On 6/7/05, Matt Fowles <ubermatt@[...].com>  wrote:
>  On 6/7/05, Ingo Blechschmidt <iblech@[...].de> wrote:
>  > Hi,
>  >
>  >   sub foo (Code $code) {
>  >     my $return_to_caller = -> $ret { return $ret };
>  >
>  >     $code($return_to_caller);
>  >     return 23;
>  >   }
>  >
>  >   sub bar (Code $return) { $return(42) }
>  >
>  >   say foo &bar; # 42 or 23?
>  >
>  > I think it should output 42, as the return() in the pointy
>  > block $return_to_caller affects &foo, not the pointy block.
>  > To leave a pointy block, one would have to use leave(), right?
>  
>  I don't like this because the function bar is getting oddly
>  prematurely halted. 

Then let's put it this way:

   sub foo () {
       for 0..10 { 
           when 6 { return 42 }
       }
       return 26;
   }

And if that didn't do it, then let's write it equivalently as:

   sub foo () {
       &map(->  $_ { return 42 }, 0..10);
       return 26;
   }

Do you see why the return binds to the sub rather than the pointy now?

Also, we're going to be avoiding the return continuation problem with:

   sub foo() { 
       return ->  { return 42 };
   }
   
   my $code = foo();
   say "Boo!";
   $code();

Says not:

   Boo
   Boo
   Boo
   ...

But:

   Boo
   Can't return from subroutine that already returned at <eval>  line 2.

Luke