Side effect of passing matched values by reference

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

Perhaps this is obvious, but the behavior of this short code snippet confused me (actually it was buried in more complicated code that wasn't working for me), until I realized the potential indirect consequences of passing by reference... I pass it along for interest since at least for me the behavior was non-obvious...

my $bar = "abc";
foo($bar);
$bar =~ /(.*)/;
my $val = $1;
foo($val);
foo($1);
print "|$1|$bar|\n";

sub foo
{
    print "$_[0]\n";
    $_[0] =~ /(.)/;
    print "$_[0]\n\n";
}
[download]

The result is:

abc
abc

abc
abc

abc
a

|abc|abc|
[download]

Comment on Side effect of passing matched values by reference Select or Download Code

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Re: Side effect of passing matched values by reference by choroba (Cardinal) on Feb 21, 2013 at 17:22 UTC
That is not a consequence of passing by reference, in fact. It is a consequence of aliasing. See perlsub for details. This would not happen if you use a different variable than @_: `sub foo { my $x = $_[0]; print "$x\n"; $x =~ /(.)/; print "$x\n\n"; }` [download] لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ	[reply] [d/l]
Re^2: Side effect of passing matched values by reference by puterboy (Scribe) on Feb 21, 2013 at 18:23 UTC
Of course - that was the solution I used. My point was just that it wasn't obvious (at least to me) that $1 acts like a global variable in that way...	[reply]
Re: Side effect of passing matched values by reference by LanX (Saint) on Feb 21, 2013 at 17:47 UTC
Please ... $1 is a global variable which should only be carefully used in local context. But your match within the sub changes this global variable and your alias in $_[0] reflects this. And after leaving the sub the implicit localization restores the old value of $1... Better stop messing with perl internals or just pass a copy of $1 or follow chorobas suggestion. Cheers Rolf	[reply]