跳至0x400f7c处,这里赋给$eax的0xcf=207就是第二个输入。故一组答案为0 207

Phase4

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
  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)
  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>
  40104d:	85 c0                	test   %eax,%eax
  40104f:	75 07                	jne    401058 <phase_4+0x4c>
  401051:	83 7c 24 0c 00       	cmpl   $0x0,0xc(%rsp)
  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   
  
0000000000400fce <func4>:
  400fce:	48 83 ec 08          	sub    $0x8,%rsp
  400fd2:	89 d0                	mov    %edx,%eax
  400fd4:	29 f0                	sub    %esi,%eax
  400fd6:	89 c1                	mov    %eax,%ecx
  400fd8:	c1 e9 1f             	shr    $0x1f,%ecx
  400fdb:	01 c8                	add    %ecx,%eax
  400fdd:	d1 f8                	sar    %eax
  400fdf:	8d 0c 30             	lea    (%rax,%rsi,1),%ecx
  400fe2:	39 f9                	cmp    %edi,%ecx
  400fe4:	7e 0c                	jle    400ff2 <func4+0x24>
  400fe6:	8d 51 ff             	lea    -0x1(%rcx),%edx
  400fe9:	e8 e0 ff ff ff       	callq  400fce <func4>
  400fee:	01 c0                	add    %eax,%eax
  400ff0:	eb 15                	jmp    401007 <func4+0x39>
  400ff2:	b8 00 00 00 00       	mov    $0x0,%eax
  400ff7:	39 f9                	cmp    %edi,%ecx
  400ff9:	7d 0c                	jge    401007 <func4+0x39>
  400ffb:	8d 71 01             	lea    0x1(%rcx),%esi
  400ffe:	e8 cb ff ff ff       	callq  400fce <func4>
  401003:	8d 44 00 01          	lea    0x1(%rax,%rax,1),%eax
  401007:	48 83 c4 08          	add    $0x8,%rsp
  40100b:	c3                   	retq   

第5行暴露了输入是两个整数(同phase3),第19行要求第二个输入必须是0(奇怪,为什么不干脆做成一个输入),第17行要求递归函数func4的返回值是0,下面开始分析func4。注意func4有3个参数,第一个参数是输入,初始调用为func4($edi,0,14).

观察可以发现,要让返回值为0必须绕开45行。唯一可以让%rax为0的返回路径是40->41->42->46,而进入40行的唯一方法是从35行跳进去。要让%ecx >= x才能从35行进入40行,而从42到45行又必须让%ecx <= x,所以只有%ecx==x时才能以%eax==0返回。

记t=%ecx,func4的3个参数分别为x(%edi),y(%esi),z(%edx)。观察27-33行不难发现:

if z>=y : 
	t=(z-y)/2+y	
else : 
	t=(z-y+1)/2+y

注意shr $0x1f, %ecx这个右移31位的指令是在取%ecx的符号位。

第一轮迭代func4(x,0,14)中t=7,于是我们令x=t=7即可。故答案为7 0

Phase5

0000000000401062 <phase_5>:
  401062:	53                   	push   %rbx
  401063:	48 83 ec 20          	sub    $0x20,%rsp
  401067:	48 89 fb             	mov    %rdi,%rbx
  40106a:	64 48 8b 04 25 28 00 	mov    %fs:0x28,%rax
  401071:	00 00 
  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
  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
  40108f:	88 0c 24             	mov    %cl,(%rsp)
  401092:	48 8b 14 24          	mov    (%rsp),%rdx
  401096:	83 e2 0f             	and    $0xf,%edx
  401099:	0f b6 92 b0 24 40 00 	movzbl 0x4024b0(%rdx),%edx
  4010a0:	88 54 04 10          	mov    %dl,0x10(%rsp,%rax,1)
  4010a4:	48 83 c0 01          	add    $0x1,%rax
  4010a8:	48 83 f8 06          	cmp    $0x6,%rax
  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
  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   

