Because =~ is the regular expression binding operator, not string equals (which is what eq). It matches the regex on its right hand side against the scalar on the left. Since you've given it a string with regular expression metacharacters in it which doesn't match itself it fails. See perlretut and perlre.

(Once you've read that try and figure out what would happen if you changed it to $b = "\Q$a\E" instead . . .)

perl -e '$a="MIN.(NERR_VOL2)"; $b="MIN.\(NERR_VOL2\)" ; $x = ( qq/$a/ 
+=~ qq/$b/ ) ; print "$a == $b ? $x \n" ;'
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$b = "$a";
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$b = $a;
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You should never assign a variable like so:

Well, not necessarily "never". In general for 99.99999% of the cases you don't want the extra quotes; however there's times when you may want to explicitly "stringify" something (say an object with an overloaded q// method). They're few and far between, but not never ever.</pedant>

Fletch--

I'm often a pedant myself, so thanks for the correction. I don't quite follow what you mean, though. Could you provide or point me to a trivial example, perhaps?

--roboticus

BTW although the '.' meta-character doesn't hurt you in this case, it does mean you are risking a match-positive when you would expect a match negative

e.g.

$a = 'hello.world';
$b = 'helloxworld';

print "matched" if $b =~ /$a/;
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You will get a match in the above example (because '.' means any single character in a regex).

Tom Melly, tom@tomandlu.co.uk

Can I ask why you're quoting $a and $b?

I'm using variables because I'm doing a context based search for a string in a file. This string changes at different places in the file.

As you rightly pointed out.. quotemeta($a) is what I was really looking for, but missed it. Too many languages in my head.. F77,lisp,skill,c,tcl,perl,csh,ksh.......etc etc. I knew I wanted to 'quote' the contents of the variable to prevent regex interpretation, but didn't see how.

And yes, everyone is correct, that in the example I chose to show, the matching is redundant and 'eq' or index() is a better choice. But once I got stuck on the problem I wanted to understand the generalities of quoting & regex with variables for future use.

Thanks to everyone, I can see clearly now ...

Let's go into details...

qq/$a/ =~ qq/$b/ is equivalent to
"$a" =~ "$b" (because qq// and "" are the same quoting operator), and to
$a =~ "$b" (because a match already forces the stringification of its left hand side), and to
$a =~ $b (because a match already forces the stringification of its right hand side if it's not a compiled regexp), and to
$a =~ /$b/ (because strings on the right hand side of a match are treated as regexp), and to
$a =~ /MIN.(NERR_VOL2)/ (because we know the value of $b).

But remember that ., ( and ) are special characters in regexps.

If you wanted to check for equality, you want
$a =~ /^\Q$b\E\z/
or the faster
$a eq $b

If you wanted to check for "$b is in $a", you want
$a =~ /\Q$b/
or the faster
index($a, $b) >= 0

By the way,
qq/$a/ eq qq/$b/ is equivalent to
"$a" eq "$b" (because qq// and "" are the same quoting operator), and to
$a eq $b (because eq already forces stringification of its args).

Please allow me to rewrite your examples in a way that maintains rough equivalence whilst hopefully making the differences more apparent.

{ my $a = "MIN.(NERR_VOL2)";
  my $b = $a;
  print "matches" if $a =~ m/$b/;
}
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{ my $a = "MIN.(NERR_VOL2)";
  my $b = $a;
  print "equals" if $a eq $b;
}
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A couple of quibbles to start:

• Please surround your code samples with <code>..>/code< tokens; this makes it easier to read
• Please do not use $a and $b as variable names; these are special

When I run you first case through the debugger, I get 'empty array' as the result of
$x =&nbsp;(qq/$a/ =~ qq/$b/)
I suspect that this is a more appropriate way of writing your use of the regex:

($n = $a) =~ /$b/;
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With the proviso (I'm repeating myself) that $a and $b should not be used as variable names except in sort { $a <=> $b} @array;

Reparented from duplicate thread Reaped: Matching quoted variables. Cant make it work.. by Arunbear

So the real issue, now, in my mind is not so much 'how regex works' but how would I use quoting to make the regex ignore the possibility that the input variables might contain a meta character, since this example is just the essence of problem and not my complete code.

A previous suggestion to use

"\Q$a\E"
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is the easiest way to avoid these kind of problems. If you want to inlclude a variable that may contain meta-characters, then \Q forces the escaping of meta-characters (i.e. puts '\' before each one). \E turns escaping off. Play with it.

Hope that helps.

Tom Melly, tom@tomandlu.co.uk