pwnable echo1

#Exploit for [email protected] #@Windcarp 2015.07.23 from pwn import * #init context(arch = 'amd64', os = 'linux') local=False if local: p = process("./echo1") libc = ELF("/lib/x86_64-linux-gnu/libc-2.19.so") else: p = remote("pwn

pwnable的echo1,一个栈溢出的漏洞,同样以后再补上分析过程和思路,先放exp 1 #!/usr/bin/env python 2 # -*- coding: utf-8 -*- 3 __Auther__ = 'M4x' 4 5 from pwn import * 6 context.arch = 'amd64' 7 context.os = 'linux' 8 context.log_level = 'debug' 9 10 elf = ELF('./echo1') 11 id_add

MarkdownPad Document html,body,div,span,applet,object,iframe,h1,h2,h3,h4,h5,h6,p,blockquote,pre,a,abbr,acronym,address,big,cite,code,del,dfn,em,img,ins,kbd,q,s,samp,small,strike,strong,sub,sup,tt,var,b,u,i,center,dl,dt,dd,ol,ul,li,fieldset,form,label

html,body,div,span,applet,object,iframe,h1,h2,h3,h4,h5,h6,p,blockquote,pre,a,abbr,acronym,address,big,cite,code,del,dfn,em,img,ins,kbd,q,s,samp,small,strike,strong,sub,sup,tt,var,b,u,i,center,dl,dt,dd,ol,ul,li,fieldset,form,label,legend,table,caption

[email protected]:~$ ls cmd1  cmd1.c  flag [email protected]:~$ cat cmd1.c #include <stdio.h> #include <string.h> int filter(char* cmd){ int r=0; r += strstr(cmd, "flag")!=0; r += strstr(cmd, "sh")!=0; r += strstr(cmd, &quo

pwnable从入门到放弃,第三题. Download : http://pwnable.kr/bin/bofDownload : http://pwnable.kr/bin/bof.c Running at : nc pwnable.kr 9000 很简单的一道栈溢出题目. 依然先读代码: #include <stdio.h> #include <string.h> #include <stdlib.h> void func(int key){ char overfl

http://www.pwnable.kr/ 逆向题目,upx脱壳: 找到flag:

pwnable echo2 linux 32位程序 涉及FSB和UAF漏洞的利用,先%x泄露地址,然后利用UAF漏洞进行利用 code:

首先查看源代码: #include <stdio.h> #include <string.h> #include <stdlib.h> #include <sys/mman.h> #include <seccomp.h> #include <sys/prctl.h> #include <fcntl.h> #include <unistd.h> #define LENGTH 128 void sandbox(){