Hi Jean-Louis,
Pattern matching is done at compile time, not run-time.
So if you call `lut(2)`, you will get `3` as an output, but if you give
it a variable input like in your example, it will use `lut(n)`.
Hope that helps.
Feel free to ask for clarification if needed!
Cheers,
Bart.
---
On 2022-01-13 15:50, Jean-Louis Paquelin 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
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