Description
csapp bomb lab. We are given an object file named bomb. We have six bombs to defuse.
command
- objdump
objdump -d bomb > bomb.dump
readelf -x .rodata bomb 可以查看object file中的data segment，包含所有的static variables及其地址
- gdb
info registers 查看所有registers的值
p /x $rax 查看%rax寄存器的值
gdb -tui bomb
(gdb) layout split 可以同时查看源码和assembly code
(gdb) b 13 在第13行设置断点
(gdb) b *0x400f51 在地址为0x400f51设置断点
(gdb) run 运行程序到下一个断点
(gdb) n 运行到下一个断点
(gdb) s 单步运行 / si 前跳 i steps
(gdb) l bomb.c:50 查看第50行开始的源码
(gdb) x/u 取地址u中的值
- vim
/text 查找text | n跳到下一个结果
h l j k 左右上下，配合数字移动对应行数 20j 上移20行
Phase_1
- 目的基本就是熟悉objdump以及gdb的各项指令
- objdump -d bomb > bomb.dump 得到object file的汇编码
- gdb 模式下layout split 对照bomb.c源码找到phase_1函数的地址为0x400ee0; 并且read_line函数从控制台读取到user输入的字符串存储到了%rax中，调用phase_1(input)之前，mov %rax %rdi, 将输入字符串作为phase_1的参数
- bomb.dump 中找到phase_1函数对应的assembly code, mov $0x402400 %esi, 将起始地址为0x402400的char*[]起始地址作为第二个参数，然后callq <strings_not_equal>,比较第一个参数(用户输入值存储在%rdi中)和第二个参数(程序中static data 0x402400)是否相等]
- readelf -x .rodata bomb找到0x402400字符串, 00000000 代表字符串结束符’\0’,查找ascii表2e对应’.';所以phase_1答案为:Border relations with Canada have never been better.
Phase_2
- phase_2调用read_six_number -> 调用scanf在phase_2开辟的栈空间地址上写入从stdin读取来的值
- 400f0a处可得(%rsp) 必须为0x1
- 400f30 ~ 400f35处限定了地址空间的范围，当%rbx 地址值 != %rbp, 每次作为计算 %rbx+=0x4
- 400f17 ~ 400f1c 将计算方法确定，(%rbx)值为 2 * (%rbx)
- mov -0x4(%rbx), %eax 地址计算完再dereferecne M[%rbx - 0x4] -> %eax 而不是 M[%rbx] - 0x4 -> %eax
- scanf返回值为成功读取值的个数
0000000000400efc <phase_2>:
400efc: 55 push %rbp #保存main函数的stack frame
400efd: 53 push %rbx
400efe: 48 83 ec 28 sub $0x28,%rsp # 开辟40bytes的栈空间
400f02: 48 89 e6 mov %rsp,%rsi # %rsp地址作为参数 传递给read_six_numbers
400f05: e8 52 05 00 00 callq 40145c <read_six_numbers> #read_six中调用scanf将数值写入栈空间地址%rsp ~ %rsp + 0x18
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp) #(%rsp) 值为0x1
400f0e: 74 20 je 400f30 <phase_2+0x34> #跳转到 400f30
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 # M[%rbx - 0x4] 中的值赋给%eax
400f1a: 01 c0 add %eax,%eax # %eax *= 2
400f1c: 39 03 cmp %eax,(%rbx) # %eax 值与 (%rbx)值做比较
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 # %rbx中地址值+4
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 # %rbx 取得栈顶+0x4的地址
400f35: 48 8d 6c 24 18 lea 0x18(%rsp),%rbp # %rbp 取得栈顶+0x18的地址 也就是对应6位数字的最后一位
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
#以phase_2传递过来的%rsp为基准，使用%rcx, %rax, %r8, %r8, read_six_number栈空间+0x8, +0x00的位置存储scanf读取到的数值，
#寄存器及栈空间存储的都是phase_2栈空间上的地址，scanf直接将keyboard输入的值写入
read_six_numbers:
40145c: 48 83 ec 18 subq $24, %rsp
401460: 48 89 f2 movq %rsi, %rdx
401463: 48 8d 4e 04 leaq 4(%rsi), %rcx
401467: 48 8d 46 14 leaq 20(%rsi), %rax
40146b: 48 89 44 24 08 movq %rax, 8(%rsp)
401470: 48 8d 46 10 leaq 16(%rsi), %rax
401474: 48 89 04 24 movq %rax, (%rsp)
401478: 4c 8d 4e 0c leaq 12(%rsi), %r9
40147c: 4c 8d 46 08 leaq 8(%rsi), %r8
