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Re: Is it possible to get all variable names which are declared? (updated)

by haukex (Abbot)
on Nov 09, 2017 at 17:20 UTC ( #1203072=note: print w/replies, xml ) Need Help??

in reply to Is it possible to get all variable names which are declared?

The fact that you want to do this means that you are either writing a debugger, or you are looking at the problem from the wrong angle (e.g. "it's stupid to use a variable as a variable name", and you should use a hash instead). If you could tell us why you want to do this, we can almost certainly provide better solutions.

PadWalker will provide you with enough rope to shoot yourself in the foot:

use warnings; use strict; use PadWalker qw/peek_my/; use Data::Dump; my $A; my $B = 0; dd peek_my(0); for my $i ( $B .. 3 ){ my $C = 3; dd peek_my(0); } __END__ { "\$A" => \undef, "\$B" => \0 } { "\$A" => \undef, "\$B" => \0, "\$C" => \3, "\$i" => \0 } { "\$A" => \undef, "\$B" => \0, "\$C" => \3, "\$i" => \1 } { "\$A" => \undef, "\$B" => \0, "\$C" => \3, "\$i" => \2 } { "\$A" => \undef, "\$B" => \0, "\$C" => \3, "\$i" => \3 }

Update: That only dumps my variables. See also the module's peek_our function, or for inspecting the Symbol Tables of packages, Devel::Symdump or my simplistic replacement here (only does scalars, arrays and hashes). When dumping symbol tables, note the caveat described in the Devel::Symdump documentation: "scalar symbol table entries are a special case. If a symbol table entry exists at all, presence of a scalar is currently unknowable, due to a feature of Perl described in "Making References" in perlref point 7. For example, this package will mark a scalar value $foo as present if any of @foo, %foo, &foo etc. have been declared or used."

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[LanX]: met a guy from Aberdeen durin holidays, liked Scotsish accent a lot ... very phonetic
[LanX]: ... understood everything after mentally write down what he said.
[choroba]: Trainspotting?
[choroba]: Ouch, trailing commas in JSON :-(
[LanX]: ... though there seems to exist a taxation for the letter 't' in Northern Bri''ain ;-)

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