Hi list,
I have an abstraction named overlap. It takes an argument, such as
first or second.
This abstraction outputs a list. I would like this list to contain the
argument. I tried with $0 from inside the abstraction to no avail. It
looks like $0 is available only to send and receive ?
Maybe
The arguments start from $1. $0 is a unique identifier per instance
of an abstraction.
The $1,...,$N strings work differently in messages and objects. $0
does nothing inside messages.
Just create an object [$1], send it a bang, and it dumps out the 1st
argument of the abstraction it's in. I
[list append] works with floats and symbols.
++
Jack
Le 15/10/2012 17:39, Charles Henry a écrit :
The arguments start from $1. $0 is a unique identifier per instance
of an abstraction.
The $1,...,$N strings work differently in messages and objects. $0
does nothing inside messages.
Just
Hi,
Charles Henry wrote:
The $1,...,$N strings work differently in messages and objects.
Ok, I think that was a missing concept in my understanding of pd.
How about [symbol $1]?
Works !
Salut Charles,
Salut Jack :),
Are you looking for something like that ? (See attached).
Yes, works
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On 2012-10-15 17:39, Charles Henry wrote:
Just create an object [$1], send it a bang, and it dumps out the
1st argument of the abstraction it's in.
this really only works for floats.
if the user calls [myabs moses], [$1] will evaluate to [moses]