401480: be c3 25 40 00 movl $4203971, %esi
401485: b8 00 00 00 00 movl $0, %eax
40148a: e8 61 f7 ff ff callq -2207 <__isoc99_sscanf@plt>
40148f: 83 f8 05 cmpl $5, %eax
401492: 7f 05 jg 5 <read_six_numbers+0x3d>
401494: e8 a1 ff ff ff callq -95 <explode_bomb>
401499: 48 83 c4 18 addq $24, %rsp
40149d: c3 retq
Phase_3
- 考察对于switch语句汇编形式的理解
- 400f51: mov $0x4025cf,%esi 使用x/s 0x4025cf 查看得到“%d %d”, 证明phase_3需要输入两个数字 ｜ 400f60: cmp $0x1,%eax 验证了这一想法，scanf返回值要大于1
- %rdx %rcx 被赋予
- 400f6a: cmpl $0x7,0x8(%rsp) 可知switch范围为 0 <= x <= 7 也是我们输入的“%d %d”的第一个数
- 400f75 jmpq *0x402470(,%rax,8) 前一步将输入的第一个数字存储到了rax中， x/8a 0x402470可以看到自0x402470起后每+0x8位置的地址，也就是jump table中的地址空间: switch(0): 0x400f7c swithc(1)0x400fb9 …
- 在查看switch语句跳转部分会将我们输入的第二值与switch分支中的值进行比较，所以phase_3 “%d %d” – switch条件(0~7), switch 语句对应的值
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 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) #switch(x) 0 <= x <= 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 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
Phase_4
- 同样先检查40101a: movl $4203983, %esi, 再次得到“%d %d”, 证明phase_4同样是输入两个数字, 401029调用scanf后的%eax值必须为2同样验证了需要两个数字
- 40102e: cmpl $0xe,0x8(%rsp), 输入的第一个数需要 <= 0xe
- 40103a ~ 401044 将0xe传入%edx(第三个参数寄存器), 0x0传入%esi(第二个参数寄存器), user输入的第一个数字传入%edi(第一个参数寄存器)，之后调用func4
- func4的小trick是400fe2: cmp %edi,%ecx，其中edi为我们输入的第一个数字，只有edi <= ecx才会跳转400ff2, 而400ff2 将return value %eax置0，之后 400ff7: cmp %edi,%ecx 400ff9: jge 401007 再次比较edi和ecx的值 并且需要 %edi >= ecx才return 0，而return 0在phase_4中40104d中满足不引爆炸弹的必要条件，所以我们输入的数值只要与ecx通过 400fce~400fdf得到的结果相同就可以，而这个值为0xe » 1 = 0x7
- 第二个值很简单401051行与0做比较，所以输入的两个值为 0x7 和 0x0
phase_4:000000000040100c <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 #返回值为2，需要两个digits
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
Phase_5
- 大概扫一遍assembly code,比较重要的几个关键点: 40107a 调用string_length 返回值需要是6，401099 中0x4024b0 string字符串的内容，4010b3 调用strings_not_equal之前传入%esi的结果
- 有了前4个phase的练习，很容易看出，phase_5需要我们输入6位字符串，然后401096 and $0xf, %edx，就是将每一个字符的bits 与 00001111做and，得到的偏移量，到0x4024b0中取出对应的字符,使得得到得结果为"flyers”.
- 此题答案也不唯一，我的解是“ionefg”
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 #正确的字符串 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
Phase_6
- 第6题感觉难度提升了很多，主要是循环的嵌套导致assembly code比较复杂
- 另外找到$0x6032d0中这个Linked-Node数据结构非常关键。
- phase_6主要拆解为 1）输入六个数字 要求在区间[1,6] 2)输入的6个数字一次被7减，得到新的序列 3)新的序列对应0x6032d0中的6个nodes，会按照新的序列的顺序一次将nodes连接起来 4)连接好的nodes 要求以value值从大到小排序
- 答案是 4 3 2 1 6 5
- 将不同function的assembly code拆分开，会更容易理解程序
phase_6:
4010f4: 41 56 pushq %r14
4010f6: 41 55 pushq %r13
4010f8: 41 54 pushq %r12
4010fa: 55 pushq %rbp
4010fb: 53 pushq %rbx
4010fc: 48 83 ec 50 subq $80, %rsp
401100: 49 89 e5 movq %rsp, %r13
401103: 48 89 e6 movq %rsp, %rsi
401106: e8 51 03 00 00 callq 849 <read_six_numbers>
40110b: 49 89 e6 movq %rsp, %r14
40110e: 41 bc 00 00 00 00 movl $0, %r12d
