Keep in mind that there's no variable y in this expression.

Presumably the 13 : evaluator uses a dummy value of one when resolving
the expression, but since 'y' is a constant rather than a variable, it
does not know that you (and 5!:5) really meant it to be a reference to
the variable.

That's what we get for trying to bend the rules...

FYI,

-- 
Raul

On Tue, Jul 4, 2017 at 8:54 PM, Nikolas Drosdek
<[email protected]> wrote:
> This is exactly what I was looking for, thanks a lot!
>
> I am slightly confused by this however:
>
>    13 : '5!:5<''y'''
> (,'1')"_
>
> as it seems to replace y with 1 instead of giving 3 : '5!:5<''y'''.
>
> Thanks,
> N.D.
>
> 2017-07-05 3:28 GMT+03:00 Raul Miller <[email protected]>:
>
>> Well, that depend on what you mean by a "tacit expression".
>>
>> If your concept of tacit is one which does not use names, then no,
>> because this expression uses names.
>>
>> If you prefer a more relaxed concept of tacit programming, however,
>> you might try something like this:
>>
>>    lrep=: 3 :'5!:5<''y'''
>>    V=: [: ". 'R=:R,' , lrep
>>
>> Thanks,
>>
>> --
>> Raul
>>
>>
>>
>> On Tue, Jul 4, 2017 at 8:14 PM, Nikolas Drosdek
>> <[email protected]> wrote:
>> > Explicitly:
>> >
>> > R=: 0
>> > V=: 3 : 'R=: R , ...'
>> >
>> > Is there a way to write V in tacit form?
>> >
>> > Thanks,
>> > N.D.
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