Re: [stack] function/object ambiguity + quotation alternative

On Feb 26, 2009, at 12:45 PM, John Nowak wrote:

>
> On Feb 26, 2009, at 12:22 PM, Stevan Apter wrote:
>
>>>  2. [dup] == [dup] first      (removing shared 'i' suffix)
>>
>> no.  dup = [dup] first.
>
> You're missing the point. I know that 'dup = [dup] first', but I can
> prove '[dup] == [dup] first'!

To be more precise, I know '[dup] first' evaluates to the function  
object 'dup'. I assume this is what you mean.

That said, the function 'dup' does *not* equal the function '[dup]  
first'. Obviously '2 dup' yields a different answer than '2 [dup]  
first'. This is the initial problem I brought up and it may affect XY;  
I know it affects Joy and Factor. The ability to prove '[dup] == [dup]  
first' is an additional issue that is XY-specific.

- John