I find ~> not visually distinct enough (otherwise, it might be my first choice). [Update: Alternately, in some fonts it doesn't even look like an arrow and so reads more like "approximately greater than". The combination of the two possible presentations makes it a particularly bad choice, IMHO.]

The name "safe de-ref" means more to me than what is being proposed here. What is being proposed is merely the more specific "undef-safe de-ref". As such, the closest representation of what it does is the &&-> choice (a syntax which I'm pretty sure I've seen in some other language, though I don't recall which).

However, it isn't actually &&-> so I'd vote for &-> because it is that much less ugly and is mnemonic for "slightly different from &&->". In case some don't quite follow... If we had an operator that was to "&&" as the new "//" is to "||", then that operator would be the ideal one to paste in front of "->" to get this new "undef-safe de-ref" operator. But we don't. Now, "&&->" naturally implies "de-ref if true" which is only slightly different from "de-ref if defined". So "&->" is about as close as we can get to "de-ref if defined" by virtue of being "slightly different from 'de-ref if true'". But we do have "&" so you could interpret "&->" as "bit-wise 'and' then de-ref" but that just makes no sense.

So my choice would be for "&->". But part of my reason for that choice (despite it being somewhat ugly) is because I would also like to have something worthy of being called "safe de-ref".

I'd like to also have "->?" added to Perl 5. Since "safe de-ref" is too vague, we should call "->?" the "de-ref if possible" operator.

So $obj->?blurg(@args); would be roughly equivalent to $obj->blurg(@args) if $obj->?can("blurg"); (note how I had to use "->?" in order to safely call can() on something that might be neither an object nor a class).

But my hopes for "->?" are even better realized if, in the process of extending Perl 5's de-ref operators, we also add the long-ago-proposed de-ref operators of "->@" and "->%". Because "is this a hash reference?" is a complete mess in Perl 5.

It started with ref which had the poor design of not thinking blessed hash refs are hash refs. Then we added ->isa("HASH") which was more accurate in several ways but rather sucked because you can't safely call ->isa(...) on some random scalar. Then chromatic's fetish for forcing everybody else to do things his way so he wouldn't have to make a minor improvement to his used-only-in-testing module got pushed as a moral stance against making isa actually convenient as well as useful so that problem just gets worse. So I've resorted to things like eval { $ref->{''} } but that gets unacceptably complicated because you might have a version of Perl that issues warnings (not suppressed by eval) when you do something that it thinks might be an attempt to use the deprecated "pseudo-hash" feature.

So $ref->?% or even defined $ref->?% would be a nice replacement for all of the prior "is a HASH?" techniques. It'd even work for blessed references to scalars that overload hash de-ref'ing.

And even if we don't get ->%, I'd still be happy to have $ref->?isa("HASH") which I don't believe would violate chromatic's edicts about how I should be allowed to use isa() (but would require overload to properly either fudge @ISA or override isa() when an object overloads a data de-ref operation if we want ->?isa to be completely accurate).

And, of course, you could do things like:

$opt = $ref->?{optName} // $ref->?[optNumb()] // ref->?();
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for when you don't really care which succeeded or why and you just want the result (or undef).

Thanks much to Tanktalus for helping me with these ideas.

Update: I'd also be happy with just implementing ->? (but my guess is that many will want "de-ref if defined" w/o having to also silently ignore the extra "de-ref isn't possible" cases). But I'd be quite sad if ->? were used to implement just "de-ref if defined".