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. -- To unsubscribe from this list: send the line "unsubscribe lartc" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
