stein@rocksanne.UUCP (Adam Stein) (01/10/87)
I am look for a Z80 to 8080 program converter. My C compiler only comes with an 8080 assembler. This is for a TRS-80 Model 4. I don't care what language it's written in. I would appreciate any help on this. Adam Stein seismo!rochester!rocksanne!stein - Work seismo!rochester!ritcv!aes0972 - School
det@herman.UUCP (Derek Terveer) (01/13/87)
In article <249@rocksanne.UUCP>, stein@rocksanne.UUCP (Adam Stein) writes: > I am look for a Z80 to 8080 program converter. My C compiler only comes with > an 8080 assembler. This is for a TRS-80 Model 4. I don't care what language > it's written in. I would appreciate any help on this. 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)
ken@rochester.ARPA (SKY) (01/14/87)
|> I am look for a Z80 to 8080 program converter. My C compiler only comes with | |I believe that z80 has a mere two instructions that 8080 doesn't have. If I am not mistaken, the poster wanted a assembler to assembler converter. You can find one on the SIMTEL20 archives as XIZI.LBR in the CPMUG collection. Sorry, don't know the archive number offhand. Ken PS: Also there are more than 2 extra instructions in the Z80. You are thinking of the 8085 perhaps.
jpn@teddy.UUCP (John P. Nelson) (01/14/87)
>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.
dcm@mimir.dmt.oz (Dennis Mills) (01/14/87)
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. > Therefore, most likely, your z80 program will run unchanged with the 8080 > assembler (I hope i'm right in this). Sorry...you're not! The Z80 has quite a considerable number of instructions in addition to the 8080 ones, a large number of which are the consequence of architectural differences such as index registers etc etc which the Z80 has and the 8080 does not. Dennis -- Dennis Mills, PHONE: +61 (03)487-9258 CSIRO Division of Manufacturing Technology, ACSNET: dcm@mimir.dmt.oz Cnr Albert and Raglan Streets, Preston Victoria, Australia, 3072.
ken@argus.UUCP (Kenneth Ng) (01/15/87)
In article <175@herman.UUCP>, det@herman.UUCP (Derek Terveer) writes: > In article <249@rocksanne.UUCP>, stein@rocksanne.UUCP (Adam Stein) writes: > > I am look for a Z80 to 8080 program converter. My C compiler only comes with > > an 8080 assembler. This is for a TRS-80 Model 4. I don't care what language > > it's written in. I would appreciate any help on this. > > 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) The z80 has a LOT more than 2 more op codes over the 8080. -- Kenneth Ng: Post office: NJIT - CCCC, Newark New Jersey 07102 uucp !ihnp4!allegra!bellcore!argus!ken *** WARNING: NOT ken@bellcore.uucp *** bitnet(prefered) ken@orion.bitnet Gillian: "Are you sure you won't change your mind?" Spock: "Is there something wrong with the one I have?"
straka@ihlpf.UUCP (01/15/87)
> > I am look for a Z80 to 8080 program converter. My C compiler only comes with > 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) If my merory serves me correctly, the Z80 also has a *pair* of register sets that the user can toggle between, whereas the 8080 only has one set of registers. Also, the Z80 has a few auto-incrementing opcodes which help in block-mode data transfers. Perhaps the previous poster was referring to the difference between the 8080 and the 8085 (which *are* almost identical for 5v-only power supply and 2 (RIM? or similar) opcodes). I'm confident that the job could be done without a real lot of effort, but Z80 code is not downward compatible with 8080 code. Hope this helps. -- Rich Straka ihnp4!ihlpf!straka
perkins@bnrmtv.UUCP (Henry Perkins) (01/15/87)
> 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) Sorry, you're thinking of the 8085, not the Z80. (The two extra instructions are RIM and SIM.) The Z80 instruction set is rather more complex than that for the 8080 (but it doesn't include RIM or SIM). -- {hplabs,amdahl,3comvax}!bnrmtv!perkins --Henry Perkins It is better never to have been born. But who among us has such luck? One in a million, perhaps.
