取消共享后的grantpt报告错误

Question

我有一个小程序,试图在取消共享后创建一个伪终端.输出是:

uid before unshare:5000
uid after unshare:0
Grant pt Error: : Permission denied

代码:

#define _GNU_SOURCE

#include <sys/mount.h>
#include <unistd.h>
#include <stdlib.h>
#include <stdio.h>
#include <fcntl.h>
#include <errno.h>
#include <sched.h>

void set_uid_map(pid_t pid, int inside_id, int outside_id, int length) {
    char path[256];
    sprintf(path, "/proc/%d/uid_map", getpid());
    FILE* uid_map = fopen(path, "w");
    fprintf(uid_map, "%d %d %d", inside_id, outside_id, length);
    fclose(uid_map);
}
void set_gid_map(pid_t pid, int inside_id, int outside_id, int length) {
    char path[256];
    sprintf(path, "/proc/%d/gid_map", getpid());
    FILE* gid_map = fopen(path, "w");
    fprintf(gid_map, "%d %d %d", inside_id, outside_id, length);
    fclose(gid_map);
}

int main(void)
{
int master;
int flag = 0;

 flag |= CLONE_NEWUSER;
 flag |= CLONE_NEWNS;
 flag |= CLONE_NEWIPC;
 flag |= CLONE_NEWNET;
 flag |= CLONE_NEWUTS;
 flag |= CLONE_NEWPID;

 printf("uid before unshare:%d \n", (int) getuid());
 unshare(flag);

 set_uid_map(getpid(), 0, 5000, 1);
 set_gid_map(getpid(), 0, 5000, 1);

 printf("uid after unshare:%d \n", (int) getuid());

 if ( ( master = posix_openpt(O_RDWR | O_NOCTTY) ) < 0)
      perror("Openpt Error: ");
 if ( grantpt(master) < 0 )
      perror("Grant pt Error: ");
 unlockpt(master);


return 0;
} // main

如果我删除flag |= CLONE_NEWUSER;,则没有报告错误.你能帮忙解释一下为什么会这样吗？提前致谢!

Answer 1

由于我遇到了同样的问题,我也研究了这个问题.以下是我的发现:

grantpt(3)尝试确保slave伪终端将其组设置为特殊tty组(或TTY_GROUP编译glibc时的任何内容):

static int tty_gid = -1;
if (__glibc_unlikely (tty_gid == -1))
  {
    char *grtmpbuf;
    struct group grbuf;
    size_t grbuflen = __sysconf (_SC_GETGR_R_SIZE_MAX);
    struct group *p;

    /* Get the group ID of the special `tty' group.  */
    if (grbuflen == (size_t) -1L)
      /* `sysconf' does not support _SC_GETGR_R_SIZE_MAX.
         Try a moderate value.  */
      grbuflen = 1024;
    grtmpbuf = (char *) __alloca (grbuflen);
    __getgrnam_r (TTY_GROUP, &grbuf, grtmpbuf, grbuflen, &p);
    if (p != NULL)
      tty_gid = p->gr_gid;
  }
gid_t gid = tty_gid == -1 ? __getgid () : tty_gid;

/* Make sure the group of the device is that special group.  */
if (st.st_gid != gid)
  {
    if (__chown (buf, uid, gid) < 0)
      goto helper;
  }

请参阅https://sourceware.org/git/?p=glibc.git;a=blob;f=sysdeps/unix/grantpt.c;h=c04c85d450f9296efa506121bcee022afda3e2dd;hb=HEAD#l137.

在我的系统上,该tty组是5.但是,该组未映射到您的用户命名空间,并且chown(2)由于GID 5不存在而失败.然后glibc回退到执行pt_chown帮助程序,它也失败了.我没有查看它失败原因的细节,但我认为这是因为除非您将root用户映射到用户命名空间,否则它是setuid nobody.这是strace输出,显示失败的操作:

[pid    30] chown("/dev/pts/36", 1000, 5) = -1 EINVAL (Invalid argument)

为您提供了几种解决此问题的方法:

• 映射所需的组(即tty),如果没有CAP_SYS_ADMIN打开用户命名空间的二进制文件,则可能无法映射
• 使用subuids和subgids一起newuidmap(1),并newgidmap(1)让这些群体提供(这可能工作,但我没有测试过).
• 进行更改以避免chown(2)调用失败,例如使用安装命名空间并将tty组的GID更改为/etc/groups用户的GID.
• 避免chown(2)通话,例如将st.st_gid != gid支票设为假; 这应该可以通过tty从目标安装命名空间中删除组来实现/etc/groups.当然,这可能会导致其他问题.