[PD] Findings regarding performance

[Mathieu Bouchard]

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.

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