BOMB LAB
很好玩的一个lab，跟解密游戏一样，层层递进，还有隐藏boss。
Answer
存在多解，详见解析。
Border relations with Canada have never been better.
1 2 4 8 16 32
0 207
3 0 DrEvil
ionefg
4 3 2 1 6 5
20Solution
Phase_1
0000000000400ee0 <phase_1>:
400ee0: 48 83 ec 08 sub $0x8,%rsp
400ee4: be 00 24 40 00 mov $0x402400,%esi # 这里的地址就是模式串
400ee9: e8 4a 04 00 00 call 401338 <strings_not_equal>
400eee: 85 c0 test %eax,%eax
400ef0: 74 05 je 400ef7 <phase_1+0x17>
400ef2: e8 43 05 00 00 call 40143a <explode_bomb>
400ef7: 48 83 c4 08 add $0x8,%rsp
400efb: c3 这个很简单，显然只是与给定字符串比较是否相等（不放心的话可以去检查一下<strings_not_equal>，防止里面偷偷搞了什么加密运算）
x/s 0x402400然后得到答案了
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 call 40145c <read_six_numbers>
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp) # a1 = 1
400f0e: 74 20 je 400f30 <phase_2+0x34>
400f10: e8 25 05 00 00 call 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 # a(i-1) * 2
400f1c: 39 03 cmp %eax,(%rbx)
400f1e: 74 05 je 400f25 <phase_2+0x29> # ai == a(i-1) * 2
400f20: e8 15 05 00 00 call 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 ret- 首先确认输入为
<read_six_numbers>，确认无误后确定这题答案为 6 个 4 字节数字。 - 找到循环，找到爆炸判定（
400f17~400f25），-0x4(%rbx)对应a[i-1]，(%rbx)对应a[i]。 - 答案为首项为 1 ，公比位 2 的等比数列
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 "%d %d"
400f56: b8 00 00 00 00 mov $0x0,%eax
400f5b: e8 90 fc ff ff call 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 call 40143a <explode_bomb>
# <phase_3+0x27>
400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp) # n1 <= 7
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 jmp *0x402470(,%rax,8)
