[PD] Findings regarding performance
matju at artengine.ca
Thu Dec 1 23:02:35 CET 2011
Le 2011-12-01 à 10:39:00, Charles Henry a écrit :
> When using [*~ 0], the inlet and outlet are borrowed. The scalar
> multiply operation is performed in place and no data transfer occurs.
What do you call « data transfer » ? multiplying in place by a constant
involves as many reads and writes as doing a (single) copy. This at least
needs to stream data from the highest-speed RAM to the CPU and back. It's
less noticeable than the copy time of very large buffers (e.g. [table] or
[pix_separator]) because those really need big RAM (which is slower), but
in any case, calling scalartimes_perf8 (or whatever) means an implicit
copy in some kind of way, just like nearly anything else does.
| Mathieu BOUCHARD ----- téléphone : +1.514.383.3801 ----- Montréal, QC
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