计算机系统基础：bomb炸弹实验

实验因为根据学号生成，和同学的都不一样，只能独立完成，不过很多大佬的博客提供的思路还是非常有参考价值的。

phase1

08048b69 <phase1>:8048b69: 55                      push   %ebp8048b6a: 89 e5                   mov    %esp,%ebp8048b6c:    83 ec 10                sub    $0x10,%esp8048b6f:   68 e4 9f 04 08          push   $0x8049fe48048b74:   ff 75 08                pushl  0x8(%ebp)8048b77:    e8 c5 04 00 00          call   8049041 <stringsnotequal>8048b7c:  83 c4 10                add    $0x10,%esp8048b7f:   85 c0                   test   %eax,%eax8048b81:    75 02                   jne    8048b85 <phase1+0x1c>8048b83: c9                      leave  8048b84: c3                      ret    8048b85: e8 b8 05 00 00          call   8049142 <explodebomb>8048b8a:  eb f7                   jmp    8048b83 <phase1+0x1a>

根据查看反汇编代码可得%eax中存储的是输入的字符串，
可得和用户输入字符串比较的字符串的存储地址0x8049fe4，可用gdb查看这个地址存储的数据内容(gdb) x/s 0x8049fe4
可得结果：Crikey! I have lost my mojo!

phase2：

08048b8c <phase2>:8048b8c: 55                      push   %ebp8048b8d: 89 e5                   mov    %esp,%ebp8048b8f:    56                      push   %esi8048b90: 53                      push   %ebx8048b91: 83 ec 28                sub    $0x28,%esp8048b94:   8d 45 e0                lea    -0x20(%ebp),%eax8048b97: 50                      push   %eax8048b98: ff 75 08                pushl  0x8(%ebp)8048b9b:    e8 ca 05 00 00          call   804916a <readsixnumbers>8048ba0:   83 c4 10                add    $0x10,%esp8048ba3:   83 7d e0 00             cmpl   $0x0,-0x20(%ebp)8048ba7: 75 06                   jne    8048baf <phase2+0x23>8048ba9: 83 7d e4 01             cmpl   $0x1,-0x1c(%ebp)8048bad: 74 05                   je     8048bb4 <phase2+0x28>8048baf: e8 8e 05 00 00          call   8049142 <explodebomb>8048bb4:  8d 5d e0                lea    -0x20(%ebp),%ebx8048bb7: 8d 75 f0                lea    -0x10(%ebp),%esi8048bba: eb 07                   jmp    8048bc3 <phase2+0x37>8048bbc: 83 c3 04                add    $0x4,%ebx8048bbf:    39 f3                   cmp    %esi,%ebx8048bc1:    74 11                   je     8048bd4 <phase2+0x48>8048bc3: 8b 43 04                mov    0x4(%ebx),%eax8048bc6:   03 03                   add    (%ebx),%eax8048bc8:  39 43 08                cmp    %eax,0x8(%ebx)8048bcb:   74 ef                   je     8048bbc <phase2+0x30>8048bcd: e8 70 05 00 00          call   8049142 <explodebomb>8048bd2:  eb e8                   jmp    8048bbc <phase2+0x30>8048bd4: 8d 65 f8                lea    -0x8(%ebp),%esp8048bd7:  5b                      pop    %ebx8048bd8: 5e                      pop    %esi8048bd9: 5d                      pop    %ebp8048bda: c3                      ret

根据调用函数readsixnumbers,查看该函数汇编代码，可得结果为6个数，根据cmpl $0x0,-0x20(%ebp)可知第一个数为0，同理根据cmpl $0x1,-0x1c(%ebp)可得第二个数为1，后面的汇编代码可以看出是一个循环，后一个数为前两个数之和，即斐波那契数列
可得结果：0 1 1 2 3 5

phase3

08048bdb <phase3>:8048bdb: 55                      push   %ebp8048bdc: 89 e5                   mov    %esp,%ebp8048bde:    83 ec 24                sub    $0x24,%esp8048be1:   8d 45 f0                lea    -0x10(%ebp),%eax8048be4: 50                      push   %eax8048be5: 8d 45 ef                lea    -0x11(%ebp),%eax8048be8: 50                      push   %eax8048be9: 8d 45 f4                lea    -0xc(%ebp),%eax8048bec:  50                      push   %eax8048bed: 68 01 a0 04 08          push   $0x804a0018048bf2:   ff 75 08                pushl  0x8(%ebp)8048bf5:    e8 66 fc ff ff          call   8048860 <_isoc99sscanf@plt>8048bfa:   83 c4 20                add    $0x20,%esp8048bfd:   83 f8 02                cmp    $0x2,%eax8048c00:    7e 14                   jle    8048c16 <phase3+0x3b>8048c02: 83 7d f4 07             cmpl   $0x7,-0xc(%ebp)8048c06:  0f 87 f0 00 00 00       ja     8048cfc <phase3+0x121>8048c0c:    8b 45 f4                mov    -0xc(%ebp),%eax8048c0f:  ff 24 85 20 a0 04 08    jmp    *0x804a020(,%eax,4)8048c16:  e8 27 05 00 00          call   8049142 <explodebomb>8048c1b:  eb e5                   jmp    8048c02 <phase3+0x27>8048c1d: b8 74 00 00 00          mov    $0x74,%eax8048c22:   81 7d f0 fb 01 00 00    cmpl   $0x1fb,-0x10(%ebp)8048c29:   0f 84 d7 00 00 00       je     8048d06 <phase3+0x12b>8048c2f:    e8 0e 05 00 00          call   8049142 <explodebomb>8048c34:  b8 74 00 00 00          mov    $0x74,%eax8048c39:   e9 c8 00 00 00          jmp    8048d06 <phase3+0x12b>8048c3e:    b8 6d 00 00 00          mov    $0x6d,%eax8048c43:   81 7d f0 76 02 00 00    cmpl   $0x276,-0x10(%ebp)8048c4a:   0f 84 b6 00 00 00       je     8048d06 <phase3+0x12b>8048c50:    e8 ed 04 00 00          call   8049142 <explodebomb>8048c55:  b8 6d 00 00 00          mov    $0x6d,%eax8048c5a:   e9 a7 00 00 00          jmp    8048d06 <phase3+0x12b>8048c5f:    b8 6e 00 00 00          mov    $0x6e,%eax8048c64:   81 7d f0 48 02 00 00    cmpl   $0x248,-0x10(%ebp)8048c6b:   0f 84 95 00 00 00       je     8048d06 <phase3+0x12b>8048c71:    e8 cc 04 00 00          call   8049142 <explodebomb>8048c76:  b8 6e 00 00 00          mov    $0x6e,%eax8048c7b:   e9 86 00 00 00          jmp    8048d06 <phase3+0x12b>8048c80:    b8 6e 00 00 00          mov    $0x6e,%eax8048c85:   81 7d f0 44 03 00 00    cmpl   $0x344,-0x10(%ebp)8048c8c:   74 78                   je     8048d06 <phase3+0x12b>8048c8e:    e8 af 04 00 00          call   8049142 <explodebomb>8048c93:  b8 6e 00 00 00          mov    $0x6e,%eax8048c98:   eb 6c                   jmp    8048d06 <phase3+0x12b>8048c9a:    b8 6e 00 00 00          mov    $0x6e,%eax8048c9f:   83 7d f0 58             cmpl   $0x58,-0x10(%ebp)8048ca3:    74 61                   je     8048d06 <phase3+0x12b>8048ca5:    e8 98 04 00 00          call   8049142 <explodebomb>8048caa:  b8 6e 00 00 00          mov    $0x6e,%eax8048caf:   eb 55                   jmp    8048d06 <phase3+0x12b>8048cb1:    b8 69 00 00 00          mov    $0x69,%eax8048cb6:   81 7d f0 53 02 00 00    cmpl   $0x253,-0x10(%ebp)8048cbd:   74 47                   je     8048d06 <phase3+0x12b>8048cbf:    e8 7e 04 00 00          call   8049142 <explodebomb>8048cc4:  b8 69 00 00 00          mov    $0x69,%eax8048cc9:   eb 3b                   jmp    8048d06 <phase3+0x12b>8048ccb:    b8 6d 00 00 00          mov    $0x6d,%eax8048cd0:   83 7d f0 52             cmpl   $0x52,-0x10(%ebp)8048cd4:    74 30                   je     8048d06 <phase3+0x12b>8048cd6:    e8 67 04 00 00          call   8049142 <explodebomb>8048cdb:  b8 6d 00 00 00          mov    $0x6d,%eax8048ce0:   eb 24                   jmp    8048d06 <phase3+0x12b>8048ce2:    b8 75 00 00 00          mov    $0x75,%eax8048ce7:   81 7d f0 21 01 00 00    cmpl   $0x121,-0x10(%ebp)8048cee:   74 16                   je     8048d06 <phase3+0x12b>8048cf0:    e8 4d 04 00 00          call   8049142 <explodebomb>8048cf5:  b8 75 00 00 00          mov    $0x75,%eax8048cfa:   eb 0a                   jmp    8048d06 <phase3+0x12b>8048cfc:    e8 41 04 00 00          call   8049142 <explodebomb>8048d01:  b8 69 00 00 00          mov    $0x69,%eax8048d06:   38 45 ef                cmp    %al,-0x11(%ebp)8048d09:  74 05                   je     8048d10 <phase3+0x135>8048d0b:    e8 32 04 00 00          call   8049142 <explodebomb>8048d10:  c9                      leave  8048d11: c3                      ret

