Return Isn't Transparent

[Howard Lovatt]

@Neal,

Building on Reinier suggestion:

	private static int m6() {
		final { boolean ==> void } return42 = { boolean b ==> if ( b ) {
return 42; } };
		final { boolean ==> void } returnNot41 = { boolean b ==> if ( !b ) {
return 41; } };
		return42.invoke( true );
		returnNot41.invoke( true );
	}

Presumably for difficult examples like this you would have no
expectation of them working, which was my point that in general you
can't have return transparency because analyzing the expressions is
too difficult for the compiler.

 -- Howrad.

On 21 July 2010 02:06, Neal Gafter <neal at gafter.com> wrote:
> On Tue, Jul 20, 2010 at 8:55 AM, Reinier Zwitserloot
> <reinier at zwitserloot.com> wrote:
>>
>> What happens when the closure isn't ===> Nothing, but ===> Void, e.g. to
>> fit the parameter type of a bggaBasedForEach() method?
>
> Invoking something that is declared to return Void is considered by the
> compiler to be capable of completing normally.  That's what you want for a
> for-each loop: a loop can complete normally even if its iterations can't,
> because it may execute zero iterations.
>
> -Neal



-- 
  -- Howard.