On Jul 29, Pine Yan said:
line1: $string3 = "bacdeabcdefghijklabcdeabcdefghijkl";
line2: $string4 = "xxyyzzbatttvv";
line3: print "\$1 = $1 [EMAIL PROTECTED],$+[0]}, \$& = $&\n"
if($string3
=~ /(a|b)*/);
line4: print "\$1 = $1 [EMAIL PROTECTED],$+[0]}, \$& = $&\n"
if($string4
=~ //);
$1 = a @{0,2}, $& = ba
$1 = @{0,0}, $& =
The regex says "match zero or more of (a or b)". In string 1, it matches
a 'b' and then an 'a' at the beginning, thus $& = 'ba'. In string 2, it
matches zero characters (because it's allowed to!) at the beginning, thus
$& eq ''.
print "\$1 = $1 [EMAIL PROTECTED],$+[0]}, \$& = $&\n" if($string3 =~
/(a|b)+/);
$1 = a @{0,2}, $& = ba
$1 = a @{6,8}, $& = ba
Now your regex says "match one or more of (a or b)". Thus in string 2,
you're matching the "ba" in the middle.
HOWEVER, you're doing something weird. You have a QUANTIFIER (the * or
the +) on a CAPTURING GROUP. Here's an example of the weirdness:
"japhy" =~ /(.)+/;
print $1;
What do you think that prints? It prints 'y'. Why? Because when you put
a quantifier on a capturing group, ONLY THE LAST REPETITION of that
capturing group gets saved. This is why you're getting only ONE letter in
$1.
--
Jeff "japhy" Pinyan % How can we ever be the sold short or
RPI Acacia Brother #734 % the cheated, we who for every service
http://japhy.perlmonk.org/ % have long ago been overpaid?
http://www.perlmonks.org/ % -- Meister Eckhart
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