ben@catnip.UUCP (01/16/87)
In article <175@herman.UUCP> det@herman.UUCP (Derek Terveer) writes: >In article <249@rocksanne.UUCP>, stein@rocksanne.UUCP (Adam Stein) writes: >> I am look for a Z80 to 8080 program converter. My C compiler only comes with >> an 8080 assembler. This is for a TRS-80 Model 4. I don't care what language >> it's written in. I would appreciate any help on this. > >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) The Z80 and 8080 use the same op-codes and very nearly compatible (except for a couple of new instructions on the Z80), but the assembly language for the two chips is quite different. I am not sure why Zilog decided to change the mnemonics (sp?), but they did. -- Ben Broder {ihnp4,decvax} !hjuxa!catnip!ben {houxm,clyde}/
zben@umd5 (Ben Cranston) (01/16/87)
In article <707@argus.UUCP> ken@argus.UUCP (Kenneth Ng) writes: > In article <175@herman.UUCP>, det@herman.UUCP (Derek Terveer) writes: >> In article <249@rocksanne.UUCP>, stein@rocksanne.UUCP (Adam Stein) writes: >>> ...Z80 to 8080 program converter. ... >> 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) > The z80 has a LOT more than 2 more op codes over the 8080. Ahem. The 8080's instructions were completely specified by the first byte of the opcode (the remaining bytes were addresses and such). The Z80 used two previously unused values of this first byte for the added instructions, but they both involved a SECOND byte to completely specify the additional instruction. Thus there could have been as many as 512 added operations. I don't think there were 512, but I KNOW there were more than two. Consider all the instructions that used the IX and IY index registers - they all had to be added opcodes... The only fault to this shoehorning of all the new instructions into but two opcodes was that all the USEFUL instructions took one additional cycle to execute... -- umd5.UUCP <= {seismo!mimsy,ihnp4!rlgvax}!cvl!umd5!zben Ben Cranston zben @ umd2.UMD.EDU Kingdom of Merryland UniSys 1100/92 umd2.BITNET "via HASP with RSCS"
ericr@hpvcla.HP.COM (Eric Ross) (01/17/87)
It is true that the Z80 --> 8080 conversion is typically transparent except for some items such as index registers, alternate registers,etc. But, the syntax between the two assemblers is radically different. Z80 uses LD, while 8080 uses MOV for instance. Eric Ross ihnp4!hpfcla!hpvcla!ericr
janm@runx.OZ (Jan Mikkelsen) (01/17/87)
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. >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) The Z80 is object compatible with the 8080, but *not* the other way around! For examply, the Z80 has two 16 bit index registers, which the 8080 does not have, IX and IY. Now if a compiler is going to ignore two 16 bit registers, it would not be a very good one. I very much double good Z80 code could run unmodified on an 8080. Maybe a macro assembler, with macros for the Z80 instructions in 8080 code? Jan Mikkelsen. ACSnet: janm@runx.oz JANET: runx.oz!janm@ukc ARPA: janm%runx.oz@seismo.css.gov CSNET: janm@runx.oz UUCP: {enea,hplabs,mcvax,prlb2,seismo,ubc-vision,ukc}!munnari!runx.oz!janm I KNOW there is something I have forgotten .... Of cource! The witty remark!
zeta@runx.OZ (Nick Andrew) (01/18/87)
> 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.... ACSnet: zeta@runx.OZ UUCP: ...!{seismo,hplabs,mcvax,ukc,nttlab}!munnari!runx.runx!zeta UUCP: ...!{seismo,hplabs,mcvax,ukc,nttlab}!munnari!runx.OZ!zeta Fidonet: Nick Andrew@[620/602] (Zeta) **** NEW NODE NUMBER **** bbs: Sysop@zeta, +61 2 627 4177, CCITT V21. Mail: P.O Box 177, Riverstone NSW 2765 Australia.