401114: 4c 89 ed movq %r13, %rbp
401117: 41 8b 45 00 movl (%r13), %eax
40111b: 83 e8 01 subl $1, %eax
40111e: 83 f8 05 cmpl $5, %eax
401121: 76 05 jbe 5 <phase_6+0x34>
401123: e8 12 03 00 00 callq 786 <explode_bomb>
401128: 41 83 c4 01 addl $1, %r12d
40112c: 41 83 fc 06 cmpl $6, %r12d
401130: 74 21 je 33 <phase_6+0x5f>
401132: 44 89 e3 movl %r12d, %ebx
401135: 48 63 c3 movslq %ebx, %rax
401138: 8b 04 84 movl (%rsp,%rax,4), %eax
40113b: 39 45 00 cmpl %eax, (%rbp)
40113e: 75 05 jne 5 <phase_6+0x51>
401140: e8 f5 02 00 00 callq 757 <explode_bomb>
401145: 83 c3 01 addl $1, %ebx
401148: 83 fb 05 cmpl $5, %ebx
40114b: 7e e8 jle -24 <phase_6+0x41>
40114d: 49 83 c5 04 addq $4, %r13
401151: eb c1 jmp -63 <phase_6+0x20>
401153: 48 8d 74 24 18 leaq 24(%rsp), %rsi
401158: 4c 89 f0 movq %r14, %rax
40115b: b9 07 00 00 00 movl $7, %ecx
401160: 89 ca movl %ecx, %edx
401162: 2b 10 subl (%rax), %edx
401164: 89 10 movl %edx, (%rax)
401166: 48 83 c0 04 addq $4, %rax
40116a: 48 39 f0 cmpq %rsi, %rax
40116d: 75 f1 jne -15 <phase_6+0x6c>
40116f: be 00 00 00 00 movl $0, %esi
401174: eb 21 jmp 33 <phase_6+0xa3>
401176: 48 8b 52 08 movq 8(%rdx), %rdx
40117a: 83 c0 01 addl $1, %eax
40117d: 39 c8 cmpl %ecx, %eax
40117f: 75 f5 jne -11 <phase_6+0x82>
401181: eb 05 jmp 5 <phase_6+0x94>
401183: ba d0 32 60 00 movl $6304464, %edx
401188: 48 89 54 74 20 movq %rdx, 32(%rsp,%rsi,2)
40118d: 48 83 c6 04 addq $4, %rsi
401191: 48 83 fe 18 cmpq $24, %rsi
401195: 74 14 je 20 <phase_6+0xb7>
401197: 8b 0c 34 movl (%rsp,%rsi), %ecx
40119a: 83 f9 01 cmpl $1, %ecx
40119d: 7e e4 jle -28 <phase_6+0x8f>
40119f: b8 01 00 00 00 movl $1, %eax
4011a4: ba d0 32 60 00 movl $6304464, %edx
4011a9: eb cb jmp -53 <phase_6+0x82>
4011ab: 48 8b 5c 24 20 movq 32(%rsp), %rbx
4011b0: 48 8d 44 24 28 leaq 40(%rsp), %rax
4011b5: 48 8d 74 24 50 leaq 80(%rsp), %rsi
4011ba: 48 89 d9 movq %rbx, %rcx
4011bd: 48 8b 10 movq (%rax), %rdx
4011c0: 48 89 51 08 movq %rdx, 8(%rcx)
4011c4: 48 83 c0 08 addq $8, %rax
4011c8: 48 39 f0 cmpq %rsi, %rax
4011cb: 74 05 je 5 <phase_6+0xde>
4011cd: 48 89 d1 movq %rdx, %rcx
4011d0: eb eb jmp -21 <phase_6+0xc9>
4011d2: 48 c7 42 08 00 00 00 00 movq $0, 8(%rdx)
4011da: bd 05 00 00 00 movl $5, %ebp
4011df: 48 8b 43 08 movq 8(%rbx), %rax
4011e3: 8b 00 movl (%rax), %eax
4011e5: 39 03 cmpl %eax, (%rbx)
4011e7: 7d 05 jge 5 <phase_6+0xfa>
4011e9: e8 4c 02 00 00 callq 588 <explode_bomb>
4011ee: 48 8b 5b 08 movq 8(%rbx), %rbx
4011f2: 83 ed 01 subl $1, %ebp
4011f5: 75 e8 jne -24 <phase_6+0xeb>
4011f7: 48 83 c4 50 addq $80, %rsp
4011fb: 5b popq %rbx
4011fc: 5d popq %rbp
4011fd: 41 5c popq %r12
4011ff: 41 5d popq %r13
401201: 41 5e popq %r14
401203: c3 retq
Summary
- Lab本身很有趣，如果对常用的汇编指令理解无误，耐心去解，把杂糅在一起的assembly code按功能切分好还是可以解出来的。
- gdb常用的打断点和x p指令查看registers 或者内存中的值非常有用，大概明白程序的各个function就输入值，打断点观察寄存器的变化，不要纯看汇编去decode。我最开始就是这么做的，很折磨也很费时间。让程序跑起来才是关键，切分成小块去decode