I can reproduce this, but before I go into it, I want to clear something
up. You say 'delimiter d', but parse2 is an adverb, and so you are
executing
(d parse2);.1
and d is not a delimiter argument to (parse2;.1) .
But I don't follow what m means in parse2, so perhaps d is a delimiter
there.
Is d correct?
Henry Rich
On 7/30/2022 2:59 PM, Raul Miller wrote:
I'm running into a problem with :.1 which, when I simplify the code, disappears.
So, here's the problem in raw form:
(TAB,' ') parseout sample1
+--+---+
|$r|1 2|
+--+---+
+---+-+
|+/d|2|
+---+-+
|assertion failure
| (#r)=+/d
parseout=: {{
d=. (=<./)y (1 i.~ -.@e.)S:0 m
r=.d parse2;.1 y
if. (#r) ~:+/d do.
echo '$r';$r
echo '+/d';+/d
assert. (#r)=+/d
end.
}}
parse2=: {{
({.y);<m parseout^:(*@#)}.y
}}
sample1=: cutLF {{)n
zeta
beta
gamma
lambda
kappa
mu
delta
alpha
theta
iota
epsilon
}}
In other words, I expect parse2 to return a list of two boxes, and I
expect that the result of parseout would have one of these pairs for
each 1 in the delimiter sequence 'd'.
But that's not happening here.
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