@James and @bart, thank you for your quick reply.
So (if I understand you well) pattern matching should be thought of as a
static rewriting tool.
And, btw, you're right
process = 2 : lut : hbargraph("B", 1, 5);
sets the bargraph to 2, while
process = lut(2) : hbargraph("B", 1, 5);
sets it to 3 (as expected).
I came with 2 ways to write the lut() function I need:
lut1(n) = 1,3,4,2,5 : ba.selectn(5, n - 1);
lut2(n) = ((n == 1) | (n == 5)) * n + (n == 2) * 3 + (n == 3) * 4 + (n == 4)
* 2;
process = hslider("A", 1, 1, 5, 1) <: (lut1 : hbargraph("B", 1, 5)), (lut2 :
hbargraph("C", 1, 5));
Both functions seem to provide the service I'm asking for. But the first
one is more readable than the second.
Any idea which one is the lightest for the CPU?
Thanks again,
jlp
Le jeu. 13 janv. 2022 à 16:41, James Mckernon <jmcker...@gmail.com> a
écrit :
> On 1/13/22, Jean-Louis Paquelin <jean-lo...@paquelin.net> wrote:
> > Hello there,
> >
> > I can't figure out how the pattern matching works in Faust.
> >
> > I've tried the following code:
> >
> > lut(2) = 3;
> > lut(3) = 4;
> > lut(4) = 2;
> > lut(n) = n;
> >
> > process = hslider("A", 1, 1, 5, 1) : int : lut : hbargraph("B", 1, 5);
> >
> > Trying this in the IDE, the B bargraph simply replicates the A slider
> > value. I was expecting a different behavior thru the lut function: the
> > output of lut should be the same as its input except when the input is
> 2, 3
> > or 4, in those cases the output of lut should be 3, 4, 2 respectively.
> >
> > I tried the alternative syntax without any success:
> >
> > lut = case {
> > (2) => 3;
> > (3) => 4;
> > (4) => 2;
> > (n) => n;
> > };
> >
> > It's as if none of the cases match except the most general one.
> >
> > Did I miss something?
> >
> > I wish you all the best for 2022, health, joy and great projects!
> >
> > Best regards,
> >
> > Jean-Louis
>
> Pattern matching is done at compile-time, so to match the case of
> lut(2) (for example), the argument needs to be fixed and known at
> compile-time to be 2. This is true of lut(2), even of expressions such
> as lut(1 + 1), but not taking lut as a function of some potentially
> time-varying expression.
>
> Also, I'm not 100% sure of this, but I think pattern-matching may only
> take effect when a function is applied with explicit, parenthesis
> syntax, not when it is chained with : (i.e. lut (x) rather than x :
> lut). If I'm wrong about this then perhaps someone can correct me; in
> any case I don't think this was the problem with your example.
>
> The way to do what you are attempting to do at run-time would probably
> be by chaining a number of select2 functions together. I'm not sure
> whether there are any helper functions in the standard library to help
> with this; it seems like the sort of thing that ought to be present,
> but I can't think of any off the top of my head.
>
> Hope that helps.
>
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