That doens't quite explain it. It the first loop, it's as if the implicit local $& is outside of the loop — It prints 'bar' on the 'baz' pass of the loop — while in the second loop, it's as is the implicit local $& is inside of the loop.

Could
print "$&\n" unless /bar/;
suffer from a problem similar to the one from which
my ... if ...;
suffers?

In reply to Re^3: The value of pattern match variable $& by ikegami
in thread The value of pattern match variable $& by jesuashok

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