根据push $0x804a001用此行代码查看输入内容：(gdb) x/s 0x804a001,可得0x804a001: “%d %c %d”,所以输入的依次是数字，字符，数字。根据cmpl $0x7,-0xc(%ebp)，可得输入的第一个数小于7.假设第一个数为0，输入代码：(gdb) p/x *0x804a001,可得下一条指令地址为8048c1d,对应汇编指令为mov $0x74,所以第二个ascll码为0x74,即字符’t’,下一条汇编语句为%eax cmpl $0x1fb,-0x10(%ebp)，所以第三个数为0x1fb,即507
可得结果：0 t 507

phase4：

08048d5d <phase4>:8048d5d: 55                      push   %ebp8048d5e: 89 e5                   mov    %esp,%ebp8048d60:    83 ec 18                sub    $0x18,%esp8048d63:   8d 45 f4                lea    -0xc(%ebp),%eax8048d66:  50                      push   %eax8048d67: 8d 45 f0                lea    -0x10(%ebp),%eax8048d6a: 50                      push   %eax8048d6b: 68 b7 a1 04 08          push   $0x804a1b78048d70:   ff 75 08                pushl  0x8(%ebp)8048d73:    e8 e8 fa ff ff          call   8048860 <_isoc99sscanf@plt>8048d78:   83 c4 10                add    $0x10,%esp8048d7b:   83 f8 02                cmp    $0x2,%eax8048d7e:    75 0b                   jne    8048d8b <phase4+0x2e>8048d80: 8b 45 f4                mov    -0xc(%ebp),%eax8048d83:  83 e8 02                sub    $0x2,%eax8048d86:    83 f8 02                cmp    $0x2,%eax8048d89:    76 05                   jbe    8048d90 <phase4+0x33>8048d8b: e8 b2 03 00 00          call   8049142 <explodebomb>8048d90:  83 ec 08                sub    $0x8,%esp8048d93:    ff 75 f4                pushl  -0xc(%ebp)8048d96:   6a 08                   push   $0x88048d98: e8 75 ff ff ff          call   8048d12 <func4>8048d9d:    83 c4 10                add    $0x10,%esp8048da0:   39 45 f0                cmp    %eax,-0x10(%ebp)8048da3: 74 05                   je     8048daa <phase4+0x4d>8048da5: e8 98 03 00 00          call   8049142 <explodebomb>8048daa:  c9                      leave  8048dab: c3                      ret

