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.

Something else to consider: "distribution" (or lack thereof) of the binding operator. Note that $_ =~ (/A/ && /B/) is not equivalent to ($_ =~ /A/) && ($_ =~ /B/). The latter is what you would want.

Regarding the original poster's code:

$count++ if $_ =~ /\b$dst[0]\b/ && /\b$service[
+0]\b/;
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This works only because $_ is being tested. Taking precedence into consideration (and the fact that m// works on $_ by default), this code is equivalent to

($_ =~ /\b$dst[0]\b/) && ($_ =~ /\b$service[+0]\b/);
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If the data were in any other variable ($var, for example), and one just replaced $_ with $var in the OP's code, it would be equivalent to

($var =~ /\b$dst[0]\b/) && ($_ =~ /\b$service[+0]\b/);
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which isn't What You Want.

Just my observations. HTH.

--

There are 10 kinds of people -- those that understand binary, and those that don't.

Operator && is logical and, not bitwise, which is '&'. Logical && is the correct choice for what you're doing. See perlop for details, but it acts [almost] like the C operator of the same name. The left argument is evaluated. If false, the right argument is ignored and the whole && expression returns false. If the left evaluates true, the right hand argument is evaluated and that result returned.

Your code is correct as far as perl goes, though the '$_ =~' is unneeded - $_ is the default argument for m//. You will get some surprises, however, when $service[0] is numeric and $_ contains a dotted quad ip matching the service you're looking for. A single regex containing more context is in order, or else perhaps a split to a predictable array.

Update: Doh! dws is on the money with the operator precedence issue.

After Compline,
Zaxo