Re: What about storing more crc32 in the unused csum size for metadata?

[Duncan <1i5t5.duncan@cox.net>]

Qu Wenruo posted on Mon, 12 Jan 2015 10:32:30 +0800 as excerpted:

> [New layout]
> Take leafsize as 4K for example.
> 
> Sectors:
> |---0---|---1---|---2---|---3---|---4---|---5---|---6---|---7---|
> Sector 0: Csum of the leaf/node (32~4K)(not changed)
> Sector 1: Csum of the first eighths of the leaf/node (32~512)
> Sector 2: Csum of the second eighths of the leaf/node (512~1024)
> ....
> Sector 7: Csum of the last eighths of the leaf/node (3584 ~ 4096)

OK, I've been up too long and should be sleeping, and it may be that's 
affecting my logic, but AFAICT either you're missing something obvious, 
or I am.

There are eight eights, and eight "sectors", but the first one is already 
used, thus leaving seven available.  You have the first eighth (ending at 
byte 512*1) in sector one (zero-based so the second sector), the second 
(ending at byte 512*2=1024) in sector two... the last (eighth) eighth 
(ending at byte 512*8=4096) in sector seven.

Of eighths 3-7 in that ..., which eighth do you skip csumming, or is one 
of them a full quarter instead of an eighth?

Of course if we could make them sevenths...  But making it four quarters 
instead of 7/8 is perhaps more reasonable, with the remaining three 
sectors remaining unused, and if only one fails csum we still save 75% of 
the leaf/node, instead of none of it, currently.

But maybe it makes complete sense as-is and I should be sleeping instead 
of trying to make sense of what is clearly not making sense to me at this 
point...

-- 
Duncan - List replies preferred.   No HTML msgs.
"Every nonfree program has a lord, a master --
and if you use the program, he is your master."  Richard Stallman