Kas,
1 => foo;
this is sane, but this isn't:
1 => foo => bar;
foo(1) is the same as 1 => foo
foo => bar is the same as bar(foo)
however, I was suggesting that I could assign the object/function foo to
reference bar this way:
foo @=> bar;
which means this:
1 => bar;
would now return 3 because bar is now a reference to the same function foo
is a reference to.
see what I mean?
additionally, given
fun fun int bing(int b)
{
fun void funk(int i) { return b+i; }
return funk;
}
this:
<<< 4 => bing(5) >>>;
should print "9" in my weird world. I'm not sure that this actually is
going to work this way, i'm just suggesting that we could get a lot of
interesting things if functions were objects. I might be lynched for saying
this, but JavaScript works this way. It's a lot easier with a dynamically
typed language in which variables are always just references to other things
(like they are in Chuck with objects, but not with primitives).
I don't think we're going to get away with this without adding some sort of
keyword to Chuck. "function" is already taken as a keyword, according to
the documentation, even though nothing using it has been implemented as far
as I know. Perhaps we can add a function type that extends object.
-Mike
2010/1/6 Kassen
Mike;
What about this:
Well;
fun int foo( int arg) { return 3; }
fun void bar( int arg) { <<<"boring">>>; }
//overload fun void bar ( fun arg) { <<<"fun">>>; }
1 => foo => bar;
Would that print "fun" or "boring"? That last line is a case where we refer to the function without parentheses yet still expect it to be called and to return under current syntax.
Yours, Kas.
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