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