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From: PerlFAQ Server on 5 May 2010 06:00
This is an excerpt from the latest version perlfaq4.pod, which
comes with the standard Perl distribution. These postings aim to
reduce the number of repeated questions as well as allow the community
to review and update the answers. The latest version of the complete
perlfaq is at http://faq.perl.org .

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4.6: Why doesn't & work the way I want it to?

The behavior of binary arithmetic operators depends on whether they're
used on numbers or strings. The operators treat a string as a series of
bits and work with that (the string "3" is the bit pattern 00110011).
The operators work with the binary form of a number (the number 3 is
treated as the bit pattern 00000011).

So, saying "11 & 3" performs the "and" operation on numbers (yielding
3). Saying "11" & "3" performs the "and" operation on strings (yielding
"1").

Most problems with "&" and "|" arise because the programmer thinks they
have a number but really it's a string. The rest arise because the
programmer says:

if ("\020\020" & "\101\101") {
# ...
}

but a string consisting of two null bytes (the result of ""\020\020" &
"\101\101"") is not a false value in Perl. You need:

if ( ("\020\020" & "\101\101") !~ /[^\000]/) {
# ...
}



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The perlfaq-workers, a group of volunteers, maintain the perlfaq. They
are not necessarily experts in every domain where Perl might show up,
so please include as much information as possible and relevant in any
corrections. The perlfaq-workers also don't have access to every
operating system or platform, so please include relevant details for
corrections to examples that do not work on particular platforms.
Working code is greatly appreciated.

If you'd like to help maintain the perlfaq, see the details in
perlfaq.pod.
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