heins@orion.UUCP (Michael Heins) (01/18/87)
In article <175@herman.UUCP> det@herman.UUCP (Derek Terveer) writes: >In article <249@rocksanne.UUCP>, stein@rocksanne.UUCP (Adam Stein) writes: >> I am look for a Z80 to 8080 program converter. My C compiler only comes with >> an 8080 assembler. This is for a TRS-80 Model 4. I don't care what language >> it's written in. I would appreciate any help on this. > >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) You are thinking of the 8085, not the Z80. The Z80 has a dozen or so more instructions than the 8080, and in addition, has nearly twice as many registers (af, bc, de, and hl are duplicated and bank-switched). The z80 ISA is a proper superset of the 8080 instructions -- with one exception. The z80 uses the parity flag in double duty (parity/overflow). In practice, this exception has rarely proved to be a problem. Normally therefore, programs compiled/assembled for the 8080 will run fine on the z80, but not the other way around. -- ...!hplabs!sdcrdcf!trwrb!orion!heins We are a way for the universe to know itself. -- Carl Sagan
rlk@chinet.UUCP (Richard Klappal) (01/19/87)
Not withstanding the 250 (+/-) object code combinations of the 8080
and the 796 combos for the Z-80, and (as I understand it) Intel
copyrighted the mnemonics for the 8080, preventing their use
by Zilog,
The Microsoft M-80 assembler will handle both sets of
mnemonics, (both in the same program, if desired), or
check a bulletin board for CP/M to find one of the
PD Z-80 assemblers (z80asm is one, I believe.) if your
need doesn't warrant buying one.
I speak for myself only, and have no financial interest in the
success or failure of Microsoft Corp.
--
---
UUCP: ..!ihnp4!chinet!uklpl!rlk || MCIMail: rklappal || Compuserve: 74106,1021
..paula@bcsaic.UUCP (01/21/87)
ben@catnip.UUCP (Bennett Broder) writes: >In article <175@herman.UUCP> det@herman.UUCP (Derek Terveer) writes: >>In article <249@rocksanne.UUCP>, stein@rocksanne.UUCP (Adam Stein) writes: >>> I am look for a Z80 to 8080 program converter. My C compiler only comes with >> >>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) > >The Z80 and 8080 use the same op-codes and very nearly compatible (except >for a couple of new instructions on the Z80), but the assembly language Since it's been a while since I looked at my Z80 Tech Ref, I'm not going to flame anybody for posting without knowing what they're talking about. But really, guys. The Z80 has at least twice as many opcodes as the 8080. Most of the new instructions were bit-manipulation, block move, and block I/O. The machine architecture included two new index registers and a duplicate set of registers for fast context-switching. There were certainly some other additional features, but I can't remember them. The Z80 was intended to be upward-compatible with the 8080. In fact, most 8080 code will run unchanged on a Z80. (There was one problem with the Z flag or something, but I don't remember the details.) In response to the original poster, a translator from Z80 to 8080 will need to output 8080 code to emulate not only the new instructions, but also the expanded register set. Probably doable, but not as trivial as some have suggested. And no, I don't know where you can get such a translator. Sorry. Paul Allen -- ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Paul L. Allen | "Look out, men! He's armed!" Boeing Advanced Technology Center | "I've got a cheese grater, POB 24346 M/S 7L-44, | and I'm not afraid to use it!" Seattle, WA, USA 98124 | "Don't make it any harder (206) 865-3207 | on yourself, kid! Drop it!" ...!uw-beaver!ssc-vax!bcsaic!paula | "Eat mozzarella, copper!"
w8sdz@brl-smoke.UUCP (01/25/87)
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GEnie Mail: W8SDZdennis@rlgvax.UUCP (Dennis Bednar) (02/04/87)
In article <445@catnip.UUCP>, ben@catnip.UUCP (Bennett Broder) writes: > In article <175@herman.UUCP> det@herman.UUCP (Derek Terveer) writes: > >In article <249@rocksanne.UUCP>, stein@rocksanne.UUCP (Adam Stein) writes: > >> I am look for a Z80 to 8080 program converter. The Z80 is a superset of the 8080 at the binary code level (that is a binary 8080 program should run on a Z80, but not the converse, since the Z80 has at least the IX index register that the 8080 doesn't have, and hence the Z80 has additional instructions that the 8080 doesn't have). The source code mneumonics for the older 8080 instructions were not preserved however. However, it is pretty easy to "hand-compile" 8080 source code into Z80 source code, once you get the hang of things (I remember doing this several years ago). Digression: The Small-C compiler written by Ron Cain (published in Dr. Dobbs Journal several years ago) was originally written to output 8080 source code. It was relatively easy to change it to output Z80 source code instead. Therefore, if you plan to convert Z80 source code to 8080 source code, you will have problems for those Z80 instructions which have no 8080 counterpart. -- FullName: Dennis Bednar UUCP: {seismo|sundc}!rlgvax!dennis USMail: CCI; 11490 Commerce Park Dr.; Reston VA 22091 Telephone: +1 703 648 3300