Re: A86: Re: Re: Questions

To: assembly-86@lists.ticalc.org
Subject: Re: A86: Re: Re: Questions
From: Cassady Roop <croop@oregontrail.net>
Date: Tue, 29 Sep 1998 20:10:56 -0700
Delivered-To: assembly-86-outgoing@towerguard.unix.edu.sollentuna.se
Organization: Slightly Immoral Technologies
References: <3.0.1.32.19980929070928.0071f560@reninet.com> <361060D7.AE068C35@oregontrail.net> <002901bdeb5d$0f095e40$67e0e6a5@Sammy.rutgers.edu> <3.0.1.32.19980929141024.0071f74c@reninet.com>
Reply-To: assembly-86@lists.ticalc.org


So at the fourth byte of a normal basic prog it would be: 00
and at the fourth byte of a locked basic prog would be:   0029

Yes?  No?  So this means I have to move everything down a byte to
edit-lock a program?  What would happen if I left out the 00 and put in
29 by itself?

Chicane wrote:
> 
> >>$28 is the asm token. _All_ asm programs are already protected.
> >If you are changing a basic program, it may not work.
> >It appears that the token for a basic program is $8e. The next byte is the
> Sorry for this mis-information. It is actually $00, not $8e.
> 
> CHICANE
> ICQ: #14727618
> chicane on EFNET #ti
> Homepage: Under Construction.

Follow-Ups:

Re: A86: Re: Re: Questions

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A86: Re: Re: Questions

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Re: A86: Re: Re: Questions

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