根据push $0x804a1b7用此行代码：(gdb) x/2s 0x804a1b7,可得0x804a1b7: “%d %d”,再根据
8048d83: 83 e8 02 sub $0x2,%eax
8048d86: 83 f8 02 cmp $0x2,%eax
8048d89: 76 05 jbe 8048d90 <phase4+0x33>
8048d8b: e8 b2 03 00 00 call 8049142
可得第二个数应该为2,3,4中的一个，对于cmp $0x2,%eax用ni和info reg指令进行查看，得到eax为3，所以第二个数为3，同理对cmp %eax,-0x10(%ebp)的操作利用ni和info reg指令查看，得到eax为162，所以第一个数为162
可得结果：3 162

phase5：

08048dac <phase5>:8048dac: 55                      push   %ebp8048dad: 89 e5                   mov    %esp,%ebp8048daf:    83 ec 18                sub    $0x18,%esp8048db2:   8d 45 f0                lea    -0x10(%ebp),%eax8048db5: 50                      push   %eax8048db6: 8d 45 f4                lea    -0xc(%ebp),%eax8048db9:  50                      push   %eax8048dba: 68 b7 a1 04 08          push   $0x804a1b78048dbf:   ff 75 08                pushl  0x8(%ebp)8048dc2:    e8 99 fa ff ff          call   8048860 <_isoc99sscanf@plt>8048dc7:   83 c4 10                add    $0x10,%esp8048dca:   83 f8 01                cmp    $0x1,%eax8048dcd:    7e 41                   jle    8048e10 <phase5+0x64>8048dcf: 8b 45 f4                mov    -0xc(%ebp),%eax8048dd2:  83 e0 0f                and    $0xf,%eax8048dd5:    89 45 f4                mov    %eax,-0xc(%ebp)8048dd8:  83 f8 0f                cmp    $0xf,%eax8048ddb:    74 2c                   je     8048e09 <phase5+0x5d>8048ddd: b9 00 00 00 00          mov    $0x0,%ecx8048de2:    ba 00 00 00 00          mov    $0x0,%edx8048de7:    83 c2 01                add    $0x1,%edx8048dea:    8b 04 85 40 a0 04 08    mov    0x804a040(,%eax,4),%eax8048df1:  01 c1                   add    %eax,%ecx8048df3:    83 f8 0f                cmp    $0xf,%eax8048df6:    75 ef                   jne    8048de7 <phase5+0x3b>8048df8: c7 45 f4 0f 00 00 00    movl   $0xf,-0xc(%ebp)8048dff:  83 fa 0f                cmp    $0xf,%edx8048e02:    75 05                   jne    8048e09 <phase5+0x5d>8048e04: 39 4d f0                cmp    %ecx,-0x10(%ebp)8048e07: 74 05                   je     8048e0e <phase5+0x62>8048e09: e8 34 03 00 00          call   8049142 <explodebomb>8048e0e:  c9                      leave  8048e0f: c3                      ret    8048e10: e8 2d 03 00 00          call   8049142 <explodebomb>8048e15:  eb b8                   jmp    8048dcf <phase5+0x23>

和上一个实验同理可得输入两个数，根据cmp $0xf,%eax，可得有一个数与0xf进行相与的操作，再根据后面的判断可得输入的数据低四位的值不能是0xf，查看汇编语句可知进行了循环，循环结束后对%edx进行了一次判断，若不等于0xf就爆炸且输入的第二个数与寄存器%ecx的值若不相等也会爆炸，所以第二个数的值等于当前ecx中的值，对循环进行具体分析可得循环进行了15次，循环退出的条件为eax=0xf，eax的初值为15，即输入的第一个数的最低四位的值为5，ecx的值为eax后面14次变化过程中所有数值的和，即115.
可得结果： 21 115

