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Re: [Scheme-reports] Proposal to add fexprs

John Cowan <cowan@x> wrote:

> Vassil Nikolov scripsit:
> >   But I think the latter is (also) because with `apply', the arguments
> >   to the function being applied have already been evaluated;
> Just so.  So when you call (apply fexpr a b), the variables a and b are
> evaluated to S-expressions, and then the fexpr may or may not decide to
> evaluate the S-expressions further in terms of some reified
> environment.

  And b had better evaluate to a list of
  S-expressions that can be evaluated by
  the fexpr.  But in any case, unlike (f a b),
  (apply f x) can work either for a function
  f or a fexpr f, but cannot be "agnostic" to
  whether it is one or the other case.
  (Apparently, the author(s) of Kernel
  simply decided to prohibit the latter

  I am not advocating fexprs at all, just
  pointing out various aspects of having
  them in a language.


Would you like your metaphors shaken or stirred?

Vassil Nikolov | Васил Николов | <vnikolov@x>

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