AS> On Wed, Apr 07, 1999 at 05:59:04PM +0200, Gabriel Paubert wrote:
>>
>>
>> On Tue, 6 Apr 1999, David Miller wrote:
>>
>> > Date: Wed, 7 Apr 1999 00:49:18 +0200 (CEST)
>> > From: Ingo Molnar <mingo@chiara.csoma.elte.hu>
>> >
>> > It should be 'inode >> 8' (which is done by the log2
>> > solution). Unless i'm misunderstanding something.
>> >
>> > Consider that:
>> >
>> > (((unsigned long) inode) >> (sizeof(struct inode) & ~ (sizeof(struct inode) - 1)))
>> >
>> > sort of approximates this and avoids the funny looking log2 macro. :-)
>>
>> May I disagree ? Compute this expression in the case sizeof(struct inode)
>> is a large power of 2. Say 0x100, the shift count becomes (0x100 & ~0xff),
>> or 0x100. Shifts by amounts larger than or equal to the word size are
>> undefined in C AFAIR (and in practice on most architectures which take
>> the shift count modulo some power of 2).
>>
AS> typo there, I guess. the >> should be an integer division. Since the divisor is
AS> a constant power of 2, the compiler will optimize it into a shift.
Actually I believe:
#define DIVISOR(x) (x & ~((x >> 1) | ~(x >> 1)))
(((unsigned long) inode) / DIVISOR(sizeof(struct inode)))
Is the magic formula.
A smart compiler can figure out the shift, and the DIVISOR macro
makes x into a power of two.
Eric
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