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From: jpn@teddy.UUCP (John P. Nelson)
Newsgroups: comp.sources.d
Subject: Re: Z80 -> 8080 converter
Message-ID: <3668@teddy.UUCP>
Date: Wed, 14-Jan-87 10:46:47 EST
Article-I.D.: teddy.3668
Posted: Wed Jan 14 10:46:47 1987
Date-Received: Thu, 15-Jan-87 00:43:21 EST
References: <249@rocksanne.UUCP> <175@herman.UUCP>
Reply-To: jpn@teddy.UUCP (John P. Nelson)
Organization: GenRad, Inc., Concord, Mass.
Lines: 21
Keywords: Z80 8080 converter

>In article <249@rocksanne.UUCP>, stein@rocksanne.UUCP (Adam Stein) writes:
>> I am look for a Z80 to 8080 program converter.

In article <175@herman.UUCP> det@herman.UUCP (Derek Terveer) writes:
>I believe that z80 has a mere two instructions that 8080 doesn't have.
>(I hope i'm right in this).

No, that's not right.  First, a "Z80" assembler usually uses different
mnemonics even when the object code instruction is identical (Z80 has
a single LD mnemonic to move data around, but the standard 8080 has
lots of mnemonics for the instruction variations).  Also, the Z80 has
extra registers, and instructions to deal with them, as well as lots of
new general instructions (bit set/clear etc.).

I have seen Z80->8080 translaters that will even translate the new
Z80 instructions to DB and DW psuedo ops - inserting constants which
just happen to correspond to the new Z80 instructions.  Not very
readable, but it will assemble a working program.

Perhaps Derek was thinking about an 8085, which is an 8080 with two
new system type instructions.