Think nothing of it. I was back in the highschool today, talking linear
algebra to the very top layer of the high-performing students, having
them write proofs on the board. It is obvious that the distinction
between "its" and "it's" is not observed in practice, even among these
kids who are tomorrow's elite. It will be gone in another 100 years.
Back when I was teaching Latin, I tried to give examples of the
difference between transitive and intransitive verbs in English. I started,
We say "I lay the book on the table: I lay it, I laid it yesterday, I
have laid it there many times."
We say "I lie down for a nap. I lie down, I lay down yesterday, I have
lain down..."
the rest of the sentence was drowned out by cries of "No!". They had
never heard such a thing. My conclusion: "lie" is dead. Write it off.
I have already written off "whom". The language evolves.
Henry Rich
On 11/20/2019 9:52 PM, Louis de Forcrand wrote:
Just to correct a mistake that I always hate making:
"... for use after _its_ application ..."
Sorry for the noise,
Louis
On 21 Nov 2019, at 03:49, Louis de Forcrand <ol...@bluewin.ch> wrote:
(a,a=.?@#) is a verb, namely (?@# , ?@#). In this expression a is set to the
_verb_ ?@# and then train (a,a) is evaluated.
In the second case a is set to the _result_ of ?@# and then (a,a) is evaluated.
To do this tacitly:
(] , ]) @ (?@#)
or more concisely
,~@?@#
or equivalently (how I would write it)
2 $ ?@#
As Henry said, to store an intermediate value in a verb's evaluation for use
_after_ it's application, you must use an explicit verb, for example:
,~ @ (3 : 'a=: y') @ (?@#)
Cheers,
Louis
On 21 Nov 2019, at 03:26, Nimp O <tr...@outlook.com> wrote:
Hello, simple question.
This behaviour surprised me.
(a,a=.?@#)'01234'
2 4
3 : 'a,a=.?@#y' '01234'
1 1
Why a is not equal to a in the first case? How can I save the roll as an
intermediate result in the tacit version?
Thanks.
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