Re: question on memory barrier

["linux-os \(Dick Johnson\)" <linux-os@analogic.com>]

On Wed, 24 Aug 2005, moreau francis wrote:

>
> --- "linux-os (Dick Johnson)" <linux-os@analogic.com> a écrit :
>
>>
>> On Wed, 24 Aug 2005, moreau francis wrote:
>>
>>> Hi,
>>>
>>> I'm currently trying to write a USB driver for Linux. The device must be
>>> configured by writing some values into the same register but I want to be
>>> sure that the writing order is respected by either the compiler and the
>> cpu.
>>>
>>> For example, here is a bit of driver's code:
>>>
>>> """
>>> #include <asm/io.h>
>>>
>>> static inline void dev_out(u32 *reg, u32 value)
>>> {
>>>        writel(value, regs);
>>> }
>>>
>>> void config_dev(void)
>>> {
>>>        dev_out(reg_a, 0x0); /* first io */
>>>        dev_out(reg_a, 0xA); /* second io */
>>> }
>>>
>>
>> This should be fine. The effect of the first bit of code
>> plus all side-effects (if any) should be complete at the
>> first effective sequence-point (;) but you need to
>
> sorry but I'm not sure to understand you...Do you mean that the first write
> into reg_a pointer will be completed before the second write because they're
> separated by a (;) ?

Yes. The compiler must make sure that every effect of all previous
code and all side-effects are complete at a "sequence-point". There
are several sequence-points and the most obvious is a ";".

The compiler is free to reorder anything in a compound statement
as long as it complies with presidence rules, but it can't re-order
the statements themselves.

In other words:

volatile unsigned int *hardware = virtual(MY_DEVICE);

    *hardware = 1;
    *hardware = 2;
    *hardware = 4;
    *hardware = 8;

.. happens exactly as shown above. If it didn't, you couldn't
write device drivers. An example of the code above:

 	.file	"xxx.c"
 	.text
.globl foo
 	.type	foo, @function
foo:
 	pushl	%ebp
 	movl	%esp, %ebp
 	subl	$4, %esp
 	movl	$305419896, -4(%ebp)	// init the pointer
 	movl	-4(%ebp), %eax		// Get pointer
 	movl	$1, (%eax)		// *hardware = 1;
 	movl	-4(%ebp), %eax		// Get pointer
 	movl	$2, (%eax)		// *hardware = 2;
 	movl	-4(%ebp), %eax		// Get pointer
 	movl	$4, (%eax)		// *hardware = 4;
 	movl	-4(%ebp), %eax		// Get pointer
 	movl	$8, (%eax)		// *hardware = 8;
 	movl	$0, %eax
 	leave
 	ret
 	.size	foo, .-foo
 	.section	.note.GNU-stack,"",@progbits
 	.ident	"GCC: (GNU) 3.3.3 20040412 (Red Hat Linux 3.3.3-7)"

Because of the necessary 'volatile' keyword, even the pointer is
obtained over again each time. This wastes CPU cycles. One
could do:

 	movl	$1, (%eax)
 	movl	$2, (%eax)
 	movl	$4, (%eax)
 	movl	$8, (%eax)

.. after the pointer is loaded. Unfortunately, 'C' doesn't have
a keyword that would accommodate that.

When communicating across a PCI/Bus, the writes are cached.
They don't necessarily occur __now__. It will take a (perhaps
dummy) read of the PCI/Bus to make them get to the hardware
now. Even then, they will still get there in order, all 4 writes,
because the interface is a FIFO. A read on the PCI/Bus will force
all pending writes to be written).

Some arcitectures do write-combining which means that, for instance
on an Intel Pentium P6, it's possible for all writes of adjacent
words may be condensed into a single quadword write. This may not be
what you want. If you have such a situation, one would execute
WBINVD after the critical writes. No not do this just to "make sure".
It wastes a lot of bandwidth.

See http://developer.intel.com/design/PentiumII/applnots/24442201.pdf



> Or because writes are encapsulated inside an inline function, therefore
> compiler
> must execute every single writes before returning from the inline function ?
>

In-line doesn't care. It's not as complicated as many expect.

> Thanks.
>
>            Francis
>

Cheers,
Dick Johnson
Penguin : Linux version 2.6.12.5 on an i686 machine (5537.79 BogoMips).
Warning : 98.36% of all statistics are fiction.

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