这是什么企图木马code呢? [英] what does this attempted trojan horse code do?
本文介绍了这是什么企图木马code呢?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
它看起来像这样只是发送一条平,但什么的地步时,你可以用ping?
/ *警告:这是在写一个恶意木马别人的企图。不要
编译和* *绝对不要安装。我增加了一个退出的
第一行,以避免事故 - MSW * /
INT主(INT ARGC,CHAR *的argv [])
{
出口(1);
unsigned int类型的pid = 0;
炭缓冲液[2];
字符* ARGS [] = {
/斌/平,
-C,
5,
空值,
空值
}; 如果(的argc!= 2)
返回0; ARGS [3] =的strdup(的argv [1]);
为(;;)
{
得到(缓冲); / * * FTW / 如果(缓冲[0] == 0x6e)
打破; 开关(PID =叉())
{
情况1:
的printf(错误分叉\\ n);
出口(255);
情况下0:
execvp(参数[0],参数);
出口(1);
默认:
打破;
}
}
返回255;
}
解决方案
这确保了平
被调用的参数 -c 5
。这是愚蠢的,因为一个shell脚本或别名会更容易阅读和快写。
It looks like this just sends a ping, but whats the point of that when you can just use ping?
/* WARNING: this is someone's attempt at writing a malware trojan. Do not
compile and *definitely* don't install. I added an exit as the
first line to avoid mishaps - msw */
int main (int argc, char *argv[])
{
exit(1);
unsigned int pid = 0;
char buffer[2];
char *args[] = {
"/bin/ping",
"-c",
"5",
NULL,
NULL
};
if (argc != 2)
return 0;
args[3] = strdup(argv[1]);
for (;;)
{
gets(buffer); /* FTW */
if (buffer[0] == 0x6e)
break;
switch (pid = fork())
{
case -1:
printf("Error Forking\n");
exit(255);
case 0:
execvp(args[0], args);
exit(1);
default:
break;
}
}
return 255;
}
解决方案
It makes sure that ping
is called with the arguments -c 5
. Which is stupid, because a shell script or alias would be easier to read and faster to write.
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