Re: LZ: A new programming question

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Re: LZ: A new programming question

Nathan Adams wrote:
> 
> At 16:18 12/18/96 -0800, you wrote:
> >There is one asm command you could use, ldir.  It means load hl into de
> >while increasing addresses until bc equals 0.  so...
> >
> >ld hl,$80E1    ;source of stuff
> >ld de,$80E0    ;destination of stuff
> >ld bc,$0003    ;amount of stuff, or bytes you want to move
> >ldir
> >
> >By the way, there is no chance of overlapping problems.  This routine is
> >especially good if you're moving lots and lots of bytes, like some
> >programs do to load pictures from graph mem to video mem.  You can have
> >the fun of working out the other things, like how to calculate bc(bytes
> >to move), and stuff like that.
> >
> >--
> >Compliments of:
> >_-_-_-_-_-_-_-_
> >  Alan Bailey
> >  mailto:bailala@mw.sisna.com
> >  IRC:Abalone
> >  Web:http://www.mw.sisna.com/users/bailala/home.htm
> >
> OK, I've been trying to get this to work but I haven't been able to yet.  I
> am writing a text editor and I need to be able to insert text.  When I call
> the subroutine for ldir, hl points to the next byte in memory to be written.
> The memory will look something like this after some text like abcde has been
> entered:
> 
> $80DF = $01
> $80E0 = a
> $80E1 = b
> $80E2 = c
> $80E3 = d
> $80E4 = e
> $80E5 = $02
> 
> $01 is the code for the start of the file
> $02 is the code for the end of the file
> 
> Now say I wanted to insert a letter between b & c. hl would point to $80E2.
> I need a routine that I could call that would increment the memory and leave
> hl pointing to the same address that it did when the routine was called.
> Usually I can figure these things out on my own but I think I'm becoming
> lazy.  Anyway, any help would be greatly appreciated.
> Nathan Adams
> nathana@goodnet.com
> "It is better to remain silent and be thought a fool,
> than to speak out and remove all doubt."

What you did last time was ldir, and this time, cause you're moving stuff 
in the other direction, you need to use yet another command...lddr.  
What you probably did, or what you might try to do, is use ldir.  Say it 
takes C out of $80E2 and puts it in $80E3.  The next thing ldir does is 
take D out of $80E3 and put it in $80E4.  But see, $80E3 was overwritten 
the previous time.  This eventually makes everything equal to C, which I 
dont' think you want.  (But it brings to mind a great way to make a large 
space of memory equal to one value)  So, lddr.  You have to start at 
$80E5 and lddr moves it toward C.  Got it?, same thing, just different 
direction.  Just remember this when there is a chance for overlappation 
(nice word) :

|when hl < de, use ldDr.	(hl is start of data, de is the end)
|when de < hl, use ldIr.

Specifically for your prog, when inserting, use ldDr, when deleting, use 
ldIr.

About keeping the same value in hl, you can just push hl before ldir or 
lddr and pop it afterwards.

Hope I didn't make a mistake :)

-- 
Compliments of:
_-_-_-_-_-_-_-_
  Alan Bailey
  mailto:bailala@mw.sisna.com
  IRC:Abalone
  Web:http://www.mw.sisna.com/users/bailala/home.htm
References: