这篇文章记录一下做bomb实验的过程。这几天封宿舍,每天除了做核酸之外就不能出去。每天拆一个炸弹,算上彩蛋刚好用时一周。
首先使用objdump -d bomb > bomb.txt
得到反汇编代码。
查看bomb.c文件看到有phase_1
到phase_6
6个输入函数,对应6个炸弹。下面从第一个开始拆除。
文章目录
phase_1
phase_1
中调用了strings_not_equal
,后者又调用了string_length
,所以从string_length
看起,翻译到C语言如下:
/*
* str in %rdi, length in %rax, p in %rdx
* 功能是返回字符串的长度
*/
int string_length(char *str) {
int length;
if (str == NULL)
return 0;
char *p = str;
while (p != NULL) {
p++;
length = p - str;
}
return length;
}
然后再看strings_not_equal
,翻译到C语言如下:
/*
* 若两个字符串不同则返回1,否则返回0
* 为了让代码更加清晰,在不改变汇编代码原意的前提下我做了些修改
* str1 in %rdi, str2 in %rsi
* str1_len in %r12, p1 in %rbx, p2 in %rbp
*/
int strings_not_equal(char *str1, char *str2) {
char *p1 = str1;
char *p2 = str2;
int str1_len = string_length(str1);
int str2_len = string_length(str2);
if (str1_len != str2_len)
return 1;
char c1 = *p1;
if (c1 == NULL)
return 0;
char c2 = *p2;
while (c1 != NULL) {
if (c1 != c2)
return 1;
// 下一个字符
p1++; p2++;
c1 = *p1; c2 = *p2;
}
return 0;
}
做好这些准备开始看phase_1
:
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 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 ret
test + je的组合意思是若%eax == 0,则跳转。phase_1很简单,就是比较以下输入的字符串和内存中的一个字符串是否相等。可以看到输入的字符串应该等于内存0x402400处的字符串,通过gdb的x/s 0x402400
查看得到该处字符串为Border relations with Canada have never been better.
phase_2
在read_six_numbers
函数中调用了sscanf
。注意到有一句mov $0x4025c3, %esi
,通过x/s 0x4025c3
查看该字符串为"%d %d %d %d %d %d"
,因此该函数读取6个整数,并且可以看到是从%rsi中依次读取。若其返回值小于6,即读取的值少于6个则会引爆炸弹。因此最终可以翻译为:
/*
* input in %rsi, arr in stack
*/
int read_six_numbers(char *input, int* arr) {
int scan_num = sscanf(input, "%d %d %d %d %d %d", arr, &arr[1], &arr[2], &arr[3], &arr[4], &arr[5]);
if (scan_num < 6)
explode_bomb();
return scan_num;
}
phase_2
代码如下:
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)
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
400f1c: 39 03 cmp %eax,(%rbx)
400f1e: 74 05 je 400f25 <phase_2+0x29>
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
调试分析phase_2
,首先需要输入6个整数,以空格隔开。然后要满足第一个数字
n
1
n_{1}
n1必须是1,且后一个数字
n
i
+
1
n_{i+1}
ni+1是前一个数字
n
i
n_{i}
ni的两倍。翻译如下:
void phase_2(char *input) {
int arr[6];
read_six_numbers(input, arr);
// 第一个数字必须是1
if (arr[0] != 1)
explode_bomb();
for (int i = 1; i < 6; i++) {
// 下一个必须是前一个的两倍
if (arr[i] != 2 * arr[i - 1])
explode_bomb();
}
}
因此答案则为1 2 4 8 16 32
phase_3
代码如下:
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 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>
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 jmp *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 call 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 call 40143a <explode_bomb>
400fc9: 48 83 c4 18 add $0x18,%rsp
400fcd: c3 ret
注意第五行,首先查看0x4025cf处的字符串为"%d %d"
,确定读入2个整数。通过调试确定第一个参数位于%rsp + 8
,第二个参数位于%rsp + 0xc
处。且第一个参数不能大于7。
对于间接跳转指令jmp *0x402470(, %rax, 8)
,使用x/wx 0x402470
查看其值为0x400f7c
,并且使用相同的命令查看switch各个情况的跳转目标地址。
注意间接跳转指令的含义是跳转到
0x402470 + 8 * %rax
存储的地址处,而非0x402470
存储的地址再加8 * %rax处。
另外,注意mov和lea的区别。mov 0x0(%rsp) %rax
