In hindsight, this existence-by-reference trick works for hashes too. Hence it's at least consistent.

The big question is now, if and what rationale there this.

Probably not really a problem at all.

  DB<1> %x=(a=>1)

  DB<2> $a= \$x{b}

  DB<3> p exists $x{b}
1
  DB<4> p exists $x{a}
1
  DB<5> p exists $x{c}

  DB<6>

[download]

Cheers Rolf
(addicted to the Perl Programming Language :)
see Wikisyntax for the Monastery

In reply to Re^7: Using exists to check element in array (existence-by-reference for hashes too) by LanX
in thread Using exists to check element in array by david

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