Re: question about Zend MM

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 <glopes@nebm.ist.utl.pt> To: Adi Mutu <adi_mutu06@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_mutu06@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