Re: Explaining =~ and .= operators

You'll find the answer to this in perldoc perlop

Update: Corrected 2nd example. (thanks lidden & bart)

=~ binds a scalar expression to a pattern match, eg:

print "Matches!" if "hello" =~ m/he/;
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The period (.) is a concatenation operator. When used in conjunction with = it concatenates the right argument to the left argument. For example..

($foo = "hello") .= " world";
# or
$foo = "hello"; $foo .= " world";
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Is the same as..

$foo = "hello world";
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Cheers,
Darren :)

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Re^2: Explaining =~ and .= operators by rinceWind (Monsignor) on Apr 04, 2006 at 10:03 UTC
=~ can also be used in list context to obtain the list of captures: $1, $2, etc. -- Oh Lord, won’t you burn me a Knoppix CD ? My friends all rate Windows, I must disagree. Your powers of persuasion will set them all free, So oh Lord, won’t you burn me a Knoppix CD ? (Missquoting Janis Joplin)	[reply]
Re^3: Explaining =~ and .= operators by revdiablo (Prior) on Apr 04, 2006 at 16:02 UTC
Not to be overly nitpicky, but that behavior is from the `m//` operator, not `=~`. I think this distinction is important to keep in mind, because `=~` can be used with `m//`, `s///` and `tr///`, which all have different behaviour.	[reply] [d/l] [select]
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