Newbie question on regex.

[SomeDeveloper]

I'm unable to understand the following statement from
http://search.cpan.org/dist/perl/pod/perlretut.pod.

"Because, for example, \d and \w are sets of characters, it is
incorrect to think of [^\d\w] as [\D\W];"

Of course, being subsumed by \w, \d is clearly redundant. But
redundancy aside, aren't [^\d\w] and [\D\W] still mathematically
equivalent?

Tia (for responding to my possible ignorance),
Some Developer.

 

[Gunnar Hjalmarsson]

I'm unable to understand the following statement from
http://search.cpan.org/dist/perl/pod/perlretut.pod.

"Because, for example, \d and \w are sets of characters, it is
incorrect to think of [^\d\w] as [\D\W];"

Of course, being subsumed by \w, \d is clearly redundant. But
redundancy aside, aren't [^\d\w] and [\D\W] still mathematically
equivalent?

No. Since \d is a subset of \w, [^\d\w] is the same as [^\w] which in
turn is the same as \W.

 

[Lars Eighner]

In our last episode,
the lovely and talented SomeDeveloper
broadcast on comp.lang.perl.misc:

I'm unable to understand the following statement from
http://search.cpan.org/dist/perl/pod/perlretut.pod.

"Because, for example, \d and \w are sets of characters, it is
incorrect to think of [^\d\w] as [\D\W];"

Of course, being subsumed by \w, \d is clearly redundant. But
redundancy aside, aren't [^\d\w] and [\D\W] still mathematically
equivalent?

No, they are not mathematically equivalent (much less "still").

[\D\W] is all the characters which are not digits plus all the
characters which are not alphanumeric. Thus [\D\W] are all the
characters except digits (i.e. [^\d] which is the same as [\D]).

[^\d\w], as you seem to understand, amounts to [^\w] which is
the same as [\W].

 

[Ben Morrow]

Quoth (e-mail address removed) (SomeDeveloper):

I'm unable to understand the following statement from
http://search.cpan.org/dist/perl/pod/perlretut.pod.

"Because, for example, \d and \w are sets of characters, it is
incorrect to think of [^\d\w] as [\D\W];"

Of course, being subsumed by \w, \d is clearly redundant. But
redundancy aside, aren't [^\d\w] and [\D\W] still mathematically
equivalent?

You are being confused by the fact that \d is a subset of \w. In general
(say, \w and \s) [\S\W] means 'not a space OR not a wordchar' (and will
match everything) whereas [^\s\d] means 'not a space AND not a
wordchar'. As you have (half-)realised, in the case where \d is a
subset of \w these two are equivalent (and also equivalent to simply
'not a wordchar').

Ben