# Re: HPACK, Draft 09, Integer Representation

```See:
https://github.com/http2/http2-spec/issues/595

I tend to think we need some text to replace it... any volunteers?

On 29 Aug 2014, at 2:12 am, Tatsuhiro Tsujikawa <tatsuhiro.t@gmail.com> wrote:

>
>
>
> On Fri, Aug 29, 2014 at 1:00 AM, Pavel Rappo <pavel.rappo@gmail.com> wrote:
> > HPACK uses little endian.
>
> I think this difference should be mentioned explicitly.
>
>
> FYI, wikipedia reference was deleted. ​
> ​https://github.com/http2/http2-spec/commit/bd8d7688a9dd8708bae543a9abe889ea7797e139
>
> ​
> > "I" is updated while loop:
> >
> >        repeat
> >            B = next octet
> >            I = I + (B & 127) * 2^M
> >
> >                      ~~~~~~~~~~~~~~~~~~~~~
> >            M = M + 7
> >        while B & 128 == 128
> >
> > So, initial "I" is not forgotten.
>
> Thanks, Tatsuhiro! It's indeed updated. I overlooked it.
>
> > "I" comes from next N bits as algorithm exactly says:
> >
> >
> > decode I from the next N bits
> >
> > HPACK integer encoding uses specific prefix bits, which is described as N
> > here in algorithm.
> > So when decoding integer encoded with 7 prefix bits, decode initial "I" from
> > next 7 bits.
>
> The only thing then, the meaning of "I" is different for the encoding
> and decoding parts. Isn't it?
> "The algorithm to represent an integer I is as follows:" and "For
> informational purpose, the algorithm to decode an integer I is as
> follows:". i.e. same "I" different meanings.
>
>
> ​If integer X is encoded to some byte string B and then B is decoded to integer Y, then X should be equal to Y, so in that sense, both "I" mean same integer.
> For decoding algorithm, "I" in the sentence you quoted means the result of the algorithm, and not just the first N bits.
>
> ​Best regards,
> Tatsuhiro Tsujikawa​
>
>
>
>

--
Mark Nottingham   https://www.mnot.net/
```

Received on Friday, 29 August 2014 00:47:37 UTC