[EMAIL PROTECTED]
Dear Olivier,
On 08.06.2006, at 10:24, Olivier Tonglet wrote:
My favorite syntax would be the following. Yet, this is no legal Oz
record syntax: a record label can not be a function (yet?).
defun Foo(X Y#42 z:Z#bla)
[X Y Z]
end
apply Foo(hi z:there) end %% -> [hi 42 there]
you can define your macro like that:
defun FunName( Args ) end
with Args being a list when calling the macro, as a varying number
of arguments is not accepted.
I hope it's what you would like to have ? Maybe I didn't understood ?
>I would like to have a record instead of a list (see above), because
>that allows for accessing arguments by their position or feature
name.
You can also use records if you like.
>How would you write a function application example with keyword
>arguments with your proposal?
I don't see what you mean. Is it passing keywords as arguments to a
macro call ? It's sad but you cannot pass keywords to my macro because
Args as to be a valid List (or Record).
I Guess I didnt answered your question with this macro :(
defun FunName( Args ) end
If you want keywords as arguments the number of arguments has to be
fixed... (there will be a meta-variable corresponding to each argument
in the macro definition, which will become like that :
fmacro defun FunName(1:Arg1 2:Arg2 3:Arg3) end
)
Sorry if I misunderstood something. The number of arguments for
functions etc is fixed. Therefore, I proposed to wrap all arguments in
some data structure (quasi a lambda list) so I can have optional and
keyword arguments.
In my first post I proposed a few ways to do so using either a record
(where I could use record features to denote my arguments by position
or by some feature) or a list (by introducing quasi lambda-list
keywords to denote, e.g., keywords for the arguments).
The last proposal then used records in a way which I would find the
most convenient. Here, Foo(X Y#42 z:Z#bla) is a record with Foo (a
function) as label. This is not allowed in Oz (yet? see
http://www.mozart-oz.org/lists/oz-users/4403.html), but perhaps it
would already be possible in the macro facility, because at the time
the macro does its processing it is not known that Foo is a function...
defun Foo(X Y#42 z:Z#bla)
[X Y Z]
end
apply Foo(hi z:there) end %% -> [hi 42 there]
Thanks!
Best,
Torsten
--
Torsten Anders
Sonic Arts Research Centre • Queen's University Belfast
Frankstr. 49 • D-50996 Köln
Tel: +49-221-3980750
www.torsten-anders.de
strasheela.sourceforge.net
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