It's not; it should give exactly the same result.
Distributing 8 u: over ::, we get (8 u: 9&u:) :: (8 u: 10&u:). The RHS of
adverse behaves the same as 8 u: 10 u: ] (from the original formulation).
The LHS decodes and then re-encodes its argument. Compare with the
original: 9&u: ] ]. This is an identity function which, as a side effect,
decodes its argument, throwing if the decode fails. But the result of the
decode is thrown away; if the input is a 1-byte char, the result will be
too.
Put another way: (9&u: ] ]) y ←→ (9 u: y) ] ]y ←→ ]y ←→ y (modulo effects).
On Tue, 22 Mar 2022, Raul Miller wrote:
On Tue, Mar 22, 2022 at 7:16 PM Elijah Stone <[email protected]> wrote:
(Alternately, slower but shorter and clearer: 8 u: (9&u:) :: (10&u:).)
And, the result is significantly more compact, which makes a
difference when working with large volumes of text.
Thanks again,
--
Raul
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