open()不设置O_CLOEXEC标志

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;
} 

当在具有内核版本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

决定暴露O_CLOEXEC标志fcntl(fd, F_GETFL)是安全漏洞.在内核3.6-rc7中通过此提交进行了更改:

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,你不应该依赖O_CLOEXEC于首先可见.