pwnable.kr-collision -Writeup

时间：2023-03-09 10:07:26
pwnable.kr-collision -Writeup

bof

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p, blockquote, ul, ol, dl, table, pre {
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body>h2:first-child, body>h1:first-child, body>h1:first-child+h2, body>h3:first-child, body>h4:first-child, body>h5:first-child, body>h6:first-child {
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h1+p, h2+p, h3+p, h4+p, h5+p, h6+p {
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a:hover {
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/* LISTS
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ul li > :first-child,
ol li > :first-child,
ul li ul:first-of-type,
ol li ol:first-of-type,
ul li ol:first-of-type,
ol li ul:first-of-type {
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ul ul, ul ol, ol ol, ol ul {
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table tr {
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table tr:nth-child(2n) {
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/* IMAGES
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img {
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-->

pwnable.kr-collision -Writeup

同第一题fd,
ssh连接，ls -l查看文件， cat col.c得到代码如下

 #include <stdio.h>

 #include <string.h>

 unsigned long hashcode = 0x21DD09EC;

 unsigned long check_password(const char* p){

     int* ip = (int*)p;

     int i;

     int res=;

     for(i=; i<; i++){

         res += ip[i];

     }

     return res;

 }

 int main(int argc, char* argv[]){

     if(argc<){

         printf("usage : %s [passcode]\n", argv[]);

         return ;

     }

     if(strlen(argv[]) != ){

         printf("passcode length should be 20 bytes\n");

         return ;

     }

     if(hashcode == check_password( argv[] )){

         system("/bin/cat flag");

         return ;

     }

     else

         printf("wrong passcode.\n");

     return ;

 }
  • 分析过程如下：
    pwnable.kr-collision -Writeup
  • 分析check_password()函数,当strlen(argv[1])==20时，要保证check_password()函数的返回值为0x21DD09EC;

    经分析，check_password()函数的作用为将长度为20的argv[1]分为5段，每段有4个字符，这些字符是以小端的形式存储的，以int *指针的形式每次读取4个字符，将5次读取的值求和

  • 最简单的想法是前16位均为\x00,后4位为\xEC\x09\xDD\x21，但经过测试答案错误，查表得\x09为制表符tab,会截断输入，同时\x00也会截断输入；
  • 因此构造另一种输入'\x01'*16+'\xE8\x05\xD9\x1D',分析过程如下：
  • 理解字节序
  • pwnable.kr-collision -Writeup
  • 则可构造语句

    python -c "print '\x01'*16+'\xE8\x05\xD9\x1D'"

\x和0x表示16进制的区别：pwnable.kr-collision -Writeup

  • 运行，结果如下,flag为daddy! I just managed to create a hash collision :)：
    pwnable.kr-collision -Writeup
    2017-2-5 0:55;56

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