Re: [stack] are concatenative languages applicative?

On Apr 24, 2008, at 6:58 PM, Don Groves wrote:

> On Apr 24, 2008, at 3:10 PM, John Nowak wrote:
>
>> Apologies for starting a new thread of sorts, but the "Object Cat"
>> thread already had run through three or four different topics, and it
>> seems like a good idea to discuss this in such a way that it'll be
>> archived properly. I'd really like an answer here.
>>
>> First off, let me say that I'm not particularly interested in if  
>> Joy's
>> 'i' is an 'eval'. This does, as Chris pointed out, seem to get us
>> bogged down in implementation details. As far as I can tell, 'eval'  
>> is
>> defined to be some procedure that works on code, be that code a  
>> string
>> or some more structured form of data. In a language like Cat, there's
>> no such code being passed around as there is in Joy, and I don't  
>> think
>> there's anyone here who would say Cat isn't concatenative.
>>
>> What I'm interested in is this "dequotation" rule:
>>
>>   [$A] i  ==  $A
>>
>> In other words, these are all equivalent:
>>
>>   1 2 [3 * *] i  ==  1 2 3 * *  ==  1 [2 3] i [* *] i
>>
>> An important thing to note here is that this translation is just a
>> rewriting of function-level code. '$A' is a function, not a value.
>> This seems fundamentally different from this rule in the combinatory
>> calculus, where 'x' is a *value*, not a function:
>>
>>   (I x)  ==  x
>>
>> (William already has said essentially the same thing with respect to
>> operators and operands.)
>>
>> I guess my questions are as follows: Are concatenative languages
>> applicative, and if so, what's the equivalent of '[$A] i == $A' in  
>> the
>> lambda calculus?
>
> As I understand it, the lambda calculus form  \X.E  represents an
> abstraction
> (\X) and an application (E). Seems to me a straightforward reading of
> Joy's
> [foo] as an abstraction and i as being/causing an application is the
> equivalence
> you seek. Joy's i and lambda calculus' dot perform the same function
> in their
> respective languages.
>
> Am I the only one here who thinks the word "function" has too many
> meanings?
> --
> don

I didn't finish answering your question: The lambda calculus equivalent
of [$A] i is \A.S where S is the stack.
--
don




>
>
>
>> ...If such a translation cannot be given, does that mean
>> concatenative languages are non-applicative? If so, is there another
>> term besides "concatenative" that can be used to describe their
>> properties?
>>
>> - John
>
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>
>