Skip site navigation (1)Skip section navigation (2)
Date:      Thu, 12 Apr 2018 21:06:10 +0200
From:      Polytropon <freebsd@edvax.de>
To:        "Ronald F. Guilmette" <rfg@tristatelogic.com>
Cc:        freebsd-questions@freebsd.org
Subject:   Re: Two questions --- SSD block sizes and buffering
Message-ID:  <20180412210610.bafc713b.freebsd@edvax.de>
In-Reply-To: <23423.1523502187@segfault.tristatelogic.com>
References:  <23423.1523502187@segfault.tristatelogic.com>

Next in thread | Previous in thread | Raw E-Mail | Index | Archive | Help
On Wed, 11 Apr 2018 20:03:07 -0700, Ronald F. Guilmette wrote:
> 
> Two rather simple questions:
> 
> 1)  I don't know much, generally speaking, but I have certainly read that
>     the underyling hardware of flash memory products (which I guess
>     includes both USB sticks and also SSDs) all effectively have
>     "physical" block sizes on the order of 128 KiB.
>     
>     So anyway, I'm just curious to know to what extent, if any, FreeBSD,
>     when running with, on, or from any such (flash-memory based) mass
>     storage device does things specifically with these larger physical
>     block sizes (of flash memory) in mind.

No, it just works(TM). :-)



>     Does FreeBSD automagically sense that it is dealing with an SSD, and
>     does it then adjust the way it operates on the relevant filesystem(s)
>     accordingly, for maximal performance?  Or must the system administrator
>     tweek something explicitly (e.g. tunefs parameters, or perhaps newfs
>     parameters) in order to get optimal performance on/with an SSD?

The fragment size is to be applied by the system administrator
when initializing the disk, depending on the block size of the
device. For example, newfs allows you to do so:

	# newfs -m 0 -i 16384 -b 16384 -f 4069 -L somelabel	\
	  -t enable -n enable -U /dev/ada0a

Note the -f flag; from "man newfs":

     -f frag-size
             The fragment size of the file system in bytes.  It must be a
             power of two ranging in value between blocksize/8 and blocksize.
             The default is 2048 bytes.

But as mentioned in many cases: Deviating from the default
values should be done for good and educated reasons, and
depending on actual requirements.

The example above uses a 4k frag size as typically used
with newer hard disks. Additionally, slices, partitions
and labels should be adjusted to match block size multiples
for better performance.



> 2)  I have written some modest C programs that output lots and lots of
>     very short little tidbits of data, one per line.  In some cases these
>     programs output millions of lines.
> 
>     For reasons that I can explain, these programs explicitly call setvbuf()
>     on stdout during startup, and they set the buffering type for stdout
>     to _IOLBF (i.e. line buffering).
> 
>     My question is just this:  Assme that one of these programs is called
>     "xyz".  Now, if I run the program thusly:
> 
>            xyz > xyz.output
> 
>      i.e. so that stdout is redirected to a file, will there be one actual
>      write to disk for each and every line that is written to stdout by xyz?
>      In other words, will my act of explicitly setting line buffering (for
>      stdout) in a case like this cause the xyz program to beat the living
>      hell out of my disk drive?

Probably not. The actual write operationg are being issued
somewhere "down the line" through the file system driver
down to the disk driver. Even a "sync" command issued by
the OS will not _immediately_ cause the drive to act.



>      I hope not, but I'd like to know if it will, or know why it won't, if
>      it won't.

Isn't all output buffered, per default?



-- 
Polytropon
Magdeburg, Germany
Happy FreeBSD user since 4.0
Andra moi ennepe, Mousa, ...



Want to link to this message? Use this URL: <http://docs.FreeBSD.org/cgi/mid.cgi?20180412210610.bafc713b.freebsd>