Re: Unknown level of hash

Andrew Pimlott wrote:

>>>(You should be able to write the first one as
>>>
>>>   ${fold_left { \${$_[0]}->{$_[1]} } \$hash, @a} = 1;
>>>
>>>but Perl complains for no reason I can see.)
>>
>>	Of course, the '{}' operator is a hash ref generator, not an 
>>	anonymous sub declarator as 'sub {}' is...
> 
> 
> But I declared fold_left with a prototype (&@), which should allow
> dropping the "sub".  (And yes, in my actual program, I put this
> declaration above the use, so the prototype is visible.)  It works for
> the "get the value back out" example, but somehow it doesn't seem to
> work inside the reference (${...}).  perlref doesn't say there are any
> restrictions on what you can put inside ${...}, so I'm mystified.

	Ok, mea culpa, I haven't seen the prototype... And in fact I didn't 
know this behavior :-) Thank you for teaching it to me...!
-- 
    \^/
  -/ O \----------------------------------------
| |/ \|       Alexandre (Midnite) Jousset      |
  -|___|----------------------------------------