Just so you get it:
fun f x:int) (y:int) => x + y;
var y = 100;
var x = 1;
var g = f x;
++x;
var k = g y;
println$ k;
~/felix>build/release/host/bin/flx --test=build/release ab
102
See? x is a val, f was inlined.
Now
fun f (var x:int) (y:int) => x + y;
var y = 100;
var x = 1;
var g = f x;
++x;
var k = g y;
println$ k;
~/felix>build/release/host/bin/flx --test=build/release ab
101
See? var forces eager evaluation.
So I think you're very confused, because g is a closure you expect eager
evaluation in both cases. What you get: the argument *to* the
closure, y, is eagerly evaluated. See this:
fun f (x:int) (y:int) () => x + y;
var y = 100;
var x = 1;
var g = f x;
++x;
var k = g y;
y = y + y;
println$ k ();
~/felix>build/release/host/bin/flx --test=build/release ab
102
See? It uses the value of y at the time the closure is applied.
A function*creating* a closure doesn't
necessarily use eager evaluation (unless the parameter is a var).
The closure itself does, however (at present).
Its very simple. When you inline a function with a var parameter
the code starts of with
var parameter = argument;
That's it. With a val, it MIGHT replace the parameter with the argument
in the body of the inlined code. In effect it says
val parameter = argument;
and you know that the word "parameter" in the body might
be replaced by the expression "argument" (lazy evaluation).
For non-inlined code we have ordinary C++ function or methods
so the evaluation is eager because it is in C/C++.
--
john skaller
skal...@users.sourceforge.net
http://felix-lang.org
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