CSAPP实验bomblab
Time: 2020-07-30 Tags: binaryLink: CSAPP实验bomblab
渊源
刚进入蓝🐋的时候👴想当②进制🖐，当时做了一次💣实验，但是当时是用ida做的，加上不会看汇编，导致lab6和sercet_lab不会做。这次会了汇编，来做做vanvan。
做了三天，主要是因为👴是蓝🐕。
获取实验并使用
从http://csapp.cs.cmu.edu/3e/labs.html获取。获取方法是点击实验后面的Self-Study Handout。
获取一个.tar文件，直接解压或者使用命令解压
tar -xvf bomb.tar
共有以下文件。
± tree
.
├── README
├── bomb
└── bomb.c
0 directories, 3 files
其中README没卵用，bomb是我们需要通关的程序，bomb.c是程序的main函数部分，由于加了-g编译参数，用gdb的时候可以看到main函数的源码。
获得反汇编
可以直接使用命令：
$ objdump -d bomb > bomb.d
然后直接在bomb.d中查看。
直接用gdb动态调试也行，当我没说。
直接拖进ida里面F5也行，也当我没说。
用法
bomb.c中已经说的很清楚了：
if (argc == 1) {
infile = stdin;
}
/* When run with one argument <file>, the bomb reads from <file>
* until EOF, and then switches to standard input. Thus, as you
* defuse each phase, you can add its defusing string to <file> and
* avoid having to retype it. */
else if (argc == 2) {
if (!(infile = fopen(argv[1], "r"))) {
printf("%s: Error: Couldn't open %s\n", argv[0], argv[1]);
exit(8);
}
}
/* You can't call the bomb with more than 1 command line argument. */
else {
printf("Usage: %s [<input_file>]\n", argv[0]);
exit(8);
}
要么直接./bomb运行，然后手动输入；要么把答案写在一个文件里，用./bomb inputfile运行。文件输入的时候，读到EOF会返回到stdin。
题解
phase_1
汇编代码：
0000000000400ee0 <phase_1>:
400ee0: 48 83 ec 08 sub $0x8,%rsp
400ee4: be 00 24 40 00 mov $0x402400,%esi # 0x402400: "Border relations with Canada have never been better."
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
比较输入的字符串和0x402400处的字符串是否一样，如果不一样💣爆炸。
查看0x402400：
pwndbg> x/s 0x402400
0x402400: "Border relations with Canada have never been better."
所以第一关的答案是：
Border relations with Canada have never been better.
phase_2
汇编代码：
0000000000400efc <phase_2>:
400efc: 55 push %rbp
400efd: 53 push %rbx
400efe: 48 83 ec 28 sub $0x28,%rsp # stack init
400f02: 48 89 e6 mov %rsp,%rsi
400f05: e8 52 05 00 00 callq 40145c <read_six_numbers> # read six numbers
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp) # first number: 1
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>
<phase_2+0x1b>
400f17: 8b 43 fc mov -0x4(%rbx),%eax
400f1a: 01 c0 add %eax,%eax # 乘2
400f1c: 39 03 cmp %eax,(%rbx) # second number: 2
400f1e: 74 05 je 400f25 <phase_2+0x29>
400f20: e8 15 05 00 00 callq 40143a <explode_bomb>
<phase_2+0x29>
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>
<phase_2+0x34>
400f30: 48 8d 5c 24 04 lea 0x4(%rsp),%rbx # the second number
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
看起来有点麻烦，改写一下。
int phase_2(int *a){
int b[6];
read_six_numbers(b);
if (b[0] != 1)
explode_bomb();
for (int i = 1;i <= 5;i++){
int t = b[i-1];
if (2 * t != b[i])
explode_bomb();
}
}
一目了然，答案：
1 2 4 8 16 32
phase_3
汇编代码：
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> # scanf("%d %d",a,b)
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>
<phase_3+0x27>
400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp) # if (a>0x7) explode_bomb();
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) # jmp 0x400fb9
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 # if (b != 0x137) explode_bomb();
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
读入两个数，第一个数不能大于7，然后直接跳转，第二个数等于0x137，结束…中间的一大片根本没用到
答案：
1 311
phase_4
汇编代码：
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> # scanf("%d %d",a,b)
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) # <=14
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> # fun4(a,0,14)
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
读入两个数，第一个数要小于等于14，第二个数要等于0。此外还调用了一个函数func4()。
0000000000400fce <func4>: func4(a,b,c)
400fce: 48 83 ec 08 sub $0x8,%rsp
// eax: t1 ecx:t2
400fd2: 89 d0 mov %edx,%eax # t1 = c
400fd4: 29 f0 sub %esi,%eax # t1 -= b
400fd6: 89 c1 mov %eax,%ecx # t2 = t1
400fd8: c1 e9 1f shr $0x1f,%ecx # t2 >>= 31
400fdb: 01 c8 add %ecx,%eax # t1 += t2
400fdd: d1 f8 sar 1,%eax # t1 >>= 1 （算术）
400fdf: 8d 0c 30 lea (%rax,%rsi,1),%ecx # t2 += b+t1
400fe2: 39 f9 cmp %edi,%ecx # a <= t2 ?
