What I've heard is that the 64-bit instruction set has wider bit fields for specifying registers, so that you can have many more of them. (The 386 had two or three I think; the 64 bit machines have dozens, depending how you count.) So one saves steps reading and writing to/from memory.
OTOH, since all pointers have to be 64 bits, one uses more memory as a whole, perhaps by a factor of 1.5 or so - I don't see why, given that memory is "the main bottleneck" most of the time, this could possibly be consistent with 64-bit architectures being faster. So basically I don't understand what's really going on.
cheers Miller
On Mon, Feb 02, 2015 at 04:25:18PM +0000, Jonathan Wilkes wrote:
Hi Miller,What do you think is causing that performance increase on the version of Pd that is compiled for the 64-bit architecture? -Jonathan