Re: Why do I (sometimes) get a REF ref and not a SCALAR ref? (silly)

Yeah, you can explain this behavior. But it is, frankly, stupid. \\$x is, indeed, a reference to a scalar and so ref should return "SCALAR". Having a special case for "reference to a scalar that happens to currently be holding a reference" is silly. It just leads to code having to check for both "SCALAR" and "REF", adding complexity. It makes about as much sense as having ref(\\\$x) return "REFREF" or ref(\\@x) return "ARRAYREF" or ref([]) return "EMPTYARRAY".

- tye

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Re^2: Why do I (sometimes) get a REF ref and not a SCALAR ref? by Anonymous Monk on Jun 23, 2016 at 19:53 UTC
The "ARRAY" references can be dereferenced via `@{ $ref }`. The "HASH" references can be dereferenced via `%{ $ref }`. The "REF" references can be dereferenced via `${ $ref }`. Silly or not, but a self-circular reference can be followed ad infinitum; you never arrive at plain "SCALAR" value.	[reply] [d/l] [select]
Re^3: Why do I (sometimes) get a REF ref and not a SCALAR ref? (cycles) by tye (Sage) on Jun 23, 2016 at 20:01 UTC
Circularity has nothing to do with whether "REF" or "SCALAR" is returned. "SCALAR" references can be dereferenced via `${ $ref }`, exactly like "REF" references, which argues for not making that distinction. Code that is going to traverse from $sv to $$sv is often also code that will traverse from checking `ref $av` to checking `ref $av->[0]` and from checking `ref $hv` to checking `ref $hv->{$key}` and so can cycle infinitely for any of those cases. - tye	[reply] [d/l] [select]
Re^3: Why do I (sometimes) get a REF ref and not a SCALAR ref? by Anonymous Monk on Jun 23, 2016 at 20:03 UTC
Uh, addendum. (Getting confused myself?) I should've said: Both "SCALAR" and "REF" can be dereferenced via `${ $ref }`, but only the "SCALAR" can be used as a honest scalar via that deref.	[reply] [d/l]


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