Re: Removing '@' ?

Yes, I expect that too.  That's where the analogy breaks down a bit.

A step without options computed from inputs is exactly like a closure.

A step with options computed from inputs transforms into a step with
additional inputs.  Within that transformation it becomes a closure.

I think we can explain that distinction as a syntactic convenience.



On Thu, Feb 18, 2016 at 4:00 PM, Henry S. Thompson <ht@inf.ed.ac.uk> wrote:
> Alex Miłowski writes:
>
>> I went there too but inputs are like function arguments and outputs
>> are like returns.
>>
>> I'll make the additional observation that our steps are more like
>> closures where we bind options to step invocations and make something
>> that operations on inputs and produces outputs.
>
> I'm with you on closures, but not on distinguishing i/o from options.  I
> expect to be able to set options from variables, whose value in turn
> comes from some pipe.
>
> ht
> --
>        Henry S. Thompson, School of Informatics, University of Edinburgh
>       10 Crichton Street, Edinburgh EH8 9AB, SCOTLAND -- (44) 131 650-4440
>                 Fax: (44) 131 650-4587, e-mail: ht@inf.ed.ac.uk
>                        URL: http://www.ltg.ed.ac.uk/~ht/
>  [mail from me _always_ has a .sig like this -- mail without it is forged spam]



-- 
--Alex Miłowski
"The excellence of grammar as a guide is proportional to the paucity of the
inflexions, i.e. to the degree of analysis effected by the language
considered."

Bertrand Russell in a footnote of Principles of Mathematics

Received on Friday, 19 February 2016 00:08:04 UTC