The "Think DeMorgan . . ." statement says it. A []-defined character class is structured like logical "or", so a character matching [\D\W] is either in \W or in \D (That's actually a degenerate example since \w contains \d). So [^\d\w] is logically "not (d or w)". DeMorgan distributes "not" over "or" as "(not d) and (not w)", while [\D\W] translates as "(not d) or (not w)".

The point is that \D is the set of all characters that do not belong to \d, so 'a', 'C', '&' all belong to \D. Now similar for \W, the set of all characters not in \w, hence '&', ' ', etc.

So far so good, but [\D\W] is the union of the sets \D and \W, hence it will contain '&' and ' ', but also 'a' and 'C' since the latter are members of \D. That's definitely not equal to the set [^\d\w] that contains none of the characters in \d or \w. The set [^\d\w] definitely doesn't contain 'a' or 'C' that are members of [\D\W].

Hope this helps, -gjb-

Think DeMorgan's laws:

### ( (! $a) && (! $b) )
[download]

### (! ( $a || $b) )
[download]

Maybe a picture will help. Try running this on a screen that is at least 108 chars wide.

#! perl -slw
use strict;

for my $re ( 
    qw[ .* 
        \d      \D     [^\d]  [^\D] 
        \w      \W     [^\w]  [^\W] 
        [\d\w] [^\d\w] [\D\W] [^\D\W] 
    ] 
) {
    printf "%8s:'%s'\n\n", $re, join '', 
        map{ /$re/ ? $_ : ' ' } map{ chr } 32 .. 126;
};;
[download]

Normally I'd post the output, but the OSU PM wrap feature doesn't like it.