Re: A86: Re: Re: Questions


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Re: A86: Re: Re: Questions




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.


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