在微博看到乌云说glibc才知道最近出现了个漏洞叫幽灵，不过对于我这种小白来说，只管给站检测然后更新就是了

检测程序：    

#include <netdb.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>

#define CANARY "in_the_coal_mine"

struct{
    char buffer[1024];
    char canary[sizeof(CANARY)];
}temp={"buffer",CANARY};

int main(void){
    struct hostent resbuf;
    struct hostent *result;
    int herrno;
    int retval;

    /*** strlen (name) = size_needed - sizeof (*host_addr) - sizeof (*h_addr_ptrs) - 1; ***/
    size_t len = sizeof(temp.buffer) - 16*sizeof(unsigned char) - 2*sizeof(char *) - 1;
    char name[sizeof(temp.buffer)];
    memset(name, '0', len);
    name[len] = '\0';  

    retval = gethostbyname_r(name, &resbuf, temp.buffer, sizeof(temp.buffer), &result, &herrno); 

    if (strcmp(temp.canary, CANARY) != 0) {
        puts("vulnerable"); 
        exit(EXIT_SUCCESS);
    }
    if (retval == ERANGE) { 
        puts("not vulnerable");  
        exit(EXIT_SUCCESS);
    }
    puts("should not happen");
    exit(EXIT_FAILURE);
}

编译并运行：gcc gistfile1.c -o CVE-2015-0235 && ./CVE-2015-0235，如果出现“vulnerable”就是存在漏洞，更新下glibc就行了。