phase6：

08048e17 <phase6>:8048e17: 55                      push   %ebp8048e18: 89 e5                   mov    %esp,%ebp8048e1a:    56                      push   %esi8048e1b: 53                      push   %ebx8048e1c: 83 ec 38                sub    $0x38,%esp8048e1f:   8d 45 e0                lea    -0x20(%ebp),%eax8048e22: 50                      push   %eax8048e23: ff 75 08                pushl  0x8(%ebp)8048e26:    e8 3f 03 00 00          call   804916a <readsixnumbers>8048e2b:   83 c4 10                add    $0x10,%esp8048e2e:   be 00 00 00 00          mov    $0x0,%esi8048e33:    8b 44 b5 e0             mov    -0x20(%ebp,%esi,4),%eax8048e37:  83 e8 01                sub    $0x1,%eax8048e3a:    83 f8 05                cmp    $0x5,%eax8048e3d:    77 0c                   ja     8048e4b <phase6+0x34>8048e3f: 83 c6 01                add    $0x1,%esi8048e42:    83 fe 06                cmp    $0x6,%esi8048e45:    74 51                   je     8048e98 <phase6+0x81>8048e47: 89 f3                   mov    %esi,%ebx8048e49:    eb 0f                   jmp    8048e5a <phase6+0x43>8048e4b: e8 f2 02 00 00          call   8049142 <explodebomb>8048e50:  eb ed                   jmp    8048e3f <phase6+0x28>8048e52: 83 c3 01                add    $0x1,%ebx8048e55:    83 fb 05                cmp    $0x5,%ebx8048e58:    7f d9                   jg     8048e33 <phase6+0x1c>8048e5a: 8b 44 9d e0             mov    -0x20(%ebp,%ebx,4),%eax8048e5e:  39 44 b5 dc             cmp    %eax,-0x24(%ebp,%esi,4)8048e62:  75 ee                   jne    8048e52 <phase6+0x3b>8048e64: e8 d9 02 00 00          call   8049142 <explodebomb>8048e69:  eb e7                   jmp    8048e52 <phase6+0x3b>8048e6b: 8b 52 08                mov    0x8(%edx),%edx8048e6e:   83 c0 01                add    $0x1,%eax8048e71:    39 c8                   cmp    %ecx,%eax8048e73:    75 f6                   jne    8048e6b <phase6+0x54>8048e75: 89 54 b5 c8             mov    %edx,-0x38(%ebp,%esi,4)8048e79:  83 c3 01                add    $0x1,%ebx8048e7c:    83 fb 06                cmp    $0x6,%ebx8048e7f:    74 1e                   je     8048e9f <phase6+0x88>8048e81: 89 de                   mov    %ebx,%esi8048e83:    8b 4c 9d e0             mov    -0x20(%ebp,%ebx,4),%ecx8048e87:  b8 01 00 00 00          mov    $0x1,%eax8048e8c:    ba 3c c1 04 08          mov    $0x804c13c,%edx8048e91:  83 f9 01                cmp    $0x1,%ecx8048e94:    7f d5                   jg     8048e6b <phase6+0x54>8048e96: eb dd                   jmp    8048e75 <phase6+0x5e>8048e98: bb 00 00 00 00          mov    $0x0,%ebx8048e9d:    eb e2                   jmp    8048e81 <phase6+0x6a>8048e9f: 8b 5d c8                mov    -0x38(%ebp),%ebx8048ea2: 8b 45 cc                mov    -0x34(%ebp),%eax8048ea5: 89 43 08                mov    %eax,0x8(%ebx)8048ea8:   8b 55 d0                mov    -0x30(%ebp),%edx8048eab: 89 50 08                mov    %edx,0x8(%eax)8048eae:   8b 45 d4                mov    -0x2c(%ebp),%eax8048eb1: 89 42 08                mov    %eax,0x8(%edx)8048eb4:   8b 55 d8                mov    -0x28(%ebp),%edx8048eb7: 89 50 08                mov    %edx,0x8(%eax)8048eba:   8b 45 dc                mov    -0x24(%ebp),%eax8048ebd: 89 42 08                mov    %eax,0x8(%edx)8048ec0:   c7 40 08 00 00 00 00    movl   $0x0,0x8(%eax)8048ec7:   be 05 00 00 00          mov    $0x5,%esi8048ecc:    eb 08                   jmp    8048ed6 <phase6+0xbf>8048ece: 8b 5b 08                mov    0x8(%ebx),%ebx8048ed1:   83 ee 01                sub    $0x1,%esi8048ed4:    74 10                   je     8048ee6 <phase6+0xcf>8048ed6: 8b 43 08                mov    0x8(%ebx),%eax8048ed9:   8b 00                   mov    (%eax),%eax8048edb:  39 03                   cmp    %eax,(%ebx)8048edd:  7e ef                   jle    8048ece <phase6+0xb7>8048edf: e8 5e 02 00 00          call   8049142 <explodebomb>8048ee4:  eb e8                   jmp    8048ece <phase6+0xb7>8048ee6: 8d 65 f8                lea    -0x8(%ebp),%esp8048ee9:  5b                      pop    %ebx8048eea: 5e                      pop    %esi8048eeb: 5d                      pop    %ebp8048eec: c3                      ret

