Consider $_ =~ /A/ && /B/; and think about whether it is equivalent to /A/ && /B/The answer depends on the precedence of =~ vs. &&
Unclear on the precedence? There's an operator precedence table in perlop.
In reply to Re: I'm confused with Bitwise Operators
by dws
in thread I'm confused with Bitwise Operators
by Anonymous Monk
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