Ah, now i got it......so it can also return size, if size is directly a
multiple of ZEND_MM_ALIGNMENT.
I was convinced that it has to be stricly > than size, i guess i was wrong.
________________________________
From: Gustavo Lopes <glo...@nebm.ist.utl.pt>
To: Adi Mutu <adi_mut...@yahoo.com>; internals@lists.php.net
Sent: Thursday, March 1, 2012 6:44 PM
Subject: Re: [PHP-DEV] question about Zend MM
On Thu, 01 Mar 2012 17:22:23 +0100, Adi Mutu <adi_mut...@yahoo.com> wrote:
> I want to understand how Zend MM works, so i'm looking trought the sources
> and i see this:
>
> #define ZEND_MM_ALIGNMENT_MASK ~(ZEND_MM_ALIGNMENT-1) #define
> ZEND_MM_ALIGNED_SIZE(size) (((size) + ZEND_MM_ALIGNMENT - 1) &
> ZEND_MM_ALIGNMENT_MASK)
>
>
> I understand that the first define will create something like 11111000 ( it
> will clear last 3 bits)
> but what does the 2nd define? Before clearing the last 3 bytes why does it
> add ZEND_MM_ALIGNMENT- 1
> to size?
>
It basically just rounds to the next multiple of ZEND_MM_ALIGNMENT, assuming
ZEND_MM_ALIGNMENT is a power of 2.
ZEND_MM_ALIGNMENT is a power of 2, so it has 1 bit set. Subtracting 1 will zero
that bit and and flip on all the other less significant bits. Then negating
flips the bits so that now the bits less significant than the
log2(ZEND_MM_ALIGNMENT)-th will be zero and the others will be one.
ZEND_MM_ALIGNED_SIZE adds ZEND_MM_ALIGNMENT - 1 and applies the mask. The
effect is that the result will be >= size and it will be a multiple of
ZEND_MM_ALIGNMENT (in particular the smallest multiple of ZEND_MM_ALIGNMENT
that's >= the argument) because the bits less significant than the
log2(ZEND_MM_ALIGNMENT)-th will be zero. "a & n-1" is the same as "a mod n"
with n being a power of 2. So if a & ZEND_MM_ALIGNMENT - 1 == 0 then a mod
ZEND_MM_ALIGNMENT == 0 and a is a multiple of ZEND_MM_ALIGNMENT.
--Gustavo Lopes
