Mailinglist Archive: radeonhd (427 mails)

[Alex Deucher <alexdeucher@xxxxxxxxx>]

2009/5/20 Rafał Miłecki <zajec5@xxxxxxxxx>:
> W dniu 20 maja 2009 15:45 użytkownik Alex Deucher
> <alexdeucher@xxxxxxxxx> napisał:
> > 2009/5/20 Rafał Miłecki <zajec5@xxxxxxxxx>:
> > > Then how does MOVE work?
> > > MOVE   work[00]   [..XX]  <-  reg[0180]  [..XX]
> > > Let's say I get following output:
> > > # rhd_dump -r 0180,0184 01:00.0
> > > 0x0180: 0x01234567
> > > 0x0184: 0x89ABCDE
> > > does it mean sth following:
> > > 0x0180: 0x01
> > > 0x0181: 0x23
> > > 0x0182: 0x45
> > > 0x0183: 0x67
> > > ?
> > > If so, what is [..XX] part of 0x0180? Shuld I treat 0x0180 as 0x0001
> > > and [..XX] means 0x01 then?
> >
> > Registers are 32 bits, and the atom ops
> > can operate on individual bytes or groups of bytes.  In this case the
> > op is operating on the lower two bytes (hence the ..XX).  In your
> > example the lower two bytes are 0x4567.
>
> You say registers are 32bits... that's fine, for example my 0x0185 is
> 0xAB, so 2*16bits = 32bits. Alright.
>
> But next you say two lower bytes (2B == 16b) of my example 0x0180
> (0x01234567) are 0x4567. Here I'm lost. 4*16bits = 64bits. That means
> 8B, not 2B :|
>
> Could you explain that one more please, so I could try to understand that?

0x180 is the offset of the regsiter in dwords.  0x180 * 4 = 0x600 =
offset in bytes since 1 dword = 4 bytes.  The register offset defines
the register's location relative to the start of register aperture. so
the register in question is 0x180 dwords from the start of the
aperture or 0x600 bytes.  Each register is 32 bits, so the lower two
bytes are 0x4567.

Alex
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