Expose functionality to get peer credentials for a Unix socket on Linux

[maghoff]

Caveat: I put this rather randomly in unix/notbsd/linux/other/mod.rs. I know nothing about mips and musl, and I have no idea if this thing also applies to Android. I would guess "yes", but I don't know how to go about testing that assumption. Superficial googling seems to indicate that the constant and struct in question are present in android linux.

Further work: There is a roughly corresponding getsockopt for freebsd and osx (and possibly more bsds?) called LOCAL_PEERCRED. It would be neat to have that supported as well.

[maghoff]

Usage example:

extern crate libc;
extern crate unix_socket;

use std::mem;
use std::io::{self, Write};
use std::os::unix::io::AsRawFd;
use unix_socket::{UnixStream, UnixListener};

fn get_peer_credentials(stream: &UnixStream) -> io::Result<libc::ucred> {
    let fd = stream.as_raw_fd();

    unsafe {
        let mut credentials: libc::ucred = mem::zeroed();
        let mut size = mem::size_of::<libc::ucred>() as libc::socklen_t;
        match libc::getsockopt(
                fd,
                libc::SOL_SOCKET,
                libc::SO_PEERCRED,
                &mut credentials as *mut _ as *mut _,
                &mut size as *mut _ as *mut _
        ) {
            val if val < 0 => Err(io::Error::last_os_error()),
            _ => Ok(credentials)
        }
    }
}

fn main() {
    let listener = UnixListener::bind("socket").unwrap();
    println!("Connect to `socket` with for example: nc -U socket");

    for maybe_stream in listener.incoming() {
        let stream = maybe_stream.unwrap();
        let ucred = get_peer_credentials(&stream).unwrap();
        println!("Connected with user {} from process {}", ucred.uid, ucred.pid);
        writeln!(&stream, "Hello {}!", ucred.uid).unwrap();
    }
}

[maghoff]

To the failing AppVeyor build, this fails similarly to how it fails on master (doesn't find kernel32.lib), which probably is, and definitely should be, unrelated to this pull request :) (But you probably already realized)

[alexcrichton]

Thanks!