On 01/25, Dario Sanfilippo wrote:
>
> Hello.
>
> I see that a function results in a different diagram when called with
> prefix having the generic signal symbol.
>
> For example, let's consider the following function:
>
> test(x, y) = (x , _ : -) ~ *(y);
>
> What I thought is that this would always force the feedback signal to be
> connected to the second input of the minus operator.
>
> If I call the function as
>
> process = test(1, 0);
>
> I get what I expect: 1 minus the feedback signal.
>
> Otherwise, if I call the function as
>
> process = 1 : test(_, 0);
>
> I get the feedback signal minus 1.
I almost forgot faust, but everything looks correct to me... You can't assume
that
something(1) is equivalent to (1 : something(_)).
Lets simplify your example:
test(x) = (x , _ : -) ~ _;
Now,
process = test(1)
results in
process = (1 , _ : -) ~ _;
note that in this case "(1 , _ : -)" has a single input.
while
process = 1 : test(_);
means
process = 1 : ((_ , _ : -) ~ _);
and in this case we have "(_ , _ : -)" with 2 inputs. the 1st one binds to
recursive
signal.
Oleg.
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