根据read six numbers可得答案为6个数字，由sub $0x1,%eax和cmp $0x5,%eax可得输入的六个数在1-6之间。
根据mov $0x804c13c,%edx，调试代码为:
(gdb) x/3x 0x804c13c
(gdb) x/3x 0x804c148
(gdb) x/3x 0x804c154
(gdb) x/3x 0x804c160
(gdb) x/3x 0x804c16c
(gdb) x/3x 0x804c178
可得结果：3 6 4 5 1 2

隐藏实验：
在第四关中phase defused函数中寻找到特殊地址查看可得需在输入第四关答案后再输入一个字符串，而这个字符串就是DrEvil,输入两个数后再输入DrEvil即可触发隐藏关

08048f41 <secretphase>:8048f41:    55                      push   %ebp8048f42: 89 e5                   mov    %esp,%ebp8048f44:    53                      push   %ebx8048f45: 83 ec 04                sub    $0x4,%esp8048f48:    e8 57 02 00 00          call   80491a4 <readline>8048f4d: 83 ec 04                sub    $0x4,%esp8048f50:    6a 0a                   push   $0xa8048f52: 6a 00                   push   $0x08048f54: 50                      push   %eax8048f55: e8 66 f9 ff ff          call   80488c0 <strtol@plt>8048f5a:  89 c3                   mov    %eax,%ebx8048f5c:    8d 40 ff                lea    -0x1(%eax),%eax8048f5f:  83 c4 10                add    $0x10,%esp8048f62:   3d e8 03 00 00          cmp    $0x3e8,%eax8048f67:  77 2f                   ja     8048f98 <secretphase+0x57>8048f69:    83 ec 08                sub    $0x8,%esp8048f6c:    53                      push   %ebx8048f6d: 68 88 c0 04 08          push   $0x804c0888048f72:   e8 76 ff ff ff          call   8048eed <fun7>8048f77: 83 c4 10                add    $0x10,%esp8048f7a:   85 c0                   test   %eax,%eax8048f7c:    75 21                   jne    8048f9f <secretphase+0x5e>8048f7e:    83 ec 0c                sub    $0xc,%esp8048f81:    68 80 a0 04 08          push   $0x804a0808048f86:   e8 85 f8 ff ff          call   8048810 <puts@plt>8048f8b:    e8 25 03 00 00          call   80492b5 <phasedefused>8048f90: 83 c4 10                add    $0x10,%esp8048f93:   8b 5d fc                mov    -0x4(%ebp),%ebx8048f96:  c9                      leave  8048f97: c3                      ret    8048f98: e8 a5 01 00 00          call   8049142 <explodebomb>8048f9d:  eb ca                   jmp    8048f69 <secretphase+0x28>8048f9f:    e8 9e 01 00 00          call   8049142 <explodebomb>8048fa4:  eb d8                   jmp    8048f7e <secretphase+0x3d>

可以看出调用了fun7函数，查看fun7函数可得这是一个递归函数，该函数会用来对一颗二叉树进行查询，我们需要输入的值就是最后一次查询到的结点储存值，
根据push $0x804c088，调试代码：
(gdb) p/x *0x804c088@3
(gdb) p/x *0x804c094@3
(gdb) p/x *0x804c0c4@3
(gdb) p/x *0x804c0e8@3
可得结果：36

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