Ok, regardless of what is happening, merlyn explained the situation. 
$string =~ /foo/;
$string =~ //;     # this repeats the last successful match

$string =~ /foo/;
$string =~ s//bar/;  # again, /foo/ is used in place of //

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japhy -- Perl and Regex Hacker

In reply to s/// tomfoolery explained (again) by japhy
in thread s/// operator and undef question... by ismail

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