Re: [hybi] frame length encoding

> Option 1 is simplest and sufficient for all use cases. There is no
> advantage
> in a more complex algorithm and non-trivial costs in complexity, and
> inefficiency for frames of 125-254 bytes if we use option 2.

At least 267% incorrect.

The Option 2 inefficiency is only 3/125 to 3/254, which is 2.4% to 1.2%.

Whereas, Option 1 is 8/127 to 8/254, which is 6.3% to 3.1%.

Also Option 1 increases inefficiency by 267% from 3/255 to 8/255 and all
values between to 2/65535 to 8/65535.

Also Option 1 causes a big discontinuity in inefficiency at 126, and that
could be a source of random performance patterns, which is hard to
quantify.

Also I want to suggest that we change Option 2 to re-use the 7 bits from
the first byte instead of adding 2 bytes, then we only add 1 byte.  The
only disadvantage is the values from 32768 to 65335 which incur a 0.002%
inefficiency.  But then our efficiency for 125 to 32767 will increase by
50%!!

So I think it is a "no brainer" that we adopt my modification for Option 2:

long long len = readByte();
switch (len) {
  case 126:
    len = ((len & 0x7f) << 8) + readByte();
    break;
  case 127:
    len = readLongLong();
    break;
}

>
> Option 1 gets my vote.
>
> -Pieter Hintjens
>
> Sent from my Android mobile phone.
>
> On Aug 20, 2010 10:49 PM, "Roderick Baier"
> <roderick.baier@hs-weingarten.de>
> wrote:
>> Willy Tarreau schrieb:
>>> Hi John,
>>>
>>> On Fri, Aug 20, 2010 at 03:08:08PM -0400, John Tamplin wrote:
>>>> Option 2
>>>> ========
>>>> Use 7 bits in the frame header, and if the value is 0-125, that is the
>>>> length. If the value is 126, then the next 2 bytes are a 16-bit
>>>> length.
> If
>>>> the value is 127, the next 8 bytes are a 63-bit length (the high bit
> must be
>>>> zero).
>>>>
>>>> long long len = readByte();
>>>> switch (len) {
>>>> case 126:
>>>> len = readUnsignedShort();
>>>> break;
>>>> case 127:
>>>> len = readLongLong();
>>>> break;
>>>> }
>>>
>>> This looks really nice and optimal to me !
>>>
>>> +1 !
>>>
>>
>> +1
>>
>>
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