Your expression $target =~ /is/ in a list context will return (1) for success (see m// in perlop). Thus, your loop is equivalent to this: 
for my $is (1) {       # one iteration with $is=1
   $target =~ s//are/;
}

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I, however, am getting different results than you under 5.6: 
thare is a test of the servise

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I get the same results by dropping the loop entirely and doing just this: 
$target =~ /is/;
$target =~ s//are/;

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Like you, I find that odd. Somehow Perl is doing the equivalent of s/is/are/;.

In reply to Re: s/// operator and undef question... by Fastolfe
in thread s/// operator and undef question... by ismail

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