Does Linux automatically clean up abstract domain sockets?

I posted this question over a year ago and was never quite satisfied with the lack of definitive documentation. I thought I'd check Linux documentation again for any updates, and was happy to see this:

Abstract sockets

Socket permissions have no meaning for abstract sockets: the process umask(2) has no effect when binding an abstract socket, and changing the ownership and permissions of the object (via fchown(2) and fchmod(2)) has no effect on the accessibility of the socket.

Abstract sockets automatically disappear when all open references to the socket are closed.

Also, The Linux Programming Interface by Michael Kerrisk covers the question (cross-posted from this other answer):

57.6 The Linux Abstract Socket Namespace

The so-called abstract namespace is a Linux-specific feature that allows us to bind a UNIX domain socket to a name without that name being created in the file system. This provides a few potential advantages:

  • We don’t need to worry about possible collisions with existing names in the file system.
  • It is not necessary to unlink the socket pathname when we have finished using the socket. The abstract name is automatically removed when the socket is closed.
  • We don’t need to create a file-system pathname for the socket. This may be useful in a chroot environment, or if we don’t have write access to a file system.

To create an abstract binding, we specify the first byte of the sun_path field as a null byte (\0). [...]

I reckon that, along with @user3188445's answer, this clears the question up very precisely.

That said, there's still an assumption made here, that processes which are SIGKILL'd will have all open sockets closed. That seems a reasonable assumption, but I don't have documentation that defines that behavior.


Yes, linux automatically "cleans up" abstract sockets to the extent that cleaning up even makes sense. Here's a minimal working example with which you can verify this:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <sys/socket.h>
#include <sys/un.h>

int
main(int argc, char **argv)
{
  int s;
  struct sockaddr_un sun;

  if (argc != 2 || strlen(argv[1]) + 1 > sizeof(sun.sun_path)) {
    fprintf(stderr, "usage: %s abstract-path\n", argv[0]);
    exit(1);
  }

  s = socket(AF_UNIX, SOCK_STREAM, 0);
  if (s < 0) {
    perror("socket");
    exit(1);
  }
  memset(&sun, 0, sizeof(sun));
  sun.sun_family = AF_UNIX;
  strcpy(sun.sun_path + 1, argv[1]);
  if (bind(s, (struct sockaddr *) &sun, sizeof(sun))) {
    perror("bind");
    exit(1);
  }
  pause();
}

Run this program as ./a.out /test-socket &, then run ss -ax | grep test-socket, and you will see the socket in use. Then kill %./a.out, and ss -ax will show the socket is gone.

However, the reason you can't find this clean-up in any documentation is that it isn't really cleaning up in the same sense that non-abstract unix-domain sockets need cleaning up. A non-abstract socket actually allocates an inode and creates an entry in a directory, which needs to be cleaned up in the underlying file system. By contrast, think of an abstract socket more like a TCP or UDP port number. Sure, if you bind a TCP port and then exit, that TCP port will be free again. But whatever 16-bit number you used still exists abstractly and always did. The namespace of port numbers is 1-65535 and never changes or needs cleaning.

So just think of the abstract socket name like a TCP or UDP port number, just picked from a much larger set of possible port numbers that happen to look like pathnames but are not. You can't bind the same port number twice (barring SO_REUSEADDR or SO_REUSEPORT). But closing the socket (explicitly or implicitly by terminating) frees the port, with nothing left to clean up.