第10行表明输入为长度为6的字符串。14-22行是一个循环,把字符串的每个字符的ascii码取后4位,然后去索引位于0x4024b0处的一个字符数组"maduiersnfotvbyl"(相当于做一个映射),得到的结果应该是0x40245e处的字符串"flyers"。所以我们依次选取的下标是9 15 14 5 6 7,一组解是0x59 0x4F 0x4E 0x45 0x46 0x47,即字符串"YONEFG"。

Phase6

00000000004010f4 <phase_6>:
  4010f4:	41 56                	push   %r14
  4010f6:	41 55                	push   %r13
  4010f8:	41 54                	push   %r12
  4010fa:	55                   	push   %rbp
  4010fb:	53                   	push   %rbx
  4010fc:	48 83 ec 50          	sub    $0x50,%rsp
  401100:	49 89 e5             	mov    %rsp,%r13
  401103:	48 89 e6             	mov    %rsp,%rsi
  401106:	e8 51 03 00 00       	callq  40145c <read_six_numbers>
  40110b:	49 89 e6             	mov    %rsp,%r14
  40110e:	41 bc 00 00 00 00    	mov    $0x0,%r12d

  401114:	4c 89 ed             	mov    %r13,%rbp
  401117:	41 8b 45 00          	mov    0x0(%r13),%eax
  40111b:	83 e8 01             	sub    $0x1,%eax
  40111e:	83 f8 05             	cmp    $0x5,%eax
  401121:	76 05                	jbe    401128 <phase_6+0x34>
  401123:	e8 12 03 00 00       	callq  40143a <explode_bomb>
  401128:	41 83 c4 01          	add    $0x1,%r12d
  40112c:	41 83 fc 06          	cmp    $0x6,%r12d
  401130:	74 21                	je     401153 <phase_6+0x5f>
  401132:	44 89 e3             	mov    %r12d,%ebx
    401135:	48 63 c3             	movslq %ebx,%rax
    401138:	8b 04 84             	mov    (%rsp,%rax,4),%eax
    40113b:	39 45 00             	cmp    %eax,0x0(%rbp)
    40113e:	75 05                	jne    401145 <phase_6+0x51>
    401140:	e8 f5 02 00 00       	callq  40143a <explode_bomb>
    401145:	83 c3 01             	add    $0x1,%ebx
    401148:	83 fb 05             	cmp    $0x5,%ebx
    40114b:	7e e8                	jle    401135 <phase_6+0x41>
  40114d:	49 83 c5 04          	add    $0x4,%r13
  401151:	eb c1                	jmp    401114 <phase_6+0x20>

  401153:	48 8d 74 24 18       	lea    0x18(%rsp),%rsi
  401158:	4c 89 f0             	mov    %r14,%rax
  40115b:	b9 07 00 00 00       	mov    $0x7,%ecx
  401160:	89 ca                	mov    %ecx,%edx
  401162:	2b 10                	sub    (%rax),%edx
  401164:	89 10                	mov    %edx,(%rax)
  401166:	48 83 c0 04          	add    $0x4,%rax
  40116a:	48 39 f0             	cmp    %rsi,%rax
  40116d:	75 f1                	jne    401160 <phase_6+0x6c>
  
  40116f:	be 00 00 00 00       	mov    $0x0,%esi
  401174:	eb 21                	jmp    401197 <phase_6+0xa3>
  
  401176:	48 8b 52 08          	mov    0x8(%rdx),%rdx
  40117a:	83 c0 01             	add    $0x1,%eax
  40117d:	39 c8                	cmp    %ecx,%eax
  40117f:	75 f5                	jne    401176 <phase_6+0x82>
  401181:	eb 05                	jmp    401188 <phase_6+0x94>
  401183:	ba d0 32 60 00       	mov    $0x6032d0,%edx
  401188:	48 89 54 74 20       	mov    %rdx,0x20(%rsp,%rsi,2)
  40118d:	48 83 c6 04          	add    $0x4,%rsi
  401191:	48 83 fe 18          	cmp    $0x18,%rsi
  401195:	74 14                	je     4011ab <phase_6+0xb7>
  401197:	8b 0c 34             	mov    (%rsp,%rsi,1),%ecx
  40119a:	83 f9 01             	cmp    $0x1,%ecx
  40119d:	7e e4                	jle    401183 <phase_6+0x8f>
  40119f:	b8 01 00 00 00       	mov    $0x1,%eax
  4011a4:	ba d0 32 60 00       	mov    $0x6032d0,%edx
  4011a9:	eb cb                	jmp    401176 <phase_6+0x82>
  
  4011ab:	48 8b 5c 24 20       	mov    0x20(%rsp),%rbx
  4011b0:	48 8d 44 24 28       	lea    0x28(%rsp),%rax
  4011b5:	48 8d 74 24 50       	lea    0x50(%rsp),%rsi
  4011ba:	48 89 d9             	mov    %rbx,%rcx
  4011bd:	48 8b 10             	mov    (%rax),%rdx
  4011c0:	48 89 51 08          	mov    %rdx,0x8(%rcx)
  4011c4:	48 83 c0 08          	add    $0x8,%rax
  4011c8:	48 39 f0             	cmp    %rsi,%rax
  4011cb:	74 05                	je     4011d2 <phase_6+0xde>
  4011cd:	48 89 d1             	mov    %rdx,%rcx
  4011d0:	eb eb                	jmp    4011bd <phase_6+0xc9>
  4011d2:	48 c7 42 08 00 00 00 	movq   $0x0,0x8(%rdx)
  4011d9:	00 
 	
  4011da:	bd 05 00 00 00       	mov    $0x5,%ebp
  4011df:	48 8b 43 08          	mov    0x8(%rbx),%rax
  4011e3:	8b 00                	mov    (%rax),%eax
  4011e5:	39 03                	cmp    %eax,(%rbx)
  4011e7:	7d 05                	jge    4011ee <phase_6+0xfa>
  4011e9:	e8 4c 02 00 00       	callq  40143a <explode_bomb>
  4011ee:	48 8b 5b 08          	mov    0x8(%rbx),%rbx
  4011f2:	83 ed 01             	sub    $0x1,%ebp
  4011f5:	75 e8                	jne    4011df <phase_6+0xeb>
  4011f7:	48 83 c4 50          	add    $0x50,%rsp
  4011fb:	5b                   	pop    %rbx
  4011fc:	5d                   	pop    %rbp
  4011fd:	41 5c                	pop    %r12
  4011ff:	41 5d                	pop    %r13
  401201:	41 5e                	pop    %r14
  401203:	c3                   	retq   

第10行看出输入是6个整数,14-33行是一个双层循环,检测6个数是否两两不同,其中16-17行保证每个数都在1-6之间。因此得出结论,输入是1 2 3 4 5 6的某种排列。35-43行的循环用7去减每个输入,放回原处。

48-63行提示我们看看0x6032d0处有什么,这里有一个结构体的链表:

struct node {
  int value;
  int index;
  node *next;
};

48-63行按照6个整数指出的index取出了6个node的地址,并放在了内存中某一位置。

65-75行按照这6个地址重新组织链表,因此,链表被按照6个整数指出的顺序重新排序了。

汇编的最后部分要求6个node的value从大到小排列,初始value为:

(gdb) x/24x 0x6032d0
0x6032d0 <node1>:       0x0000014c      0x00000001      0x006032e0      0x00000000
0x6032e0 <node2>:       0x000000a8      0x00000002      0x006032f0      0x00000000
0x6032f0 <node3>:       0x0000039c      0x00000003      0x00603300      0x00000000
0x603300 <node4>:       0x000002b3      0x00000004      0x00603310      0x00000000
0x603310 <node5>:       0x000001dd      0x00000005      0x00603320      0x00000000
0x603320 <node6>:       0x000001bb      0x00000006      0x00000000      0x00000000

