Hi Derek,
On 2017-08-13 17:33, Derek Holzer via Pd-list wrote:
Dear list,
I am transcribing some code from C to Pd expressions, and I am
wondering how I might best get the tri function working since it's not
native to expr.
My favourite triangle function is more expensive (division, branching)
but has a variable shape:
[expr~ if ($v1 < $v2, $v1 / $v2, (1 - $v1) / (1 - $v2))]
all of $v1, $v2, output are in [0..1), with $v2=0.5 for symmetrical
triangle
I have been using this expression as my triangle function:
(min($v1, 1- $v1) - 0.25) * 4
So for this example:
wave_x = .5-triangle((phasor-9/16)/(2/16))*.5;
here you have 0.5 - (... * 0.5)
I have tried the following, which doesn't function correctly compared
with the original C:
[phasor~ 200]
|
[expr~ ($v1-9./16.)/(2./16.)]
|
[expr~ (0.5 - (min($v1, 1.- $v1) - 0.25) * 4.) * 0.5]
here you have (0.5 - ...) * 0.5, which isn't the same
|
[s~ wave_x]
I am quite sure I have made some errors in the syntax somewhere, and I
did trial-and-error many other combinations before arriving at this
one. Would someone more math-literate than me care to help work this
out?
SIDENOTE: when is it necessary to force expr~ to consider a number as
a float by using a decimal?
(9/16) doesn't work but (9./16.) does.
I think int/int -> int with truncation is the most common occasion where
it matters in practice (except possibly overflow/wrapping with large
ints?). If at least one argument is a float it should promote the other
to float too and give a float result.
((min($v1, 1- $v1) - 0.25) * 4) works regardless of whether the 1 and
the 4 have a decimal point following them or not.
$v1 is a float so the 1 in 1-$v1 is promoted from int to float.
min(float, float) is a float so the 4 is promoted to float too.
Thanks much!
Derek
hth,
Claude
--
https://mathr.co.uk
_______________________________________________
[email protected] mailing list
UNSUBSCRIBE and account-management ->
https://lists.puredata.info/listinfo/pd-list
