Using standardized SI prefixes - Debian

This is a discussion on Using standardized SI prefixes - Debian ; On Friday 22 June 2007 07:29, Ivan Jager wrote: > > CD-ROMs have 2304 byte raw sectors. > > 2048 + 256 for ECC, both of which are powers of two. Even if you use the > 2304 raw bytes, ...

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Thread: Using standardized SI prefixes

  1. Re: Using standardized SI prefixes

    On Friday 22 June 2007 07:29, Ivan Jager wrote:
    > > CD-ROMs have 2304 byte raw sectors.

    >
    > 2048 + 256 for ECC, both of which are powers of two. Even if you use the
    > 2304 raw bytes, that is a multiple of 2^8 bytes, and not even divisible by
    > 10^1.


    Powers of 2 are everywhere. I have 8+2 toes, both of which are powers of two.
    How did humans even start counting in base 10 when it's obvious that there
    are 8+2 digits to count with (and that's both powers of 2). :-#

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  2. Re: Using standardized SI prefixes

    On Thu, Jun 21, 2007 at 05:29:47PM -0400, Ivan Jager wrote:
    > On Thu, 21 Jun 2007, Hamish Moffatt wrote:
    > >On Wed, Jun 20, 2007 at 08:11:23PM -0400, Ivan Jager wrote:
    > >You seem to claim that binary units (ie powers of 2) are natural
    > >everywhere related to computers, but I disagree.

    >
    > Not everywhere related to computers. Only when the unit is bytes.


    Wow, what a concession!

    > >It's natural for
    > >memory and structures like it, but not for bitstream quantities like
    > >network traffic.

    >
    > Yes, for network traffic both are just as natural.


    Except that our decimal prefixes (10^N) are part of our language and
    therefore win by default.

    > >Most NAND FLASH chips have 2062 byte
    > >blocks, which even throws the memory device argument out the window.

    >
    > I have no idea about this, but I would expect
    > http://www.google.com/search?hl=en&q...nd&btnG=Search
    > to have more results where the 2062 is a block size...


    Sorry, I meant 2112.

    > You forgot about ECC SDRAM which is 72 bits wide. So when you buy a 1GB
    > (72x128M) DIMM, you're actually getting 1207959552 bytes of raw storage.


    Actually the controllers don't memory-map the extra 8 bits per 64. The
    existence of the extra bits is totally hidden between the RAM and the
    controller.

    For NAND flash however the whole 2112 byte blocks are memory mapped.
    After every 2112 bytes there's a gap until the next 4K boundary.

    > But even then, the powers of two are more natural than the powers of 10.


    Yes for memory structures, I agree. You failed to address my point about
    bitstream quantities like network traffic.

    Hamish
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