顺序应为3 4 5 6 1 2,这是用7减之后的结果。所以答案为4 3 2 1 6 5

隐藏阶段

二进制里有一个符号secret_phase,搜索发现它出现在函数phase_defused里。

00000000004015c4 <phase_defused>:
  4015c4:	48 83 ec 78          	sub    $0x78,%rsp
  4015c8:	64 48 8b 04 25 28 00 	mov    %fs:0x28,%rax
  4015cf:	00 00 
  4015d1:	48 89 44 24 68       	mov    %rax,0x68(%rsp)
  4015d6:	31 c0                	xor    %eax,%eax
  4015d8:	83 3d 81 21 20 00 06 	cmpl   $0x6,0x202181(%rip)        # 603760 <num_input_strings>
  4015df:	75 5e                	jne    40163f <phase_defused+0x7b>
  4015e1:	4c 8d 44 24 10       	lea    0x10(%rsp),%r8
  4015e6:	48 8d 4c 24 0c       	lea    0xc(%rsp),%rcx
  4015eb:	48 8d 54 24 08       	lea    0x8(%rsp),%rdx
  4015f0:	be 19 26 40 00       	mov    $0x402619,%esi
  4015f5:	bf 70 38 60 00       	mov    $0x603870,%edi
  4015fa:	e8 f1 f5 ff ff       	callq  400bf0 <__isoc99_sscanf@plt>
  4015ff:	83 f8 03             	cmp    $0x3,%eax
  401602:	75 31                	jne    401635 <phase_defused+0x71>
  401604:	be 22 26 40 00       	mov    $0x402622,%esi
  401609:	48 8d 7c 24 10       	lea    0x10(%rsp),%rdi
  40160e:	e8 25 fd ff ff       	callq  401338 <strings_not_equal>
  401613:	85 c0                	test   %eax,%eax
  401615:	75 1e                	jne    401635 <phase_defused+0x71>
  401617:	bf f8 24 40 00       	mov    $0x4024f8,%edi
  40161c:	e8 ef f4 ff ff       	callq  400b10 <puts@plt>
  401621:	bf 20 25 40 00       	mov    $0x402520,%edi
  401626:	e8 e5 f4 ff ff       	callq  400b10 <puts@plt>
  40162b:	b8 00 00 00 00       	mov    $0x0,%eax
  401630:	e8 0d fc ff ff       	callq  401242 <secret_phase>
  401635:	bf 58 25 40 00       	mov    $0x402558,%edi
  40163a:	e8 d1 f4 ff ff       	callq  400b10 <puts@plt>
  40163f:	48 8b 44 24 68       	mov    0x68(%rsp),%rax
  401644:	64 48 33 04 25 28 00 	xor    %fs:0x28,%rax
  40164b:	00 00 
  40164d:	74 05                	je     401654 <phase_defused+0x90>
  40164f:	e8 dc f4 ff ff       	callq  400b30 <__stack_chk_fail@plt>
  401654:	48 83 c4 78          	add    $0x78,%rsp
  401658:	c3                   	retq   
  401659:	90                   	nop
  40165a:	90                   	nop
  40165b:	90                   	nop
  40165c:	90                   	nop
  40165d:	90                   	nop
  40165e:	90                   	nop
  40165f:	90                   	nop

第12行位置的字符串是"%d %d %s",第13行位置的字符串是阶段4的输入"7 0",看起来需要在阶段4输入后面补一个字符串进入隐藏阶段。这个字符串就是strings_not_equal的参数,位于0x402622处的DrEvil.