是把%rsp处的数据给%rax,而lea 0x0(%rsp) %rax
是把%rsp的值给%rax。
下面是对用的C语言:
void phase_3(char *input) {
int a, b;
int scan_num = sscanf(input, "%d %d", &a, &b);
if (scan_num < 2)
explode_bomb();
int eax;
if (a > 7 || a < 0)
explode_bomb();
switch (a) {
// 0x400f7c in 0x402470
case 0:
eax = 0xcf;
break;
// 0x400fb9 in 0x402478
case 1:
eax = 0x137;
break;
// 0x400f83 in 0x402480
case 2:
eax = 0x2c3;
break;
// 0x400f8a in 0x402488
case 3:
eax = 0x100;
break;
// 0x400f91 in 0x402490
case 4:
eax = 0x185;
break;
// 0x400f98 in 0x402498
case 5:
eax = 0xce;
break;
// 0x400f9f in 0x4024a0
case 6:
eax = 0x2aa;
break;
// 0x400fa6 in 0x4024a8
case 7:
eax = 0x147;
break;
}
if (b != eax)
explode_bomb();
}
因此以下答案都是正确的:(0, 207) (1, 311) (2, 707) (3, 256) (4, 389) (5, 206) (6, 682) (7, 327)
phase_4
phase_4
调用了func4
:
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 call 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 call 400fce <func4>
401003: 8d 44 00 01 lea 0x1(%rax,%rax,1),%eax
401007: 48 83 c4 08 add $0x8,%rsp
40100b: c3 ret
注意第6行shr是逻辑右移指令,在C语言实现时应该先转为unsigned
类型再右移。另外,sar %eax
表示的是sar 1 %eax
。对应的C语言函数如下:
/*
* 函数功能是二分查找
* 关于符号位那里,如果出现lo > hi的情况,经过处理后mid将会正确地表示中点。很巧妙的一种处理方法。
* hi in %edx, lo in %esi, a in %edi
*/
int func4(int hi, int lo, int a) {
int result = hi - lo;
int mid = (unsigned)result >> 0x1f; // 逻辑右移31位获得符号位
// 若a-b非负,则result不变,否则+1
result += mid;
result >>= 1; // sar %eax
// mid = b~a的中点
mid = lo + result;
if (mid <= a) {
result = 0;
if (mid >= a)
return result;
// 更新下界
lo = mid + 1;
result = func4(hi, lo, a);
return (2 * result + 1);
}
else {
// 更新上界
hi = mid - 1;
result = func4(hi, lo, a);
return 2 * result;
}
}
下面再看phase_4
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 call 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 call 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 call 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 call 40143a <explode_bomb>
40105d: 48 83 c4 18 add $0x18,%rsp
401061: c3 ret
和phase_3
一样,先读取两个整数。
下面是对应的C语言:
void phase_4(char *input) {
int a, b;
int result = 0;
int scan_num = sscanf(input, "%d %d", &a, &b);
if (scan_num != 2)
explode_bomb();
if (a >= 14 || a < 0)
explode_bomb();
result = func4(14, 0, a);
if (result != 0 || b != 0)
explode_bomb();
}
注意到func4里只有在更新下界的时候会让result+1,因此输入的第一个参数a在满足边界[0, 13)的前提下,只要满足调用的所有(包括递归)func4都不调整下界即可。最显而易见的答案是7,因为phase_4中是这样调用的:func4(14, 0, a)
,一次找到就不会调整下界。
因此最终a可选的值包括:0, 1, 3, 7;而b只能取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 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
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 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
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
对于4010b3处的指令,查看内存0x40245e处的字符串为flyers。
对于401099处的指令,查看内存0x4024b0处的字符串为maduiersnfotvbylSo you think you can stop the bomb with ctrl-c, do you?,由于只有低四位是可变的,因此用到的只有从头开始的16个字符,即maduiersnfotvbyl。
另外,在运行Bomb时使用ctrl-c命令结束进程时会显示后面那句话,不过Dr.evil犹豫一会儿就会让我们成功结束,他会说:“Well…OK.