Getting name of a lexical variable without PadWalker?

LanX has asked for the wisdom of the Perl Monks concerning the following question:

it's possible to use PadWalker::var_name( LEVEL, VAR_REF ) to get the original name of a reference in the callers scope.

Question: can I use the B back-end to achieve the same by calling methods with pure Perl (aka without dependency to a° XS module)?

ehm ... turns out the monastery is my virtual rubber duck

B::Xref is in pure Perl and is capable to inspect the lexicals of a sub. So the answer should be yes, provided I'm able to compare the refs too.

Cheers Rolf
_{(addicted to the Perl Programming Language :)

Wikisyntax for the Monastery}

°) non-core

Comment on Getting name of a lexical variable without PadWalker?

Replies are listed 'Best First'.
Re: Getting name of a lexical variable without PadWalker? by dave_the_m (Monsignor) on Jul 15, 2020 at 21:10 UTC
Erm, B::Xref is dependent on B, which is an XS module (albeit bundled with the perl distribution). Dave.	[reply]
Re^2: Getting name of a lexical variable without PadWalker? by LanX (Saint) on Jul 15, 2020 at 21:38 UTC
> bundled with the perl distribution yep, correct, this I meant. hence > > without dependency to a non-core* XS module* updated Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply]
Re^2: Getting name of a lexical variable without PadWalker? by LanX (Saint) on Jul 16, 2020 at 15:40 UTC
> Erm, B::Xref is dependent on B, B::Xref itself is written in Perl, necessary methods are imported from B > which is an XS module (albeit bundled with the perl distribution). B is advertised as "Perl's Compiler Backend" and bundled in core. Hence no dependency risk, using it is far safer as recompiling an XS module. PadWalker has bugs, a Pure Perl solution could be updated more often. Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply]