Context is important here.
In
lfold=: {{ ]F..(u~) }}
The u for lfold is the v for F..
The J parser is not going to interpret u or v according to some
referenced definition -- it's going to interpret u or v according to
the definition it's currently interpreting.
The same holds for x and y.
Consider this example:
thing1=: {{ y thing2 x }}
thing2=: ,&<
thing1/i.3
+-----+-+
|+-+-+|0|
||2|1|| |
|+-+-+| |
+-----+-+
thing2/i.3
+-+-----+
|0|+-+-+|
| ||1|2||
| |+-+-+|
+-+-----+
I hope this makes sense...
--
Raul
On Tue, Dec 6, 2022 at 5:24 PM Brian Schott <[email protected]> wrote:
>
> Don't you mean v instead of u.
> And even so, that may be the backward stuff that was discussed for a while
> in the early Fold stages?
>
> On Tue, Dec 6, 2022 at 5:21 PM Raul Miller <[email protected]> wrote:
>
> > You should probably spend a few minutes studying the diagram at
> >
> > https://code.jsoftware.com/wiki/Vocabulary/fcap#How_should_I_define_u_and_v.3F
> >
> > The short form, however, might be summarized as:
> >
> > lfold=: {{ ]F..(u~) }}
> > rfold=: {{ ]F.:(u~) }}
> >
> > I hope this makes sense,
> >
> > --
> > Raul
> >
> >
> >
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
