why * is not like unquote (scheme)
kaz at ashi.footprints.net
Thu Mar 20 17:52:12 CET 2003
"Amir Hadar" <amir at clockwise3d.com> wrote in message news:<b5c8uj$s78$1 at news2.netvision.net.il>...
> I checked a simpler case:
> and this is also syntax error.
> but I can path parameters to a function as follow:
> def f(a,b):
> print a,b
> l = ("a","b")
> So why doesn't it work in all cases?
This syntactic sugar is nothing like the splicing unquote ,@ unquote
in Lisp or Scheme. In a parameter list, it's more like the consing dot
(or the &rest parameter in Common Lisp).
In a function call, it's like apply.
(defun f (a &rest b) def f(a, *b)
(apply 'g 3 list) g(3, *list)
Why it doesn't work in all cases is because * is a syntactic sugar
which has an entirely context-dependent meaning. It means whatever the
parser wants it to mean in the context of a given grammar production.
Function application, and trailing parameters, have little to do with
the backquote syntax and unquote. The backquote is a convenient
notation for constructing lists.
In Lisp if you wanted to (ab)use unquote to do the job of apply, you
would also have to use eval:
(eval `(my-function , at my-list))
Each time this is evaluated, it constructs a new variation of the
source code to a function call, and then hands that source code to
eval. I hope you don't do this in your Scheme or Lisp programming
unless there is an excellent reason for it! ;)
Unlike the backquote approach, the apply function does not generate a
new function call with varying parameters by rewriting source code; it
just does whatever it takes to split a list into individual parameters
to be passed to the function. This has nothing to do at all with
splicing extra arguments into the function call *syntax* itself.
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