Hi Jay,
thanks, fixed in SVN 267.
/// Jürgen
On 05/14/2014 05:44 PM, Jay Foad wrote:
That's because of a bug in GNU APL:
x←(1 2)(3 4)
(a b)←x
a≡1 2
0
:-(
Jay.
On 14 May 2014 15:24, Blake McBride <blake1...@gmail.com> wrote:
Your unbox doesn't work. The following does:
(s r)←⊃x ⋄ z←(⊃s)⍴⊃r
On Wed, May 14, 2014 at 3:43 AM, Jay Foad <jay.f...@gmail.com> wrote:
On 13 May 2014 15:00, Blake McBride <blake1...@gmail.com> wrote:
Here are the functions, examples to follow:
∇box[⎕]∇
[0] z←box x
[1] z←⊂(⊂⍴x),⊂,x
∇unbox[⎕]∇
[0] z←unbox x
[1] z←(⊃x[⎕IO])⍴⊃(x←⊃x)[⎕IO+1]
FYI you can write your box as: z←⊂(⍴x)(,x)
and unbox as: (s r)←⊃x ⋄ z←s⍴r
Jay.
