Re: Is 1111 1110 10 equal to 0xfe80 or 0x3fa?

["Bless, Roland (TM)" <[e-mail]>]

Hi Alex,

Am 07.06.19 um 07:50 schrieb Alexandre Petrescu:
> Le 06/06/2019 à 23:41, Fred Baker a écrit :
>>
>>
>>> On Jun 6, 2019, at 10:18 AM, Alexandre Petrescu
>>> <[e-mail]> wrote:
>>>
>>> I maintain that 1111 1110 10 equals 0x3fa, because that's what my
>>> Windows Calculator says: I type 1111 1110 10 and it converts to
>>> 0x3fa.
>>
>> You're correct if those are the only bits. However, in 1111 1110 10
>> you have to also include 00 0000, because it's a /10 prefix.
> 
> ?
> 
> 1111 1110 10 are 10 bits and they are the only bits.

Gee, it's only binary arithmetic. We use a nibble-wise textual
representation with hex digits, so you have to fill it up to
a multiple of 4 bits, i.e., adding two bits to the right gives you
12 bits, which nicely converts to 0xfe8. If you want 0xfe80 you
need to add 00 0000 as Fred pointed out.

> I would indeed add the 00 0000 if they were 16bits, and then I would
> write fe80::/16.
> 
> But, since a few people say differently, I think I will need to look at
> it deeper.  MAybe it is just a matter of language expression.

Sorry, but fe80:: doesn't imply fe80::/16, that's exactly the reason
why we have the prefix notation. fe80::/10 means to expand to 128 bit
and then use the leftmost 10 bits only. That's unambiguous.

Regards
 Roland