Return Isn't Transparent

[Neal Gafter]

One more thing.  This example is indeed "transparent" because it behaves the
same as the non-lambda version:

       private static int m6() {
               if ( true ) { return 42; }
               if ( !true ) { return 41; }
       }

Cheers,
Neal

On Tue, Jul 20, 2010 at 5:58 PM, Howard Lovatt <howard.lovatt at gmail.com>wrote:

> @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.
>