Re: Implicit outputs

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Norman Walsh writes:

> In other words, does this paragraph:
>
>   Additionally, if a compound step has no declared outputs and the
>   last step in its subpipeline has an unbound primary output, then an
>   implicit primary output port will be added to the compound step (and
>   consequently the last step's primary output will be bound to it).
>   This implicit output port has no name. It inherits the sequence
>   property of the port bound to it.
>
> apply to p:declare-step?
>
> We don't say it doesn't, but I'm not sure it should.

Hmm, seems to me it would surprise people if it didn't.  What downside
are you worrying about?

ht
- -- 
       Henry S. Thompson, School of Informatics, University of Edinburgh
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Received on Tuesday, 23 June 2009 18:02:37 UTC