Re: [Bug-apl] box and unbox that work uniformly and without exceptions

[Blake McBride]

That fixed it for me.  Thanks!

On Wed, May 14, 2014 at 1:14 PM, Juergen Sauermann <address@hidden> wrote:

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 <address@hidden> 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 <address@hidden> wrote:

On 13 May 2014 15:00, Blake McBride <address@hidden> 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.