TCP hashing function

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

Re: TCP hashing function

[Neil Spring]

On Thu, Mar 28, 2002 at 08:11:27PM -0500, Marek Zawadzki wrote:
> Hello,

Hello.   Please leave the mailing list you're posting to
in the To: field of your message.

> 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 you're a student, you should probably try to figure
this out for yourself; it's the only way to learn.  

> [...]
> 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!
> [...]

This comment makes an assumption about the architecture
of the machine, and the way IP addresses are stored
into __u32s.

Consider the following code fragment:
  printf("%x %x\n",
         inet_addr("1.0.0.0"),
         inet_addr("0.0.0.1"));

-neil

Re: TCP hashing function

[andrew may]

On Fri, Mar 29, 2002 at 12:08:18AM -0800, Neil Spring wrote:
> On Thu, Mar 28, 2002 at 08:11:27PM -0500, Marek Zawadzki wrote:
> > Hello,
> 
> >  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 you're a student, you should probably try to figure
> this out for yourself; it's the only way to learn.  

Add one to the homework on the list tally.

> > 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!
> > [...]

Look at the size and see why it doesn't matter.