Hi,

Matt Fowles wrote:
>  On 6/7/05, Ingo Blechschmidt <iblech@[...].de> wrote:
> >   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.  If bar had read
>  
>  sub bar(Code $moo) {
>      $moo(13);
>      save_the_world();
>  }
>  
>  it would not have gotten to save the world.  One might argue that $moo
>  could throw an exception, but bar has a way to catch that.

yep. $moo(13) will never return.

But this is not specific to pointy blocks: Consider
  bar &return;

>  It seems to me that what you are asking for has the potential to cause
>  some vary large unexpected jumps down the stack.

Yep.

>  so maybe this is just one of those things that one has to be ware of.

I think the reponsibility is at the user using &return or other
evil Codes (like, as in the example, ->  $val { return $val }), not the
innocent subroutine programmer (&bar)) -- if you play with
continuations, you know what might happen. But they can be very useful,
too! :)


--Ingo

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