Re: [PATCH 03/10] tcp: Partially unify IPv4 and IPv6 paths in tcp_hash_match()

[Stefano Brivio <sbrivio@redhat.com>]

On Fri,  4 Nov 2022 19:43:26 +1100
David Gibson <david@gibson.dropbear.id.au> wrote:

> When given an IPv4 address tcp_hash_match() checks if the connection
> stores an IPv4-mapped IPv6 address, and if so compares the mapped part of
> that address to the given address.  This is equivalent to converting a
> given IPv4 address to an IPv4-mapped IPv6 address then comparing it to the
> connection address using the existing IPv6 logic.
> 
> Convert to this slightly simpler form, which will also allow some further
> simplifications in future.
> 
> Signed-off-by: David Gibson <david@gibson.dropbear.id.au>
> ---
>  tcp.c | 13 +++++++------
>  1 file changed, 7 insertions(+), 6 deletions(-)
> 
> diff --git a/tcp.c b/tcp.c
> index 26dd268..508d6e9 100644
> --- a/tcp.c
> +++ b/tcp.c
> @@ -1285,13 +1285,14 @@ static int tcp_opt_get(const char *opts, size_t len, uint8_t type_find,
>  static int tcp_hash_match(const struct tcp_conn *conn, int af, const void *addr,
>  			  in_port_t tap_port, in_port_t sock_port)
>  {
> -	if (af == AF_INET && CONN_V4(conn)			&&
> -	    !memcmp(&conn->a.a4.a, addr, sizeof(conn->a.a4.a))	&&
> -	    conn->tap_port == tap_port && conn->sock_port == sock_port)
> -		return 1;
> +	struct in6_addr v4mapped;
> +
> +	if (af == AF_INET) {
> +		encode_ip4mapped_ip6(&v4mapped, addr);
> +		addr = &v4mapped;
> +	}

It's probably worth the simplification, just mind that this replaces a
32-bit comparison with 128 bits of writes.

>  
> -	if (af == AF_INET6					&&
> -	    IN6_ARE_ADDR_EQUAL(&conn->a.a6, addr)		&&
> +	if (IN6_ARE_ADDR_EQUAL(&conn->a.a6, addr)		&&
>  	    conn->tap_port == tap_port && conn->sock_port == sock_port)
>  		return 1;
>  

-- 
Stefano