Subject: [rohc] Explanation on F,H and P bits

From: [email-masked]
Date: Tue, 28 May 2002 18:00:28 -0500

Can anyone explain how these bits are used for Input instructions? 
Say, if the INPUT_BITS instruction is encountered and the current bits in the dispatcher are 11000011 and the instruction requested 3 bits.

Case P=0, 
     resultant Integer = 6 (bits 110).
    How does F bit affect the way this integer is stored at the destination address? My understanding is that ,
    case F=0,
                 the integer stored at destination address = (6)110
    case F=1,
                the integer stored at destination address = (3) 011

Case P=1,
    resultant integer = 3(011)
    Case F=0,
              the integer stored at destination address = 3 (011)
    Case F=1,
              the integer stored at destination address = 6 (110).

  Please correct me if my interpretation is wrong?

thanks,
Narendra

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From: "West, Mark (ITN)" <[email-masked]>
Date: Thu, 30 May 2002 13:37:31 +0100

Hi Narenda,

My understanding (and I'm willing to be corrected by a sigcomp author, 
if one wishes to stick his head above the parapet...) is as follows.

(Cribbing heavily from -06 in the process)

"The P-bit controls the order in which bits are passed from the
    dispatcher to the INPUT instructions. If set to 0, it indicates that
    the bits within an individual byte are passed to the INPUT
    instructions in MSB to LSB order. If it is set to 1, the bits are
    passed in LSB to MSB order."

So, given 11000011, then the P bit means that reading a sequence of 3 
bits will give:

P = 0, gives the sequence 1, 1, 0 (reading the byte 'left to right')
P = 1, gives the sequence 1, 1, 0 (reading the byte 'right to left')

So, in this case, the string of bits produced is the same, because your 
input byte is palindromic.

Then:

"If the F-bit is set to 0, the INPUT-BITS instruction interprets the
    received bits as arriving MSBs first, and if it is set to 1 it
    interprets the bits as arriving LSBs first. "

Given the above example, where the string of bits is 1, 1, 0 in any 
case, then the F-bit will mean:
F = 0, the bits are seen as running MSB -> LSB giving 110
F = 1, the bits are seen as running LSB -> MSB giving 011

Does that help?

Cheers,

Mark.

[email-masked] wrote:

 > Can anyone explain how these bits are used for Input instructions?  Say,
 >  if the INPUT_BITS instruction is encountered and the current bits
 > in the dispatcher are 11000011 and the instruction requested 3 bits.
 >
 > Case P=0,  resultant Integer = 6 (bits 110). How does F bit affect
 > the way this integer is stored at the destination address? My
 > understanding is that , case F=0, the integer stored at destination
 >  address = (6)110 case F=1, the integer stored at destination
 > address = (3) 011
 >
 > Case P=1, resultant integer = 3(011) Case F=0, the integer stored at
 >  destination address = 3 (011) Case F=1, the integer stored at
 > destination address = 6 (110).
 >
 > thanks, Narendra
 > _______________________________________________ Rohc mailing list 
[email-masked]

-- 
Mark A. West, Consultant Engineer
Roke Manor Research Ltd., Romsey, Hants.  SO51 0ZN
Phone +44 (0)1794 833311   Fax  +44 (0)1794 833433

(Yes, I do know that my disclaimer is in an attachment.  And, no, I
didn't ask for it to be that way)