8086 is source code compatible with 8080. Zilog Z80 extended Intel 8080 with:

  • An enhanced instruction set including bit manipulation, block move, block I/O, and byte search instructions
  • New IX and IY index registers with instructions for direct base+offset addressing
  • A better interrupt system...[1]

Is it possible to translate Z80 asm into 8086, even with different register layout? There was a translator but it left some opcode untranslated and needed human intervention.

PS: What about 8085's extention to 8080? Can these features be adapted for Z80 or 8086?

Illustration of four syntaxes, using samples of equivalent, or (for 8086) very similar, load and store instructions.

Intel 8008
Datapoint 2200
Intel 8080
Intel 8085
Zilog Z80 Intel 8086
Intel 8088
before ca. 1973 ca. 1974 1976 1978
-- STAX D LD (DE),A MOV [DX],AL[x]
LDI 56 MVI D,56 LD D,56 MOV DL,56
LMI 56 MVI M,56 LD (HL),56 MOV byte ptr [BP],56
-- LDA 1234 LD A,(1234) MOV AL,[1234]
-- STA 1234 LD (1234),A MOV [1234],AL
-- -- LD B,(IX+56) MOV BL,[SI+56]
-- -- LD (IX+56),C MOV [SI+56],CL
-- -- LD (IY+56),78 MOV byte ptr [DI+56],78
-- LXI B,1234 LD BC,1234 MOV BX,1234
-- LXI H,1234 LD HL,1234 MOV BP,1234
-- SHLD 1234 LD (1234),HL MOV [1234],BP
-- LHLD 1234 LD HL,(1234) MOV BP,[1234]
-- -- LD BC,(1234) MOV BX,[1234]
-- -- LD IX,(1234) MOV SI,[1234]
  • 1
    If you try to make LD A,(BC) translate to MOV AL,[BX] you've got the problem that in the Z80 case the address would have been loaded into B and C, and in the 8086/8 case the address would have therefore been loaded into BL and CL, not BX – user7761803 Apr 14 at 12:11
  • 1
    @user7761803 Due that reason the registers aren't assigned like you assumed. It's basically like this: A-> AL; B/C -> CX; D/E -> DX; H/L -> BX; SP -> SP. Now all works as expected, doesn't it? – Raffzahn Apr 14 at 22:22
  • Here's an example of somone translating GW BASIC the other way 8086->Z80 tia.mat.br/posts/2020/06/21/converting-gwbasic-to-z80.html . However, this is mostly 8080 code, so maybe doesn't count. – Mark Williams Apr 15 at 8:15
  • I'd say that '8086 source code compatible to 8080' is actually a vast marketoid exaggeration. It is, in fact, not, examples are given below in the answers and above in the comments. Of course, one can translate almost each assembler to another, then even optimize it (given rather intellectual translator, either JIT or source-code-based, or just manually), but does that makes them all 'source code compatible'? – lvd Apr 15 at 19:44

I'm going to say "No" simply because the 8086 doesn't support the alternate registers of the Z80. That was a fairly important concept that you can not directly mimic on the 8086.

Mind, if you're willing to dedicate memory and whatnot to support it, then, "sure". Replace the Z80 functionality with a macro, say. But now you're stretching it. Your instruction count was going to change.

Also 8086 MOVSB is not the same as the Z80 LDIR, it works on different registers.

There may well be other differences.

The idea was that you could translate 8080 to 8086, one instruction for one instruction. But the Z80 was not an Intel chip, it was a Zilog chip. As much as Intel may have wanted to perhaps support it, they may have chosen not too.

So, a 1:1 Z80 ⟶ 8086 translation could not be done perfectly. You could get quite a ways with a simple macro assembler, but there would absolutely be differences that would best be hand-checked in the end.

  • 4
    "The idea was that you could translate 8080 to 8086, one instruction for one instruction" - not quite; there were a few 8080 instructions that translated into multiple 8086 instructions; for example, PUSH PSW -> LAHF; PUSH AX, and RZ -> JNZ skip; RET; skip: – poncho Apr 14 at 13:28
  • 1
    I wonder how often the LAHF/PUSH AX represented the behavior that a program would actually need, and how often PUSHF/PUSH AX would be better despite the extra stack space. An interrupt handler that trashes AH would be highly disruptive, for example, unless none of the main line code used AH at all--not even as part of an LAHF/PUSH AX sequence. – supercat Apr 14 at 22:41

Sure, a lot can be done. Source code translation always offers the possibility to replace one instruction by a sequence - like Intel already did for a few. This would as well solve the issue of incompatible hardware, like simply exchanging all registers with a copy in memory when the alternate register set is selected.

Just, who should do this?

Intel had only interest to support their customers. Taking on all the support pitfalls for a rare chance to win over customers that have already left Intel? With a CPU that, despite all improvements, still cannot compete with the Z80 in the important area of interrupt handling? Interrupt handling was the main reason for professional customers to switch to Zilog. Not funny instructions.

Zilog had even less motivation to support porting Z80 code to x86 as that would mean loosing customers.

Lastly, there would be software companies, but their need was also rather limited. The ones writing machine-independent software did not use Z80 extensions, as they preferred to have a single binary for all of the CP/M world. Producers of machine-dependent software on the other hand were much more about performance; any automatic translation would have helped only at the curiosity level.

For the 8085, Intel simply resolved that issue by not making any of the new instructions officially available. The only official new ones were RIM/SIM and these are part of interrupt management, something that had to be rewritten anyway when porting to x86.

Bottom line

While it would be possible, there was no real need to do so.

  • I wonder if it's feasible that z80 be inserted into and considered a part of 8008-8080-8086 evolving tree especially for a 3rd party manufacturer such as the USSR. – Schezuk Apr 14 at 5:40
  • @Schezuk Would it matter? The switch happened anyway, no matter if one did put an emphasis on the 8080, like the USSR did, or used the Z80 instead, like the GDR did. Both ended up going for 8086 later on. GDR even somewhat faster IIRC. – Raffzahn Apr 14 at 5:42
  • 1
    8086 in the USSR? I can't think of any such computer. – OmarL Apr 14 at 10:20
  • @OmarL see wiki K1810VM86 – Schezuk Apr 14 at 15:05
  • 2
    @Schezuk Isn't that more like another question? In general, look at the Interrupt response time. The great advantage of the Z80 over the 8080 was it's direct(er) interrupt handling and second register set, saving many cycles when used for this. Speeding response time up quite a lot. While interrupt handling was eased on the 86 by having more vectors, it was slower as well due long vectors and register saving. So Z80 could, at similar clockspeeds, still run circles around an 8086. – Raffzahn Apr 15 at 10:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.