# 0
400f7c: b8 cf 00 00 00 mov $0xcf,%eax
400f81: eb 3b jmp 400fbe <phase_3+0x7b>
# 2
400f83: b8 c3 02 00 00 mov $0x2c3,%eax
400f88: eb 34 jmp 400fbe <phase_3+0x7b>
# 3
400f8a: b8 00 01 00 00 mov $0x100,%eax
400f8f: eb 2d jmp 400fbe <phase_3+0x7b>
# 4
400f91: b8 85 01 00 00 mov $0x185,%eax
400f96: eb 26 jmp 400fbe <phase_3+0x7b>
# 5
400f98: b8 ce 00 00 00 mov $0xce,%eax
400f9d: eb 1f jmp 400fbe <phase_3+0x7b>
# 6
400f9f: b8 aa 02 00 00 mov $0x2aa,%eax
400fa4: eb 18 jmp 400fbe <phase_3+0x7b>
# 7
400fa6: b8 47 01 00 00 mov $0x147,%eax
400fab: eb 11 jmp 400fbe <phase_3+0x7b>
# <phase_3+0x6a>
400fad: e8 88 04 00 00 call 40143a <explode_bomb>
# can't reach here???
400fb2: b8 00 00 00 00 mov $0x0,%eax
400fb7: eb 05 jmp 400fbe # ??hyw
# 1
400fb9: b8 37 01 00 00 mov $0x137,%eax
# <phase_3+0x7b>
400fbe: 3b 44 24 0c cmp 0xc(%rsp),%eax # n2 == (n1 jmp to)
400fc2: 74 05 je 400fc9 <phase_3+0x86>
400fc4: e8 71 04 00 00 call 40143a <explode_bomb>
400fc9: 48 83 c4 18 add $0x18,%rsp
400fcd: c3 ret
- 首先在
400f51处，可以看出这里的地址是sscanf的输入，读出来这里的字符串是"%d %d"，得知这题输入是两个 4 字节整形。 - 关键在于
400f75处的间接跳转，可以想到这里的地址里存的应该是一个pc位置的数组，根据后面的比较，我们知道答案应该是 第一个数对应偏移量，第二个数是对应跳转的分支上的值
(gdb) x/8gx 0x402470
0x402470: 0x0000000000400f7c 0x0000000000400fb9
0x402480: 0x0000000000400f83 0x0000000000400f8a
0x402490: 0x0000000000400f91 0x0000000000400f98
0x4024a0: 0x0000000000400f9f 0x0000000000400fa6所有可行的答案：0 207 1 311 2 707 3 256 4 392 5 206 6 682 7 327
Phase_4
这里phase_4函数很好分析，就是输入两个数，在 0 ~ 14 范围内，需要第一个数经过func4运算后为 0，第二个数恒为 0。
这里其实也可以不管
func4，就 15 个数，总有一个输出是 0，一个个试也能试出来。但这是，弱者的思维.jpg
0000000000400fce <func4>:
400fce: 48 83 ec 08 sub $0x8,%rsp
400fd2: 89 d0 mov %edx,%eax
400fd4: 29 f0 sub %esi,%eax # r - l
400fd6: 89 c1 mov %eax,%ecx # copy
400fd8: c1 e9 1f shr $0x1f,%ecx # get sign(0)
400fdb: 01 c8 add %ecx,%eax # r - l + sign
400fdd: d1 f8 sar $1,%eax # (r - l + sign) / 2
400fdf: 8d 0c 30 lea (%rax,%rsi,1),%ecx # (l + r) / 2
400fe2: 39 f9 cmp %edi,%ecx
400fe4: 7e 0c jle 400ff2 <func4+0x24>
# n1 < mid -> (l ~ mid - 1)
400fe6: 8d 51 ff lea -0x1(%rcx),%edx
400fe9: e8 e0 ff ff ff call 400fce <func4>
400fee: 01 c0 add %eax,%eax
400ff0: eb 15 jmp 401007 <func4+0x39>
# <func4+0x24> mid <= n1
400ff2: b8 00 00 00 00 mov $0x0,%eax
400ff7: 39 f9 cmp %edi,%ecx # mid == n1 -> ret 0
400ff9: 7d 0c jge 401007 <func4+0x39>
# (mid + 1 ~ r)
400ffb: 8d 71 01 lea 0x1(%rcx),%esi
400ffe: e8 cb ff ff ff call 400fce <func4>
401003: 8d 44 00 01 lea 0x1(%rax,%rax,1),%eax # ret * 2 + 1
# <func4+0x39>
401007: 48 83 c4 08 add $0x8,%rsp
40100b: c3 ret其实这就是一个类似二分查找的递归，下面是可能的源码，功能一致。
int func4(int x, int l, int r)
{
int mid = l + (r - l) / 2;
if (x < mid) return 2 * func4(x, l, mid - 1);
else if (x == mid) return 0;
else return 2 * func4(x, mid + 1, r) + 1;
}抽象为下面的树形结构
flowchart TD
A1["key: 7 ret: 0"]
B1["key: 3 ret: 0"]
B2["key: 11 ret: 1"]
C1["key: 1 ret: 0"]
C2["key: 5 ret: 2"]
C3["key: 9 ret: 1"]
C4["key: 13 ret: 3"]
D1["key: 0 ret: 0"]
D2["key: 2 ret: 4"]
D3["key: 4 ret: 2"]
D4["key: 6 ret: 6"]
D5["key: 8 ret: 1"]
D6["key: 10 ret: 5"]
D7["key: 12 ret: 3"]
D8["key: 14 ret: 7"]
A1 --> B1
A1 --> B2
B1 --> C1
B1 --> C2
B2 --> C3
B2 --> C4
C1 --> D1
C1 --> D2
C2 --> D3
C2 --> D4
C3 --> D5
C3 --> D6
C4 --> D7
C4 --> D8显然第一个数只能取 0 1 3 7。
Phase_5
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 call 40131b <string_length>
40107f: 83 f8 06 cmp $0x6,%eax
401082: 74 4e je 4010d2 <phase_5+0x70>
401084: e8 b1 03 00 00 call 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 # 取低4位（0~15）作为偏移量
