Re: Couple of questions

[Adrian McMenamin <adrian@mcmen.demon.co.uk>]

On Fri, 2006-03-17 at 21:29 +0100, Takashi Iwai wrote:
> At Fri, 17 Mar 2006 20:27:32 +0000,
> Adrian McMenamin wrote:
> > 
> > On Fri, 2006-03-17 at 11:13 +0100, Takashi Iwai wrote:
> > 
> > > 
> > > The period size is same, but the copy is not done like LLLLRRRR.
> > > 
> > > As default, the whole buffer is divided to number of channels.  For
> > > example, using two channels,
> > > 
> > > 	|LLLLLLL..|RRRRRRR....|
> > > 	|<--- buffer_size --->|
> > > 
> > > and using six channels,
> > > 
> > > 	|1111....|2222....|.....|6666....|
> > > 	|<-------- buffer_size --------->|
> > > 
> > > where each channel has a length of (period_size * periods) samples.
> > > Unless the buffer size is changed, this assignment of each channel
> > > also isn't changed.
> > > 
> > > The above is the default behavior.  You can change the assignment as
> > > you like by defining proper copy and silence callback, too.  For
> > > mmap, you need to define CHANNEL_INFO ioctl, too.
> > > 
> > 
> > Presumably this is what happens at the start, but when doing a single
> > period update presumably it is split LLLLRRRR on non interleaved
> 
> No, as I wrote in the above, the assignment of each channel isn't
> changed, i.e. the buffer position of each channel remains.
> 
> The middle layer handles the whole buffer simply as a bundle of
> multiple mono streams.
> 

So, just to be clear. When I am transferring a "period" I should
actually stream half a period's worth of bytes into the channel playing
the left and half a period into the right - and can advance each dma
transfer through the buffer by half a period length on both the left and
the right side of the buffer



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