400fe4: 7e 0c jle 400ff2 <func4+0x24>
400fe6: 8d 51 ff lea -0x1(%rcx),%edx #
400fe9: e8 e0 ff ff ff callq 400fce <func4> # func4(a,b,c-1)
400fee: 01 c0 add %eax,%eax # t1 *= 2
400ff0: eb 15 jmp 401007 <func4+0x39> # break
400ff2: b8 00 00 00 00 mov $0x0,%eax # t1 = 0
400ff7: 39 f9 cmp %edi,%ecx # a >= t2 ?
400ff9: 7d 0c jge 401007 <func4+0x39> # break
400ffb: 8d 71 01 lea 0x1(%rcx),%esi #
400ffe: e8 cb ff ff ff callq 400fce <func4> # func4(a,b+1,c)
401003: 8d 44 00 01 lea 0x1(%rax,%rax,1),%eax # t1 *= 2
401007: 48 83 c4 08 add $0x8,%rsp
40100b: c3 retq
很nt的一个递归函数。有很多运算，改写一下得到c代码。
int func4(a,b,c){
int t1,t2;
t1 = c - b; // 14
t2 = t1 >> 31; // 0
t1 += t2; // 14
t1 >>= 1;(算术) // 7
t2 += b + t1; // 7
if (a <= t2){
t1 = 0;
if (a >= t2)
return t1;
func4(a,b+1,c);
t1 *= 2;
return t1;
}
else {
func4(a,b,c-1);
t1 *= 2;
return t1;
}
}
由于第一次调用时有b=0c=14，可以直接算出部分临时变量的值。根据 40104d:test %eax,%eax和40104f:jne 401058可以判断我们需要返回一个0。根据我们算出来的值，进入if (a <= t2)，再进入if (a >= t2)，即返回0，而满足此条件只要a=7即可。
所以，答案是：
7 0
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 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 # x/s 0x4024b0
4010a0: 88 54 04 10 mov %dl,0x10(%rsp,%rax,1) # 0x4024b0 <array>: "maduiersnfotvbyl"
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 # string: "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>
<phase_5+0x70>
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
这个函数有大约三个部分。401073~401082部分，读入一个长度必须为6的字符串，否则💣爆炸。40108b~4010ac部分将读入字符进行操作&0xf，即取低4个二进制位，根据低四位的值在0x4024b0 <array>: "maduiersnfotvbyl"处的字符串取出字符。4010ae~4010d0部分，根据取出的字符和0x40245e处的字符串”flyers”比较。
所以我们需要的字符串必须满足低四位分别为：0x9 0xf 0xe 0x5 0x6 0x7
于是我选择的字符串是：
9on567
phase_6
汇编代码：
00000000004010f4 <phase_6>: // 4 3 2 1 6 5
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>
// 上面这段401114~401151的功能是检测输入的数字是否有重复并且是否满足 0<x<=6
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>
// 401160~40116d
for (int i = 0; i <= 5; i++){
a[i] = 7 - a[i];
}
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>
// 上面这段根据a[i] = 7 - a[i]之后的序号将node填入栈中
typedef struct node{
int num;
int order;
struct node *next;
}
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>
// 根据刚才入栈的顺序，改变链表
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>
// 检测是否是降序排列的
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
//
pwndbg> 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
注释已经写的比较清楚了，不想写了。说实话，这一关挺好玩的。
答案：
4 3 2 1 6 5
sercet_phase
这关有点难。首先直接ctrl+f查找”sercet_phase”，在phase_defused中找到了此函数的调用。
401630: e8 0d fc ff ff callq 401242 <secret_phase>
直接对这个函数进行逆向。
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
cmpl $0x6,0x202181(%rip)其中0x202181+%rip是0x603760。记录的是你拆完💣的数量，到达6个的话就能绕过jne 40163f。执行scanf。格式符是”%d %d %s”。查看地址0x603870，发现是phase_4的输入。因此缺少一个%s。因为后面有strings_not_equal，猜测0x402622就是目标字符串地址。0x402622: "DrEvil"。所以，在7 0之后加上一个”DrEvil”即可进入secret_phase。
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
401293: 90 nop
401294: 90 nop
401295: 90 nop
401296: 90 nop
401297: 90 nop
401298: 90 nop
401299: 90 nop
40129a: 90 nop
40129b: 90 nop
40129c: 90 nop
40129d: 90 nop
40129e: 90 nop
40129f: 90 nop
使用strtol将读取的字符转化成数字，然后判断这个数小于0x3e8，之后将一个二叉树的根节点和所输入的数作为参数调用func7，目标是返回一个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
一顿分析，得到c代码
int func7(int *a,int b){
if (a == 0){
return -1;
}
if (*a > b){
a = a->left;
func7(a,b);
return result*2;
}
else{
result = 0;
if (*a == b){
return 0;
}
a = a->right;
result = func7(a,b);
return result*2+1;
}
}
这样的话就非常清晰了。
二叉树结构如下。
要想返回一个2的话，我们需要进入if (*a > b)这个分支，并在下一层func7中得到1，那么我们的b应该小于24。
要想返回一个1的话，我们需要进入else分支，并在下一层func7中得到一个0，那么我们的b应该大于8。
要想返回一个0的话，我们需要进入else分支，并进入if (*a == b)这个分支，所以我们的b应该等于16。
所以，答案是：
16
总结
CSAPP的第二个实验。个人感觉难度没有datalab高。也没有第一次的那个版本难度高。可能是因为本人做pwn题目比较多对汇编比较熟悉。
通过本次实验，感受到了静态分析的局限性，动态调试可以有效的观察各个寄存器和内存的变化，是一种值得学习的调试方法。