0000000000401242 <secret_phase>:
  401242:	53                   	push   %rbx
  401243:	e8 56 02 00 00       	callq  40149e <read_line>
  401248:	ba 0a 00 00 00       	mov    $0xa,%edx
  40124d:	be 00 00 00 00       	mov    $0x0,%esi
  401252:	48 89 c7             	mov    %rax,%rdi
  401255:	e8 76 f9 ff ff       	callq  400bd0 <strtol@plt>
  40125a:	48 89 c3             	mov    %rax,%rbx
  40125d:	8d 40 ff             	lea    -0x1(%rax),%eax
  401260:	3d e8 03 00 00       	cmp    $0x3e8,%eax
  401265:	76 05                	jbe    40126c <secret_phase+0x2a>
  401267:	e8 ce 01 00 00       	callq  40143a <explode_bomb>
  40126c:	89 de                	mov    %ebx,%esi
  40126e:	bf f0 30 60 00       	mov    $0x6030f0,%edi
  401273:	e8 8c ff ff ff       	callq  401204 <fun7>
  401278:	83 f8 02             	cmp    $0x2,%eax
  40127b:	74 05                	je     401282 <secret_phase+0x40>
  40127d:	e8 b8 01 00 00       	callq  40143a <explode_bomb>
  401282:	bf 38 24 40 00       	mov    $0x402438,%edi
  401287:	e8 84 f8 ff ff       	callq  400b10 <puts@plt>
  40128c:	e8 33 03 00 00       	callq  4015c4 <phase_defused>
  401291:	5b                   	pop    %rbx
  401292:	c3                   	retq   

这段汇编要求输入是小于1000的正整数,并用其作为第二个参数调用fun7,还要求fun7返回2。

0000000000401204 <fun7>:
  401204:	48 83 ec 08          	sub    $0x8,%rsp
  401208:	48 85 ff             	test   %rdi,%rdi
  40120b:	74 2b                	je     401238 <fun7+0x34>
  40120d:	8b 17                	mov    (%rdi),%edx
  40120f:	39 f2                	cmp    %esi,%edx
  401211:	7e 0d                	jle    401220 <fun7+0x1c>
  401213:	48 8b 7f 08          	mov    0x8(%rdi),%rdi
  401217:	e8 e8 ff ff ff       	callq  401204 <fun7>
  40121c:	01 c0                	add    %eax,%eax
  40121e:	eb 1d                	jmp    40123d <fun7+0x39>
  401220:	b8 00 00 00 00       	mov    $0x0,%eax
  401225:	39 f2                	cmp    %esi,%edx
  401227:	74 14                	je     40123d <fun7+0x39>
  401229:	48 8b 7f 10          	mov    0x10(%rdi),%rdi
  40122d:	e8 d2 ff ff ff       	callq  401204 <fun7>
  401232:	8d 44 00 01          	lea    0x1(%rax,%rax,1),%eax
  401236:	eb 05                	jmp    40123d <fun7+0x39>
  401238:	b8 ff ff ff ff       	mov    $0xffffffff,%eax
  40123d:	48 83 c4 08          	add    $0x8,%rsp
  401241:	c3                   	retq  

fun7的第一个参数是一个指针,看看这个指针的初始值处存了什么:

(gdb) x/30x 0x6030f0
0x6030f0 <n1>:  0x00000024      0x00000000      0x00603110      0x00000000
0x603100 <n1+16>:       0x00603130      0x00000000      0x00000000      0x00000000
0x603110 <n21>: 0x00000008      0x00000000      0x00603190      0x00000000
0x603120 <n21+16>:      0x00603150      0x00000000      0x00000000      0x00000000
0x603130 <n22>: 0x00000032      0x00000000      0x00603170      0x00000000
0x603140 <n22+16>:      0x006031b0      0x00000000      0x00000000      0x00000000
0x603150 <n32>: 0x00000016      0x00000000      0x00603270      0x00000000
0x603160 <n32+16>:      0x00603230      0x00000000

每个node有一个value和两个指针,自然猜想到这里存了一棵二叉树。知道了这一点,不难看出fun7是二叉搜索树上的查找过程,且从右子树返回时对返回值2+1,从左子树返回时对返回值2,在当前节点匹配时返回0。要让返回值是2,需要向左子树查询一次,再向右子树查询一次。于是从根节点开始向左走一步(0x603110),再向右走一步(0x603150),该节点的value为0x16,故答案为22