Ipv4Addr ordering and endianness

[hroi]

Sorting Ipv4Addrs results in unexpected ordering. Ordering is based on the internal representation of the Ipv4Addr without regard to "network order".

I tried this code on little-endian architecture:

use std::net::Ipv4Addr;

fn main() {
    let mut ips = vec![Ipv4Addr::new(100,64,3,1),
                       Ipv4Addr::new(100,64,3,3),
                       Ipv4Addr::new(192,0,2,2),
                       Ipv4Addr::new(192,0,2,4),];

    ips.sort();
    for ip in ips.iter() {
        println!("ip: {}, u32: {}", ip, u32::from(*ip));
    }
}

I expected to see this happen:

ip: 100.64.3.1, u32: 1681916673
ip: 100.64.3.3, u32: 1681916675
ip: 192.0.2.2, u32: 3221225986
ip: 192.0.2.4, u32: 3221225988

Instead, this happened:

ip: 100.64.3.1, u32: 1681916673
ip: 192.0.2.2, u32: 3221225986
ip: 100.64.3.3, u32: 1681916675
ip: 192.0.2.4, u32: 3221225988

Rust playground demonstration: http://is.gd/hEjytb.

[hroi]

Ipv6Addr has a similar issue:

use std::net::Ipv6Addr;
use std::str::FromStr;

fn main() {
    let mut ips = vec![Ipv6Addr::from_str("2001:db8:f00::1002").unwrap(),
                       Ipv6Addr::from_str("2001:db8:f00::1004").unwrap(),
                       Ipv6Addr::from_str("2001:db8:f00::2001").unwrap(),
                       Ipv6Addr::from_str("2001:db8:f00::2003").unwrap(),
                       Ipv6Addr::from_str("2001:db8:baa::1002").unwrap(),
                       Ipv6Addr::from_str("2001:db8:baa::1004").unwrap(),
                       Ipv6Addr::from_str("2001:db8:baa::2001").unwrap(),
                       Ipv6Addr::from_str("2001:db8:baa::2003").unwrap(),];

    ips.sort();
    for ip in ips.iter() {
        println!("ip: {}", ip);
    }
}

outputs:

ip: 2001:db8:f00::2001
ip: 2001:db8:f00::1002
ip: 2001:db8:f00::2003
ip: 2001:db8:f00::1004
ip: 2001:db8:baa::2001
ip: 2001:db8:baa::1002
ip: 2001:db8:baa::2003
ip: 2001:db8:baa::1004