- From: Pavel Rappo <pavel.rappo@gmail.com>
- Date: Thu, 28 Aug 2014 16:25:30 +0100
- To: ietf-http-wg@w3.org

Hi everyone, I have several questions regarding Integer Representation, if you don't mind. (1) First of all, the section "6.1. Integer Representation" contains a link to Wikipedia's VLQ article. The article states that: "...The VLQ octets are arranged most significant first in a stream..." (Big Endian from the byte stream point of view). On the other hand (from what I understand), encoding/decoding algorithms described in the HPACK draft works exactly the opposite -- they assume Little Endian: decode I from the next N bits if I < 2^N - 1, return I else M = 0 repeat B = next octet I = I + (B & 127) * 2^M (*) M = M + 7 while B & 128 == 128 return I In other words there is: I = I + (B & 127) * 2^M instead of I = I* 2^M + (B & 127). Either there's an error in the draft or I'm not getting it right. So which one is it? (2) The second question is this. Does the decoding algorithm "forget" to add (2^N - 1) to the result in the end? ... while B & 128 == 128 return I + 2^N - 1 though encoding one "remembers" to do this: if I < 2^N - 1, encode I on N bits else (*) encode (2^N - 1) on N bits I = I - (2^N - 1) while I >= 128 encode (I % 128 + 128) on 8 bits I = I / 128 encode I on 8 bits (3) And the third question is a minor one. When decoding algorithm mentions "I", which "I" exactly is that? decode I from the next N bits if I < 2^N - 1, return I (*) ... If I'm not mistaken, we don't know it in that point. The whole purpose of this code snippet is to get the "I". -Pavel

Received on Thursday, 28 August 2014 15:26:16 UTC