On Jun 26, 2006, at 11:42 AM, tom arnall wrote:
On Monday 26 June 2006 10:49 am, Peter Cornelius wrote:
...
x|a #followed by an 'x' or an 'a'
aren't the alternates (1) everything to the left of '|' and (2)
'a'? thus - to
take another example - the following:
$_='abcde'
print /(abx|c)/
produces:
c
not:
abc
Yes, I think you're right. I guess my internal parser hiccuped :-)
do you have any idea why:
$_ = " x11x x22x a ";
$re1 = qr/x.*?\d\dx|a/;
$re2 = qr/($re1\s)?$re1/;
($_) = /($re2)/;
print $_;
doesn't produce 'x11x' ? (note btw that if you insert '\n' between
the first
two tokens of the target string, the result >does become 'x11x'.
note also
that if you drop '|a' from $re1 you also get 'x11x'.)
i read this example as follows:
$re1 = qr/
x #find an 'x'
.*? #find whatever of whatever length
\d\d #find two digits
x #find an 'x'
| #or, instead of all the
foregoing,
a #find an 'a'
/x;
$re2 = qr/
(
$re1 #find $re1
\s #and whitespace
)? #or maybe none of the foregoing
$re1 #find for sure $re1
#in sum, find $re1 possibly
preceded by $re1+whitespace
/x;
Check this out:
#!/usr/bin/perl
$_ = " x11x x22x a ";
#with '.*?'
$re1 = qr/x.*?\d\dx|a/;
($j) = /($re1\s)?/;
($k) = /($re1\s){1}/;
print "j:$j:$/";
print "k:$k:$/";
Gives you:
j::
k:x11x :
I think this lines up with Tom's comments about the optional '?'
operator making the first occurrence be the empty string. What's
confusing me now is this next bit.
#!/usr/bin/perl
$_ = " x11x x22x a ";
#with '.*?'
$re1 = qr/x.*?\d\dx|a/;
($g,$h) = /($re1\s)?($re1)/;
print "g:$g:$/";
print "h:$h:$/";
Gives you:
g:x11x x22x :
h:a:
I guess I expected it to give me something like
g::
h:x11x :
where the first match was the empty string and the second was the
pattern match.
Hrmmm,
Peter Cornelius
Sr. Applications Engineer
LiveOps http://www.liveops.com
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