Re: [stack] What does "concatenative" actually mean?

>
> I never claimed anyone ever claimed Joy was lazy. My claim was
> essentially that, if Joy were purely functional in the same sense
> Haskell was, you *could* evaluate it lazily if you wanted to. Given
> that you cannot do so, I don't think it is correct to call it "purely
> functional". The reason for this is that it does not have the property
> that you can replace an expression with the result of evaluating it as
> Haskell does (i.e. referential transparency). Such a property is what
> enables sensible lazy evaluation.
>

i think referential transparency doesn't mean that.
it means that a function should always return the same output given  
the same input.
in that sense, joy is referentially transparent.

and joy could be lazy (just as enchilada)
> - John
>
- robbert