TCP hashing function

[Marek Zawadzki]

Hello,

In a transport protocol I'm implementing, I've adapted this TCP hashing
function:

 static __inline__ int dcp_hashfn(__u32 laddr, __u16 lport,
                                  __u32 faddr, __u16 fport)
 {
         int h = ((laddr ^ lport) ^ (faddr ^ fport));
         h ^= h>>16;
         h ^= h>>8;
         /* make it always < size : */
         return h & (MY_HTABLE_SIZE - 1);   /* MY_HT... = 128 */
 }

Although I am treating it as a blackbox and it works fine for me, my
professor pointed the following about this function:
[...]
If both IP addresses have the same upper 16 bits (like 155.246.10.5 and
155.246.120.30), then the 1st 4-way XOR will put 16 bits of zero in h.
Then "h ^= h>>16" will preserve the upper 16 bits as zero.  Then
"h ^= h>>8" will preserve the upper 24 bits!
[...]
Also, he says:
[...]
Having the same class B network number seems like a common case.  In
that case, I don't see much difference between doing the last two
XORs and leaving them out.
[...]

It is indeed true that in mentioned case one will end up with many zeros.
Does it mean this function is in-efficient? Having many sockets linked to
the same hash entry means long lookups... Or is there another reason for
this function being implemented this way?
What are yours opinions?

-marek