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Re: return() in pointy blocks

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Ingo Blechschmidt
June 7, 2005 10:11
Re: return() in pointy blocks
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Matt Fowles wrote:
> On 6/7/05, Ingo Blechschmidt <> 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.


> 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! :)


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