WHCTF-babyre
首先执行file命令得到如下信息
ELF 64-bit LSB executable, x86-64
尝试用IDA64打开,定位到关键函数main发现无法F5,尝试了修复无果,于是用gdb动态调试一发。
在scanf处下断点
b *0x0400660
输入12346789之后next
=> 0x40066c <main+102>: call 0x4004c0 <strlen@plt>
0x400671 <main+107>: mov DWORD PTR [rbp-0x8],eax
0x400674 <main+110>: cmp DWORD PTR [rbp-0x8],0xe
可以看到调用了strlen函数,然后将长度和0xe比较,于是知道flag长度为0xe。
再次运行
Please input flag:abcdefghijklmn
通过长度验证之后可以看到
0x40067a <main+116>: mov edx,0x600b00
0x40067f <main+121>: lea rax,[rbp-0x20]
0x400683 <main+125>: mov rdi,rax
0x400686 <main+128>: call rdx
接着调用了0x600b00处的内容,于是
b *0x600b00
c(ontinue)
然后看到一连串的数
0x600b08 <judge+8>: mov BYTE PTR [rbp-0x20],0x66
0x600b0c <judge+12>: mov BYTE PTR [rbp-0x1f],0x6d
0x600b10 <judge+16>: mov BYTE PTR [rbp-0x1e],0x63
=> 0x600b14 <judge+20>: mov BYTE PTR [rbp-0x1d],0x64
0x600b18 <judge+24>: mov BYTE PTR [rbp-0x1c],0x7f
0x600b1c <judge+28>: mov BYTE PTR [rbp-0x1b],0x6b
0x600b20 <judge+32>: mov BYTE PTR [rbp-0x1a],0x37
0x600b24 <judge+36>: mov BYTE PTR [rbp-0x19],0x64
0x600b28 <judge+40>: mov BYTE PTR [rbp-0x18],0x3b
0x600b2c <judge+44>: mov BYTE PTR [rbp-0x17],0x56
0x600b30 <judge+48>: mov BYTE PTR [rbp-0x16],0x60
=> 0x600b34 <judge+52>: mov BYTE PTR [rbp-0x15],0x3b
0x600b38 <judge+56>: mov BYTE PTR [rbp-0x14],0x6e
0x600b3c <judge+60>: mov BYTE PTR [rbp-0x13],0x70
0x600b40 <judge+64>: mov DWORD PTR [rbp-0x4],0x0
0x600b47 <judge+71>: jmp 0x600b71 <judge+113>
这串数字在后面会用到,接着调试
RAX: 0x7fffffffe2c0 ("abcdefghijklmn")
RBX: 0x0
RCX: 0x0
RDX: 0x61 (‘a‘)
EFLAGS: 0x206 (carry PARITY adjust zero sign trap INTERRUPT direction overflow)
[-------------------------------------code-------------------------------------]
0x600b60 <judge+96>: add rdx,rcx
0x600b63 <judge+99>: movzx edx,BYTE PTR [rdx]
0x600b66 <judge+102>: mov ecx,DWORD PTR [rbp-0x4]
=> 0x600b69 <judge+105>: xor edx,ecx
可以看到,rdx=0x61(‘a‘),rcx=0x0,然后rdx=rdx^rcx。于是大致知道是异或加密了flag。
接着调试发现在下一次,rdx=0x62(‘b‘),rcx=0x1。
于是知道flag[i] = key[i] ^ i && 0 <= i < strlen(flag)
而这里的key便是上面的一连串数字,脚本输出flag
#!/usr/bin/python
# -*- coding: utf-8 -*-
__Author__ = "LB@10.0.0.55"
key = [0x66,0x6d,0x63,0x64,0x7f,0x6b,0x37,0x64,0x3b,0x56,0x60,0x3b,0x6e,0x70]
flag = ‘‘
for i in range(len(key)):
flag += chr(key[i]^i)
print flag
#flag{n1c3_j0b}
作者: LB919
出处:http://www.cnblogs.com/L1B0/
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