On Fri, Jun 05, 2020 at 05:55:26PM +0200, Igor Raits wrote:
> On Fri, 2020-06-05 at 15:11 +0100, Richard W.M. Jones wrote:
> > I'm unclear: that ~50M is still in RAM?  Or it's compressed on a disk
> > somewhere?
> 
> IIUC, it takes some part of the RAM and just compresses it on the fly.
> For example, I have 32G memory on my laptop and when I enable zram
> device, I basically have 23G listed in free as memory and 11G as the
> swap. And that 11G is supposed to be compressed memory (in avg cases
> with 2:1 ration).
> 
> Of course, as part of this change, it will be made that it can't take
> more than 4G no matter what by default.

Thanks - the fact that the primary swap partition now lives in RAM is
what I was missing.

But let's say we also add a lower priority disk swap, then my next
question ...

> > Also does the swap partition on disk contain compressed pages, or
> > uncompressed pages, or a mix of both?
> 
> With zram there is no partition on disk, or was this question about
> something else?

... means: Does this secondary swap partition (on disk) contain
swapped out zram pages?  Or uncompressed pages?  (Or maybe the
question just makes no sense, I don't really know.)

> > Also what is the compression algorithm?  zlib or zstd or something
> > else?
> 
> zramctl shows ALGORITHM
> 
> NAME       ALGORITHM DISKSIZE DATA COMPR TOTAL STREAMS MOUNTPOINT
> /dev/zram0 lzo-rle      11.7G   4K   74B   12K       8 [SWAP]
> 
> So it is lzo-rle by default, but it should be possible to override this
> algorithm. There is an RFE for this already at zram-generator github.

Interesting, thanks.  It looks like zstd is an alternative, although I
don't know which will be better/faster.

Rich.

-- 
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