C - open() 不设置O_CLOEXEC标志

Question

我尝试使用open（）设置O_CLOEXEC标志，但没有成功。

考虑以下微观测试：

#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 = %dn", fd);
  printf("ret = %x, O_CLOEXEC = %xn", ret, O_CLOEXEC);
  return 0;
} 

当在内核版本为2.6的linux上运行时，测试成功并打印出"OK！"，但在3.8或3.9内核上失败。

怎么了？谢谢

Answer 1

决定将O_CLOEXEC标志暴露给fcntl(fd, F_GETFL)是安全泄漏。此提交在内核3.6-rc7:中进行了更改

commit c6f3d81115989e274c42a852222b80d2e14ced6f
Author: Al Viro <[e-mail]>
Date:   Sun Aug 26 11:01:04 2012 -0400
don't leak O_CLOEXEC into ->f_flags
Signed-off-by: Al Viro <[e-mail]>

换句话说，您不应该首先依赖于O_CLOEXEC是可见的。

Answer 2

fcntl调用参数F_GETFD，标志为FD_CLOEXEC，O_CLOEXEC支持出现在2.6中。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) {
...
}