On 31/01/2012 19:04, Mathieu Bouchard wrote:
Le 2012-01-31 à 18:33:00, Lorenzo Sutton a écrit :
On 31/01/2012 16:07, Jonghyun Kim wrote:
expr $f1; if ($f1 == 1, 1, 0); if ($f1 == 5, 2, 0); if ($f1 == 9, 3, 0); if ($f1 == 13, 4, 0); if ($f1 == 17, 5, 0); if ($f1 == 21, 6, 0); if ($f1 == 25, 7, 0); if ($f1 == 29, 8, 0); if ($f1 == 33, 9, 0); if ($f1 == 37, 10, 0); if ($f1 == 41, 11, 0); if ($f1 == 45, 12, 0);
You could also use a [select] instead of [expr]. Something like | [sel 1 5 9 ... ] | | | ... | [1( [2( [3( ... [t b] | [0(
Now that I think of it, with GridFlow (any version), you have this :
[listfind 1 5 9 13 17 21 25 29 33 37 41 45] | [+ 1]
To get a single index from 1 to 12, or 0 if not found.
If you really need to send a zero for each time something is not found, however, you can do this :
[#outer == (1 5 9 13 17 21 25 29 33 37 41 45)] | [# * (1 2 3 4 5 6 7 8 9 10 11 12)] | [#unpack 12] ||||||||||||
But because of the pattern of numbers involved (equally spaced), you could use [mod 4] and [div 4] to separate $f1 into two parts, one part that should be 1, and the other that should be between 1 and 12. That's quite a shortcut.
Ah true! I hadn't seen the pattern initially (I saw prime numbers for some reason...) So maybe one could simply use two counters with a mod 4 'driving' the second one... No? I mean:
| [f 0] X [+ 1] | [mod 4] | [sel 1] | [f 1] X [+ 1]