Question

我尝试使用open（）设置O_CLOEXEC标志并且没有成功。

考虑以下microtest：

#include <stdio.h>
#include <fcntl.h>

int main() {
  int fd = open("test.c", O_RDONLY | O_CLOEXEC);
  int ret = fcntl(fd, F_GETFL);
  if(ret & O_CLOEXEC) {
    printf("OK!\n");
  } else {
    printf("FAIL!\n");
  }
  printf("fd = %d\n", fd);
  printf("ret = %x, O_CLOEXEC = %x\n", ret, O_CLOEXEC);
  return 0;
}

在内核版本2.6的linux上运行时，测试成功并打印“OK！”，但是使用3.8或3.9内核失败。

怎么了？谢谢！

Answer 1

决定将O_CLOEXEC标志暴露给fcntl(fd, F_GETFL)是安全漏洞。内核3.6-rc7中的this commit进行了更改：

commit c6f3d81115989e274c42a852222b80d2e14ced6f
Author: Al Viro <viro@zeniv.linux.org.uk>
Date:   Sun Aug 26 11:01:04 2012 -0400

don't leak O_CLOEXEC into ->f_flags

Signed-off-by: Al Viro <viro@zeniv.linux.org.uk>

换句话说，您不应该依赖O_CLOEXEC首先可见。

Answer 2

fcntl调用参数F_GETFD，标记为FD_CLOEXEC，O_CLOEXEC支持位于2.6。23。阅读手册：

 File descriptor flags
   The following commands manipulate the flags associated with a file descriptor.
   Currently,  only one such flag is defined: FD_CLOEXEC, the close-on-exec flag.
   If the FD_CLOEXEC bit is 0, the file descriptor will  remain  open  across  an
   execve(2), otherwise it will be closed.

   F_GETFD (void)
          Read the file descriptor flags; arg is ignored.

   F_SETFD (int)
          Set the file descriptor flags to the value specified by arg.

Answer 3

你做错了。你应该这样做：

int ret = fcntl(fd, F_GETFD);
if (ret & FD_CLOEXEC) {
...
}

open（）不设置O_CLOEXEC标志

3 个答案: