[phobos] byte alignment for arrays

[Steve Schveighoffer]

Thanks, this information helps a lot!

I will make the change to 16-byte aligned.  I'm already using 8 bytes for a 4 byte length.  Using 16 bytes isn't much different, especially when the block size is 4096+ bytes.

One final question -- I currently use sizeof(size_t) * 2, which could now be sizeof(size_t) * 4, but of course, this changes to 32 bytes on 64-bit dmd.  Would it make sense to just use 16 instead of some multiple of size_t?

-Steve


----- Original Message ----
> From: Jason Spencer <spencer8 at sbcglobal.net>
> To: Discuss the phobos library for D <phobos at puremagic.com>
> Sent: Mon, June 28, 2010 4:09:01 PM
> Subject: Re: [phobos] byte alignment for arrays
> 
> Sorry, I forgot to address the every-other-one concern.

The MMX registers 
> are 64-bits, so you can only do 1 double at a time.  Those instructions 
> only require 8-byte aligned memory.  The SSE instructions use 128-bit 
> registers, so they take 2 doubles at a time.  As long as the first one is 
> 16-byte aligned, you can iterate through on 16-byte (128 bits) chunks, and 
> you'll be good.  That's why element 0 should be 128-aligned.

If it's 
> not, the processor will either have an alignment fault (in the instruction 
> requires alignment) or will do a bunch of split-loads across cache lines, which 
> kill performance.  

One other thought:  If you wanted to be 
> tricky, you could do a general, 4-byte allocation and based on the address you 
> get, assign your storage pointer to the next 128-aligned address.  But 
> you're offloading to run-time lot's of housekeeping.  Again, maybe 
> tolerable for just these large arrays.  But it starts to add a lot of 
> corner cases.  Walter might have some good suggestions 
> here.

Jason




----- Original Message ----
> From: 
> Steve Schveighoffer <

> A question then --  let's say you have 
> an array of 
> doubles, which are 8 bytes wide, and you want to 
> 
> use these SSE instructions.  Even if the first one is aligned on a 16-byte 
> 
> boundary, wouldn't every other double be 
> 
> misaligned?
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