Pine Yan wrote:
A script like this:
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
=~ //);
Run and gett result:
$1 = a @{0,2}, $& = ba
$1 = @{0,0}, $& =
line 3: you are matching the last single letter of the first match
[ ^ba ] as $1.
@+ and @- merely point to the starting points on that string $string3
being characters 1 and 3 (in array-speak it's 0,2) and from perldocs you
can find what $& is all about.
If you wanted to match either 'ab' or 'ba' you would do it this way:
(ab|ba) to require two letters.
If I change the code of line3 to:
print "\$1 = $1 [EMAIL PROTECTED],$+[0]}, \$& = $&\n" if($string3 =~
/(a|b)+/);
and keep everything else the same, I will get:
$1 = a @{0,2}, $& = ba
$1 = a @{6,8}, $& = ba
I think you are right in that this doesn't make sense right away.
It appears that the second expression is returning a match for the
previous regex and not a regex of //.
So I don't think that the statement of $string =~ // is going to return
the first element '', rather it returns the last regex that was applied.
RUN THIS
$string3="This is my favorite day.";
$string4="Some days are better than others.";
print $-[0],"\n" if $string4 =~ //;
print $-[0],"\n" if $string4 =~ /day/;
print $-[0],"\n" if $string3 =~ //;
print $-[0],"\n" if $string3 =~ /day/;
and you get
0
5
20
20
0 because there is no regex expression defined.
but the first '20' acts as if you were matching the last expression run
(/day/).
Question: Bug or Feature?
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