<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[orange_cat__的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/orange_cat__</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; orange_cat__]]></copyright><item><title><![CDATA[buuctf pwn 部分exp（无解析，持续更新）]]></title><link>https://blog.youkuaiyun.com/orange_cat__/article/details/127901774</link><guid>https://blog.youkuaiyun.com/orange_cat__/article/details/127901774</guid><author>orange_cat__</author><pubDate>Thu, 17 Nov 2022 12:55:22 +0800</pubDate><description><![CDATA[【代码】buuctf pwn 部分exp（无解析，持续更新）]]></description><category></category></item><item><title><![CDATA[ctf pwn buuctf ciscn_2019_n_1]]></title><link>https://blog.youkuaiyun.com/orange_cat__/article/details/127883114</link><guid>https://blog.youkuaiyun.com/orange_cat__/article/details/127883114</guid><author>orange_cat__</author><pubDate>Wed, 16 Nov 2022 13:03:42 +0800</pubDate><description><![CDATA[主要思路就是覆盖v2的地址为11.28125，利用gets函数输入v1时存在的栈溢出漏洞。图上箭头指的为11.28125的16进制。调用了func（）函数，进去后分析。v1距离v2为16×3-4=44.四，打开靶机，写payload。三，找payload所需条件。NX保护，堆栈不可执行。五，打通获得flag。]]></description><category></category></item><item><title><![CDATA[CTF misc 0宽度字节隐写]]></title><link>https://blog.youkuaiyun.com/orange_cat__/article/details/127579216</link><guid>https://blog.youkuaiyun.com/orange_cat__/article/details/127579216</guid><author>orange_cat__</author><pubDate>Fri, 28 Oct 2022 21:43:14 +0800</pubDate><description><![CDATA[CTF 0宽度字节隐写。]]></description><category></category></item></channel></rss>