This now works:
var x = 2;
var y = (x + 1 as k) + k;
println$ y;
var a = (x + 1 as var q) + q;
println$ a;
var b = (var x + 1) + 2;
println$ b;
var c = (var b + 1);
++b;
println$ c,b;
The "as k" construction creates a new value which can be used in the expression
(before or after definition!).
var y = (x + 1 as k) + k;
is equivalent to:
val k = x + 1;
var y = k + k;
"as var" is the same except it creates a variable.
Writing just:
(var b + 1)
create an anonymous variable.
The primary intent of the latter construction is to ensure eager evaluation.
--
john skaller
skal...@users.sourceforge.net
http://felix-lang.org
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