From mboxrd@z Thu Jan 1 00:00:00 1970 From: Nicolas Goaziou Subject: Re: [RFC] Alternative to sub/superscript regexp Date: Tue, 26 Nov 2013 17:46:02 +0100 Message-ID: <87k3fv862d.fsf@gmail.com> References: <87wqjw8kuk.fsf@gmail.com> <7CF22FEC-70F1-46A5-BDE0-A07832F9EDAF@gmail.com> Mime-Version: 1.0 Content-Type: text/plain Return-path: Received: from eggs.gnu.org ([2001:4830:134:3::10]:48591) by lists.gnu.org with esmtp (Exim 4.71) (envelope-from ) id 1VlLlq-0001gV-Dm for emacs-orgmode@gnu.org; Tue, 26 Nov 2013 11:46:02 -0500 Received: from Debian-exim by eggs.gnu.org with spam-scanned (Exim 4.71) (envelope-from ) id 1VlLlh-0001ut-Qt for emacs-orgmode@gnu.org; Tue, 26 Nov 2013 11:45:54 -0500 Received: from mail-ea0-x235.google.com ([2a00:1450:4013:c01::235]:51975) by eggs.gnu.org with esmtp (Exim 4.71) (envelope-from ) id 1VlLlh-0001um-HI for emacs-orgmode@gnu.org; Tue, 26 Nov 2013 11:45:45 -0500 Received: by mail-ea0-f181.google.com with SMTP id m10so3824593eaj.12 for ; Tue, 26 Nov 2013 08:45:44 -0800 (PST) In-Reply-To: <7CF22FEC-70F1-46A5-BDE0-A07832F9EDAF@gmail.com> (Carsten Dominik's message of "Tue, 26 Nov 2013 10:20:20 +0100") List-Id: "General discussions about Org-mode." List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , Errors-To: emacs-orgmode-bounces+geo-emacs-orgmode=m.gmane.org@gnu.org Sender: emacs-orgmode-bounces+geo-emacs-orgmode=m.gmane.org@gnu.org To: Carsten Dominik Cc: Org Mode List Hello, Carsten Dominik writes: > I have tested this a bit, and it does pretty much what I want. Great. > Just to be sure: We will also support expressions with braces, right? Of course, we will. Curly braces are not ambiguous so I'm no suggesting to change this part of the syntax (even though it would be best to parse them with an appropriate syntax table rather than a regexp). OTOH, while we're at it, I wonder if we should keep grouping with parenthesis. I guess that if you want to use parenthesis, you're going to need math mode anyway, so there's no real need to make it easy to type, e.g. : x^(2-i) which will become x$^\text{(2-i)}$ anyway. What do you think? Regards, -- Nicolas Goaziou