linux-kernel - Re: [PATCH] NVMe: Fix compilation on architecturs without readq/writeq

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Message-ID: <CA+55aFyxbieV2-WpFNrEtwQ6gXbW7y_X9oW5N6FcEcxmQKA38g@mail.gmail.com>
Date:	Mon, 30 Jan 2012 19:03:48 -0800
From:	Linus Torvalds <torvalds@...ux-foundation.org>
To:	Hitoshi Mitake <h.mitake@...il.com>
Cc:	Ingo Molnar <mingo@...e.hu>,
	Matthew Wilcox <matthew.r.wilcox@...el.com>,
	Roland Dreier <roland@...estorage.com>,
	Andrew Morton <akpm@...ux-foundation.org>,
	James Bottomley <James.Bottomley@...allels.com>,
	linux-kernel@...r.kernel.org, hpa@...ux.intel.com
Subject: Re: [PATCH] NVMe: Fix compilation on architecturs without readq/writeq

On Sun, Jan 29, 2012 at 12:02 AM, Hitoshi Mitake <h.mitake@...il.com> wrote:
>
> I don't know about the minor architectures, but some of them,
> like alpha, seems to do reordering of memory access agressively.
>
> Is the reordering is applied to io rw?
> Should memory barriers be placed between two readl/writel?

No need to place barriers - the "readl/writel()" functions are ordered
in themselves. There are non-ordered versions in theory
("writel_relaxed()") for things like frame buffers etc that actively
want the ordering, but that's a separate issue entirely.

You do want to make sure that they aren't in the same C expression, so
that the compiler doesn't re-order the expression. IOW, if you just do

  return (readl(addr+4) << 32) | readl(addr);

then that doesn't have any ordering at all simply because there is
none at the C level. But

  u64 val;
  val = readl(addr);
  val |= readl(addr+4) << 32;

is well-defined and must read the low word first - both at the C level
*and* at the CPU level. Anything else would be a bug in the
architecture "readl()" implementation or the hardware.

(On x86, for example, a "readl()" is just a memory access, but while
x86 can re-order reads to regular memory in hardware, that is *not*
true of IO memory accesses. On architectures like POWER, 'readl()'
implies synchronization instructions)

                   Linus
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