Messages in this thread | | | Date | Thu, 25 Aug 2005 11:14:03 +0200 (CEST) | From | moreau francis <> | Subject | Re: question on memory barrier |
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--- Andy Isaacson <adi@hexapodia.org> a écrit :
> > The first register write will be completed before the second register > write because you use writel, which is defined to have the semantics you > want. (It uses a platform-specific method to guarantee this, possibly > "volatile" or "asm("eieio")" or whatever method your platform requires.)
I'm compiling Linux kernel for a MIPS32 cpu. On my platform, writel seems expand to:
static inline writel(u32 val, volatile void __iomem *mem) { volatile u32 *__mem; u32 __val;
__mem = (void *)((unsigned long)(mem)); __val = val;
*__mem = __val; }
I don't see the magic in it since "volatile" keyword do not handle memory ordering constraints...Linus wrote on volatile keyword, see http://www.ussg.iu.edu/hypermail/linux/kernel/0401.0/1387.html
> > The sequence points, by themselves, do not make any requirements on the > externally visible behavior of the program. Nor does the fact that > there's an inline function involved. It's just the writel() that > contains the magic to force in-order execution. > > You might benefit by running your source code through gcc -E and seeing > what the writel() expands to. (I do something like "rm drivers/mydev.o; > make V=1" and then copy-n-paste the gcc line, replacing the "-c -o mydev.o" > options with -E.)
make drivers/mydev.i should do the job but preprocessor doesn't expand inline functions. So I won't be able to see the expanded writel function.
> > The sequence point argument is obviously wrong, BTW - if it were the > case that a mere sequence point required the compiler to have completed > all externally-visible side effects, then almost every optimization that > gcc does with -O2 would be impossible. CSE, loop splitting, etc. > > -andy >
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