ERRATA: ...Boolean and a scalar. (erase "and a scalar".)

Paul Gauthier
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Paul Gauthier/NJ/CheckFree
03/14/2007 03:23 PM

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Re: [Jgeneral] definition of a:





the reason I used 42=$0 is twofold:
1) It highlights that the argument of weather to use $0 or 0$0 was a no 
man's land debate.
2) It refers to the movie "hitchhiker's guide to the galaxy" where as 
everyone knows, the meaning of life is 42.

As a footnote, I remember early discussions of "jot" being the enclose of 
the empty vector as in: <i.0 
As I see now, the only difference is that a: is actually Boolean and a 
scalar.

I was thinking that Roger could explain the evolution from "jot" to the 
actual a: and make comments if required.
But I suppose I was ignored and misinterpreted (as is usually the case).
My only excuse is to be from a different cultural background...


Paul Gauthier
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Dan Bron <[EMAIL PROTECTED]> 
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03/14/2007 03:01 PM
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Re: [Jgeneral] definition of a:






Why is everyone quoting  3!:3  all of a sudden? 

What is  3!:3]42=$0  supposed to demonstrate?  What about  3!:3]'Q'=$0  ? 
The only interesting byte I see in either result is 2nd row, 2nd column, 
which indicates the result type is boolean, which is what you would expect 
from a logical comparison like  =  . 

By the way, in the context of the current discussion, it would be more 
useful to quote  3!:0  which gives the datatype of the argument.  All the 
rest of the output of  3!:3  is just noise to that signal.

>The result's domain as a side effect of a loose type language seems to be 

>at the center of the argument... 

What is the argument?  That empty arrays of type A should not match empty 
arrays of type B when A isn't B?

To me, the crux of THAT argument lies in  1 = *./''  .  For the details of 
my thoughts on this, see:  
http://www.jsoftware.com/pipermail/general/2005-November/025629.html 

If the argument is whether  a:  should be documented as  <$0  or <0$0  , 
then:

                    datatype $0
                 integer
 
                    datatype 0$0
                 boolean
 
                    datatype >a:
                 boolean
 
 
But, since  ($0)-:0$0  (and even  ($0) -:&{. 0$0), REGARDLESS of the 
reasons, then  <$0   is not wrong; so what matters it? 

-Dan
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