From: Will Deacon

Sent: 28 March 2018 09:54

...

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I don't think so. My reading of memory-barriers.txt says that writeX =

might

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expand to outX, and outX is not ordered with respect to other types o=

f

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memory.

Ugh ?

My understanding of HW at least is the exact opposite. outX is *more*
ordered if anything, than any other accessors. IO space is completely
synchronous, non posted and ordered afaik.

=20
I'm just going by memory-barriers.txt:
=20
=20
 (*) inX(), outX():
=20
     [...]
=20
     They are guaranteed to be fully ordered with respect to each other.
=20
     They are not guaranteed to be fully ordered with respect to other ty=

pes of

     memory and I/O operation.

A long time ago there was a document from Intel that said that inb/outb wer=
en't
necessarily synchronised wrt memory accesses.
(Might be P-pro era).
However no processors actually behaved that way and more recent docs
say that inb/outb are fully ordered.

	David