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From: zeta@runx.OZ (Nick Andrew)
Newsgroups: comp.sources.d
Subject: Re: Z80 -> 8080 converter
Message-ID: <606@runx.OZ>
Date: Sat, 17-Jan-87 23:57:28 EST
Article-I.D.: runx.606
Posted: Sat Jan 17 23:57:28 1987
Date-Received: Sat, 24-Jan-87 13:48:16 EST
References: <249@rocksanne.UUCP> <175@herman.UUCP>
Organization: RUNX Un*x Timeshare.  Sydney, Australia.
Lines: 29


    
>   I believe that z80 has a mere two instructions that 8080 doesn't have.
>   Therefore, most likely, your z80 program will run unchanged with the 8080
>   assembler (I hope i'm right in this).  (As long as the z80 code doesn't use
>   those two instructions that is)

   Good heavens, protect us from totally uninformed opinions.

   The Z80 contains many improvements on the 8080. Some of these are:

- Relative jumps (8 bit displacement)
- Alternate register set
- Additional Index registers (IX & IY) (one pair)
- Block-statements (block move, block IO)
- Rationalised assembler mnemonics and instruction formats

   If you have a good 8080 macro assembler you could write macros to handle
the conversion (eg change all jump relatives to jump absolutes), but at the
same time there will probably need to be some syntax changes to the Z80
source so it will comply with the requirements of the macro assembler.


   You can mail me at....
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