401099: 0f b6 92 b0 24 40 00 movzbl 0x4024b0(%rdx),%edx
# (gdb) x/2s 0x4024b0
# 0x4024b0 <array.3449>: "maduiersnfotvbylSo you think you can stop the
# bomb with ctrl-c, do you?"
# 0x4024f8: "Curses, you've found the secret phase!"
# "m a d u i e r s n f o t v b y l"
# "0 1 2 3 4 5 6 7 8 9 a b c d e f"
# "flyers" <= "0x(6/4)" + "9 f e 5 6 7" <= "ionefg"
4010a0: 88 54 04 10 mov %dl,0x10(%rsp,%rax,1) # new string
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 # "flyers"
4010b8: 48 8d 7c 24 10 lea 0x10(%rsp),%rdi
4010bd: e8 76 02 00 00 call 401338 <strings_not_equal>
4010c2: 85 c0 test %eax,%eax
4010c4: 74 13 je 4010d9 <phase_5+0x77>
4010c6: e8 6f 03 00 00 call 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 # eax 循环计数器
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 call 400b30 <__stack_chk_fail@plt>
4010ee: 48 83 c4 20 add $0x20,%rsp
4010f2: 5b pop %rbx
4010f3: c3 ret看上面注释，这里就是对 "flyers" 的一个加密，"maduiersnfotvbyl"作为映射使用。
注意这里用掩码只截取了后四bit，也就是说这个答案是大小写不敏感的，非一一映射。
Phase_6
00000000004010f4 <phase_6>:
4010f4: 41 56 push %r14 # 0
4010f6: 41 55 push %r13 # 0
4010f8: 41 54 push %r12 # 2
4010fa: 55 push %rbp # 0x7fffffffd780
4010fb: 53 push %rbx # 0x7fffffffd808
# 被调用者保存寄存器
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 call 40145c <read_six_numbers>
40110b: 49 89 e6 mov %rsp,%r14
40110e: 41 bc 00 00 00 00 mov $0x0,%r12d
# Part1: O(n^2) 双层循环, 判断是否为1~6的排列
# <phase_6+0x20>
401114: 4c 89 ed mov %r13,%rbp
401117: 41 8b 45 00 mov 0x0(%r13),%eax # a0
40111b: 83 e8 01 sub $0x1,%eax # a0-1 # 细节 -1 排除 0
40111e: 83 f8 05 cmp $0x5,%eax
401121: 76 05 jbe 401128 <phase_6+0x34> #如果有0, -1后溢出,无符号比较下不成立
401123: e8 12 03 00 00 call 40143a <explode_bomb>
# a0 <= 5
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
# <phase_6+0x41>
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 call 40143a <explode_bomb>
# <phase_6+0x51>
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>
# Part2: 7 - a_n
# <phase_6+0x5f>
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
# <phase_6+0x6c>
401160: 89 ca mov %ecx,%edx
401162: 2b 10 sub (%rax),%edx
401164: 89 10 mov %edx,(%rax) # 7 - a_n
401166: 48 83 c0 04 add $0x4,%rax
40116a: 48 39 f0 cmp %rsi,%rax
40116d: 75 f1 jne 401160 <phase_6+0x6c>
# Part2: 用结构体(链表)按重构顺序排序
40116f: be 00 00 00 00 mov $0x0,%esi
401174: eb 21 jmp 401197 <phase_6+0xa3>
# <phase_6+0x82>
401176: 48 8b 52 08 mov 0x8(%rdx),%rdx # p = p->nxt
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>
# <phase_6+0x8f>
401183: ba d0 32 60 00 mov $0x6032d0,%edx
/*
struct node
{
int value;
int key;
node *nxt;
};
*/
0x6032d0 <node1>: 332 1 nxt
0x6032e0 <node2>: 168 2 nxt
0x6032f0 <node3>: 924 3 nxt
0x603300 <node4>: 691 4 nxt
0x603310 <node5>: 477 5 nxt
0x603320 <node6>: 443 6 nxt
3 4 5 6 1 2
4 3 2 1 6 5
*/
# <phase_6+0x94>
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>
# <phase_6+0xa3>
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>
# Part3: 重构链表
# <phase_6+0xb7>
4011ab: 48 8b 5c 24 20 mov 0x20(%rsp),%rbx # p->value
4011b0: 48 8d 44 24 28 lea 0x28(%rsp),%rax # q = p->next
4011b5: 48 8d 74 24 50 lea 0x50(%rsp),%rsi # end
4011ba: 48 89 d9 mov %rbx,%rcx
# <phase_6+0xc9>
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>
# Part4: 降序检查
# <phase_6+0xde>
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 call 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 ret
上面的注释已经做了大致的功能分析。
这题的重点在于识别链表形式的结构体，以及链表的一些基础写法。
struct node
{
int value;
int key;
node *nxt;
};secret_phase
搜索一下文件中的 <secret_phase>，发现调用在<phase_defused>中，以及主角<fun7>。
0000000004015c4 <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
# 需先完成6个phase
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 # "%d %d %s"
4015f5: bf 70 38 60 00 mov $0x603870,%edi # "input of phase 4"
4015fa: e8 f1 f5 ff ff call 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 # "DrEvil"
401609: 48 8d 7c 24 10 lea 0x10(%rsp),%rdi
40160e: e8 25 fd ff ff call 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 # "Curses, you've found the secret phase!"
40161c: e8 ef f4 ff ff call 400b10 <puts@plt>
401621: bf 20 25 40 00 mov $0x402520,%edi # "But finding it and solving it are quite different..."
401626: e8 e5 f4 ff ff call 400b10 <puts@plt>
40162b: b8 00 00 00 00 mov $0x0,%eax
401630: e8 0d fc ff ff call 401242 <secret_phase>
401635: bf 58 25 40 00 mov $0x402558,%edi
40163a: e8 d1 f4 ff ff call 400b10 <puts@plt>
# <phase_defused+0x7b>
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 call 400b30 <__stack_chk_fail@plt>
401654: 48 83 c4 78 add $0x78,%rsp
401658: c3 ret
401659: 90 nop
40165a: 90 nop
40165b: 90 nop
40165c: 90 nop
40165d: 90 nop
40165e: 90 nop
40165f: 90 nop- 注意到这里在 6 个phase解完后的
<phase_defuse>会解析第四次输入是否为"%d %d %s"，也就是隐藏进入的方法为在第四次输入后面添加一个字符串。读取内存可以知道这个串是"DrEvil"邪恶Deltarune <secret_phase>读取一个数字字符串，并转化为 8 字节整型，之后经过<fun7>运算结果为 2。以上为该题要求。
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>
# x < 36
401213: 48 8b 7f 08 mov 0x8(%rdi),%rdi
401217: e8 e8 ff ff ff call 401204 <fun7>
40121c: 01 c0 add %eax,%eax
40121e: eb 1d jmp 40123d <fun7+0x39>
# <fun7+0x1c> x >= 36
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 call 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
# <fun7+0x39>
40123d: 48 83 c4 08 add $0x8,%rsp
401241: c3 ret<fun7>其实和<func4>很像。这里可以看出是一个类似二叉搜索树的树形结构，对应位置相同的节点返回值相同，但 key 值不同。与<func4>不同的地方在于，<func4>的树形结构、key、val都是由二分与递归天然形成的，而<fun7>的树形结构是由参数传入的，如下。
40126e: bf f0 30 60 00 mov $0x6030f0,%edi
401273: e8 8c ff ff ff call 401204 <fun7>- 接下来我们对这个地址储存的内容解析，这是一个很典型的二叉树。而且我们发现 key 大小的规律正好符合一颗搜索二叉树
(gdb) x/64dg 0x6030f0
0x6030f0 <n1>: 36 6304016
0x603100 <n1+16>: 6304048 0
0x603110 <n21>: 8 6304144
0x603120 <n21+16>: 6304080 0
0x603130 <n22>: 50 6304112
0x603140 <n22+16>: 6304176 0
0x603150 <n32>: 22 6304368
0x603160 <n32+16>: 6304304 0
0x603170 <n33>: 45 6304208
0x603180 <n33+16>: 6304400 0
0x603190 <n31>: 6 6304240
0x6031a0 <n31+16>: 6304336 0
0x6031b0 <n34>: 107 6304272
0x6031c0 <n34+16>: 6304432 0
0x6031d0 <n45>: 40 0
0x6031e0 <n45+16>: 0 0
0x6031f0 <n41>: 1 0
0x603200 <n41+16>: 0 0
0x603210 <n47>: 99 0
0x603220 <n47+16>: 0 0
0x603230 <n44>: 35 0
0x603240 <n44+16>: 0 0
0x603250 <n42>: 7 0
0x603260 <n42+16>: 0 0
0x603270 <n43>: 20 0
0x603280 <n43+16>: 0 0
0x603290 <n46>: 47 0
0x6032a0 <n46+16>: 0 0
0x6032b0 <n48>: 1001 0
0x6032c0 <n48+16>: 0 0结构体大概这样
struct TreeNode
{
long key;
TreeNode *left_child;
TreeNode *right_child;
};对应的结构为
flowchart TD
A1["key: 36 ret: 0"]
B1["key: 8 ret: 0"]
B2["key: 50 ret: 1"]
C1["key: 6 ret: 0"]
C2["key: 22 ret: 2"]
C3["key: 45 ret: 1"]
C4["key: 107 ret: 3"]
D1["key: 1 ret: 0"]
D2["key: 7 ret: 4"]
D3["key: 20 ret: 2"]
D4["key: 35 ret: 6"]
D5["key: 40 ret: 1"]
D6["key: 47 ret: 5"]
D7["key: 99 ret: 3"]
D8["key: 1001 ret: 7"]
A1 --> B1
A1 --> B2
B1 --> C1
B1 --> C2
B2 --> C3
B2 --> C4
C1 --> D1
C1 --> D2
C2 --> D3
C2 --> D4
C3 --> D5
C3 --> D6
C4 --> D7
C4 --> D8这次需要返回值为 2，对应的key值只有 20 22。
异常控制流彩蛋
按^C有惊喜，至于怎么捕获的等我学成归来在填充这部分内容。