From owner-freebsd-questions@FreeBSD.ORG Wed Jun 22 06:48:05 2005 Return-Path: X-Original-To: questions@freebsd.org Delivered-To: freebsd-questions@FreeBSD.ORG Received: from mx1.FreeBSD.org (mx1.freebsd.org [216.136.204.125]) by hub.freebsd.org (Postfix) with ESMTP id E226C16A41C for ; Wed, 22 Jun 2005 06:48:05 +0000 (GMT) (envelope-from on@cs.ait.ac.th) Received: from mail.cs.ait.ac.th (mail.cs.ait.ac.th [192.41.170.16]) by mx1.FreeBSD.org (Postfix) with ESMTP id 03E6243D49 for ; Wed, 22 Jun 2005 06:48:04 +0000 (GMT) (envelope-from on@cs.ait.ac.th) Received: from banyan.cs.ait.ac.th (banyan.cs.ait.ac.th [192.41.170.5]) by mail.cs.ait.ac.th (8.12.11/8.12.11) with ESMTP id j5M6m0N9022465 (version=TLSv1/SSLv3 cipher=DHE-RSA-AES256-SHA bits=256 verify=NO); Wed, 22 Jun 2005 13:48:00 +0700 (ICT) Received: (from on@localhost) by banyan.cs.ait.ac.th (8.13.1/8.12.11) id j5M6m0XI089322; Wed, 22 Jun 2005 13:48:00 +0700 (ICT) Date: Wed, 22 Jun 2005 13:48:00 +0700 (ICT) Message-Id: <200506220648.j5M6m0XI089322@banyan.cs.ait.ac.th> From: Olivier Nicole To: on@cs.ait.ac.th In-reply-to: <200506220411.j5M4BQ6G087888@banyan.cs.ait.ac.th> (message from Olivier Nicole on Wed, 22 Jun 2005 11:11:26 +0700 (ICT)) References: <200506220411.j5M4BQ6G087888@banyan.cs.ait.ac.th> X-Virus-Scanned: on CSIM by amavisd-milter (http://www.amavis.org/) Cc: questions@freebsd.org Subject: Re: Using regex(3) X-BeenThere: freebsd-questions@freebsd.org X-Mailman-Version: 2.1.5 Precedence: list List-Id: User questions List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , X-List-Received-Date: Wed, 22 Jun 2005 06:48:06 -0000 > I must missunderstand how to use regex(3). To add a bit, running the same program on Linux gives the expected results: regexpr=a(.)c number of substrings=1 return from regexec=0 nmatch=0 p0.so=0 p0.eo=0 p1.so=0 p1.eo=0 p2.so=0 p2.eo=0 p3.so=0 p3.eo=0 return from regexec=0 nmatch=1 p0.so=0 p0.eo=3 p1.so=0 p1.eo=0 p2.so=0 p2.eo=0 p3.so=0 p3.eo=0 return from regexec=0 nmatch=2 p0.so=0 p0.eo=3 p1.so=1 p1.eo=2 p2.so=0 p2.eo=0 p3.so=0 p3.eo=0 Olivier