Re: [PATCH] libuuid: improve uuid_unparse() performance

["Aurélien Lajoie" <orel@melix.net>]

On Wed, Mar 25, 2020 at 12:11 PM Karel Zak <kzak@redhat.com> wrote:
>
> On Tue, Mar 24, 2020 at 10:26:25PM +0100, Aurelien LAJOIE wrote:
> > There is 2 improvements:
> >
> >  * remove useless uuid_unpack,
> >  * directly print the hexa format from memory without using printf
> >    we can do this as the bytes order is the network byte order
>
> I'm not sure, but are you sure that whole UUID is in big-endian order?
> I think that last part (aka "node", 6 bytes) is not subject to swapping.
> It seems uuid_unpack() does nothing with the last part of the UUID.
>
> But your patch works on LE as well as on BE, so I probably miss
> something :-)
The RFC is quite clear on this "with each field encoded with the Most
Significant Byte first"
https://tools.ietf.org/html/rfc4122#section-4.1.2

I agree this is not clear for the node part
>From the RFC node should be an unsigned 48 bit integer

The parsing is done byte per byte
        for (i=0; i < 6; i++) {
                buf[0] = *cp++;
                buf[1] = *cp++;
                uuid.node[i] = strtoul(buf, NULL, 16);
        }
Then the sprintf is also done byte per byte.
So Big Endian and the swapping are hidden by the manipulation byte per byte
I cannot find any usage of the node field to set specific value only random.
As long the node is handled byte per byte it will work.

>
> > before took 382623 us
> > after  took  36740 us
> >
> > Big Endian OpenBSD:
> > before took 3138172 us
> > after  took  180116 us
>
> I guess all this is about sprintf(), another way would be to use
> uuid_unpack() but avoid sprintf().

Using uuid_unpack to fill a struct uuid, will trigger to handle the
endianness to print it
whereas uuid_t is matching the order needed to print it.
>
>     Karel
>
> --
>  Karel Zak  <kzak@redhat.com>
>  http://karelzak.blogspot.com
>