Re: IHL vs shift 6 vs accurate location in upper layers

jsl6uy js16uy wrote:
> Hello all hope all is well
> 
> I looked around and have tried to get this sorted on multiple
> occasions, but just can't get it
> how does dividing by 64/2^6, ~ shift 6, ultimately get you to
> something like the proper location for a src port in layer 4?
> I get, after reading again and again and looking at ip packet header
> diagrams, offset at 0 and the mask used to extract the IHL from the
> 16bit word.
> But then comes 'ol shift 6 ... right-shifting by 6 eliminates the
> offset of the field and at the same time converts the value into byte
> unit ....and I am lost again
> IHL # 32 bit words I guess dividing by 64 bits will get you bytes
> multiples 8 and then I have nothing
> 
> googling around verified binary math and this is dividing by 64, but
> that's where it ends, anything else is people just using it. I figure
> I must be missing the deal.
> Would like to understand it to fully use

If you read the first 16 bits and mask to leave just IHL then you get
(assuming normal IHL = 5)

0000 0101 0000 0000 this is not 5 (101 in binary) to get 5 you would
need to right shift by 8 = 0000 0000 0000 0101.

Your examples want IHL in bytes (octets) and since IHL is in units of
32 bits you need to multiply by 4 to get bytes.

left shift by 2 is x4 so instead of right shift 8, right shift 6 gives
the IHL x4 = bytes

0000 0000 0001 0100 = 20.

Either way you know the header length and for tcp or udp the next 16
bits will be the source port.