Re: [Bug-apl] defining new operator

[Fausto Saporito]

Hi Jürgen,

thanks... my fault. I wrote without space after the right parenthesis and the interpret gave me an error. I.e. ∇z←(F scan)x;y

I didn't notice the blank space was mandatory.

regards,

Fausto

2015-04-14 12:58 GMT+02:00 Juergen Sauermann <address@hidden>:

Hi Fausto,

page 30 (Defined Functions and Operators) explains it.
In your example below F is expected to be a function because it
is inside () in the header while the variable(s) are outside ().

/// Jürgen

On 04/14/2015 12:42 PM, Fausto Saporito wrote:

Hello all,

sorry if I bother you again, but I tried to find some hints in the APL2 Language Reference Manual without luck.

In the Sullivan's paper, there's the reference to a "scan" operator quite fast more suited to be used with his multi precision package.
This is its definition:

∇ z←(F scan)x;y

z←⊂y←⊃x
∆1:!(0=⍴x←1"x)/0
z,←⊂y←y F⊃x 

!∆1 

the "!" is the branch arrow.

Now the problem is with GNU APL I cannot define this operator, cause I don't know how to specify F is a function not a variable.

is there a way to do that ? 

thanks,

fausto