thanks jurgen, i havent had time to pull latest and compile, but i
appreciate the fix.
- Rowan
On Tue, Apr 7, 2020, 5:58 AM Blake McBride wrote:
> Thanks. I saw that after I sent the email. When I used APL, there was no
> APL2. So, I've never used APL2 professionally.
>
> I also see that the
Thanks. I saw that after I sent the email. When I used APL, there was no
APL2. So, I've never used APL2 professionally.
I also see that the latest repo version has a fix for the original report.
Thanks!
Blake
On Tue, Apr 7, 2020 at 4:37 AM Dr. Jürgen Sauermann <
mail@jürgen-sauermann.de> wr
Hi,
actually it assigns a new value (6 7 in the example) to a part
of x.
Even though ⊃⊃x is
a temporarym
the assignment changes the original x.
I suppose IBM APL2 does the same.
BR, Jürgen
On 4/7/20 1:28 AM,
Without assigning it to a variable, I don't see what sense the
statement even makes.
(⊃⊃x)
produces an intermediate value. I can't think of what sense making an
assignment to an intermediate value even does.
I think APL should just throw a syntax error as in:
(+/10 10⍴⍳100)[2]←4
SYNTAX
hello y'all, hope everyone is staying safe.
x ← 1 (2 3 (4 5))
(⊃⊃x)[2;3;] ← 6 7
===
SEGMENTATION FAULT
-- Stack trace at main.cc:88
0x7F032D7BABBB __libc
