Side effect of passing matched values by reference

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
Re: Side effect of passing matched values by reference by choroba (Prior) 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 (Beadle) 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 (Monsignor) 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]


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Side effect of passing matched values by reference

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