470,614 Members | 1,458 Online
Bytes | Developer Community
New Post

Home Posts Topics Members FAQ

Post your question to a community of 470,614 developers. It's quick & easy.

Regex doesn't match - what am I doing wrong?

Hi,

I am having trouble matching a regex that combines a negated character
class and an anchor ($). Basically, I want to match all strings that
don't end in a digit. So I tried:

bash-2.05a@bermuda:15$perl -e 'while (<STDIN>) { if (/[^0-9]$/) {
print;}}'
skdsklds
skdsklds
sklskl2 <== why does this match? it ends in a digit.
sklskl2

This matched all strings regardless of whether or not they ended in a
digit. But the complemented regex seems to work fine:

bash-2.05a@bermuda:13$perl -e 'while (<STDIN>) { if (/[0-9]$/) {
print;}}'
sdkldsklds2
sdkldsklds2
sdsk2
sdsk2
sks <==== doesn't match as expected

I replaced [0-9] with [\d] but got the same results.

On the other hand, grep works as expected:

bash-2.05a@bermuda:9$grep '[0-9]$'
sdksdjk2
sdksdjk2
22221
22221
sdjkdsjk <== doesn't match as expected

bash-2.05a@bermuda:11$grep '[^0-9]$'
sdklsdklds
sdklsdklds
sdkldslk2 <== doesn't match as expected

What am I doing wrong? Here's the perl version info:

bash-2.05a@bermuda:17$perl -V
Summary of my perl5 (revision 5.0 version 6 subversion 1)
configuration:
Platform:
osname=linux, osvers=2.4.17-0.13smp, archname=i386-linux
uname='linux daffy.perf.redhat.com 2.4.17-0.13smp #1 smp fri feb 1
10:30:48 est 2002 i686 unknown '
config_args='-des -Doptimize=-O2 -march=i386 -mcpu=i686 -Dcc=gcc
-Dcf_by=Red Hat, Inc. -Dcccdlflags=-fPIC -Dinstallprefix=/usr
-Dprefix=/usr -Darchname=i386-linux -
Dvendorprefix=/usr -Dsiteprefix=/usr -Uusethreads -Uuseithreads
-Uuselargefiles -Dd_dosuid -Dd_semctl_semun -Di_db -Di_ndbm -Di_gdbm
-Di_shadow -Di_syslog -Dman3ext=3pm
'
hint=recommended, useposix=true, d_sigaction=define
usethreads=undef use5005threads=undef useithreads=undef
usemultiplicity=undef
useperlio=undef d_sfio=undef uselargefiles=undef usesocks=undef
use64bitint=undef use64bitall=undef uselongdouble=undef
Compiler:
cc='gcc', ccflags ='-fno-strict-aliasing -I/usr/local/include',
optimize='-O2 -march=i386 -mcpu=i686',
cppflags='-fno-strict-aliasing -I/usr/local/include'
ccversion='', gccversion='2.96 20000731 (Red Hat Linux 7.2
2.96-109)', gccosandvers=''
intsize=4, longsize=4, ptrsize=4, doublesize=8, byteorder=1234
d_longlong=define, longlongsize=8, d_longdbl=define,
longdblsize=12
ivtype='long', ivsize=4, nvtype='double', nvsize=8, Off_t='off_t',
lseeksize=4
alignbytes=4, usemymalloc=n, prototype=define
Linker and Libraries:
ld='gcc', ldflags =' -L/usr/local/lib'
libpth=/usr/local/lib /lib /usr/lib
libs=-lnsl -ldl -lm -lc -lcrypt -lutil
perllibs=-lnsl -ldl -lm -lc -lcrypt -lutil
libc=/lib/libc-2.2.5.so, so=so, useshrplib=false,
libperl=libperl.a
Dynamic Linking:
dlsrc=dl_dlopen.xs, dlext=so, d_dlsymun=undef,
ccdlflags='-rdynamic'
cccdlflags='-fPIC', lddlflags='-shared -L/usr/local/lib'
Characteristics of this binary (from libperl):
Compile-time options:
Built under linux
Compiled at Apr 1 2002 12:23:22
@INC:
/usr/lib/perl5/5.6.1/i386-linux
/usr/lib/perl5/5.6.1
/usr/lib/perl5/site_perl/5.6.1/i386-linux
/usr/lib/perl5/site_perl/5.6.1
/usr/lib/perl5/site_perl/5.6.0
/usr/lib/perl5/site_perl
/usr/lib/perl5/vendor_perl/5.6.1/i386-linux
/usr/lib/perl5/vendor_perl/5.6.1
/usr/lib/perl5/vendor_perl
Jul 19 '05 #1
2 4276
On Wed, 10 Mar 2004 14:47:40 -0800, Sriram wrote:

The string 'sklskl2' does not end in a digit, it ends in a character
return. That's probably why everything always matches. If you chop the $_
first, like

perl -e 'while (<stdin>){chop $_; if (/[^0-9]$/) {print } }'

results will look more promising.

Hope this helps.

Hi,

I am having trouble matching a regex that combines a negated character
class and an anchor ($). Basically, I want to match all strings that
don't end in a digit. So I tried:

bash-2.05a@bermuda:15$perl -e 'while (<STDIN>) { if (/[^0-9]$/) {
print;}}'
skdsklds
skdsklds
sklskl2 <== why does this match? it ends in a digit.
sklskl2


Jul 19 '05 #2
Hi,

Thanks! That works much better; so does '/[^0-9]\n/' or !/[0-9]$/.

My Perl book says "The $ and \Z assertions can match not only at the
end of the string, but also one character earlier than that, if the
last character of the string happens to be a newline." This is also
how I'm used to regexs working in grep/sed etc.

I'm baffled that $ behaves differently when used in conjunction with a
negated character class. The complemented regex /[0-9]$/ works without
chop. Why the subtle difference?

Sriram

"Arno H.P. Reuser" <bi************@xs4all.nl> wrote in message news:<pa****************************@xs4all.nl>...
On Wed, 10 Mar 2004 14:47:40 -0800, Sriram wrote:

The string 'sklskl2' does not end in a digit, it ends in a character
return. That's probably why everything always matches. If you chop the $_
first, like

perl -e 'while (<stdin>){chop $_; if (/[^0-9]$/) {print } }'

results will look more promising.

Hope this helps.

Hi,

I am having trouble matching a regex that combines a negated character
class and an anchor ($). Basically, I want to match all strings that
don't end in a digit. So I tried:

bash-2.05a@bermuda:15$perl -e 'while (<STDIN>) { if (/[^0-9]$/) {
print;}}'
skdsklds
skdsklds
sklskl2 <== why does this match? it ends in a digit.
sklskl2

Jul 19 '05 #3

This discussion thread is closed

Replies have been disabled for this discussion.

Similar topics

4 posts views Thread by Masahiro Ito | last post: by
8 posts views Thread by Bibe | last post: by
2 posts views Thread by D | last post: by
2 posts views Thread by Michael R. Pierotti | last post: by
4 posts views Thread by sklett | last post: by
6 posts views Thread by Gary Bond | last post: by
By using this site, you agree to our Privacy Policy and Terms of Use.