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The Z80 indirect I/O instructions are a little quirky: the instructions with the standard mnemonic in r, (c) (where r is a stand-in for any of the normal registers) actually puts the register pair bc on the address bus, not just c as implied in the mnemonic, and likewise for out (c), r.

This capability is rarely used, but there are some instances (e.g. the keyboards of the ZX81 and ZX Spectrum use the upper byte to select a row to scan) where it was used to good effect.

Is this behaviour intentional, in order to allow the Z80 to expand beyond the 256 port limit of the 8080 -- in which case, is the reason it wasn't documented known? -- or is it a result of a bug in the processor implementation (presumably a microcode error that wasn't detected prior to release)?

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Based on the fact that when you build a computer that evaluates the upper 8 address lines on an I/O cycle (and can use 16-bit I/O addresses), you can then unfortunately no longer use the block input and block output commands (as B is used as a repeat count), I think we can safely assume it was not intended that port address evaluation would take A8-A15 into account when the chip was designed.

In case a designer would decide to actually evaluate the upper address lines on an I/O cycle, the fact that OUT <PORT>,A puts the contents of the accumulator on A8-A15 is not very useful, as A would hold the byte to output. OUT (C),A could be useful and actually is used sometimes, like on the ZX-81 and on the Amstrad CPC range that use 16-bit I/O addresses (thus generally inhibiting any use of the former, as well as the block I/O commands). The sad fact that the CPC doesn't really make good use of its "enhanced" 16-bit I/O port address range is a different story, though)

The Zilog Z80 manual of 2016 (!) clearly states

In all Register Indirect input output instructions, including block I/O transfers, the contents of the C Register are transferred to the lower half of the address bus (device address) while the contents of Register B are transferred to the upper half of the address bus.

The same manual also very clearly states in the description of the /IOREQ signal (emphasis mine):

IORQ. Input/Output Request (output, active Low, tristate). IORQ indicates that the lower half of the address bus holds a valid I/O address for an I/O read or write operation.

The same manual shows A0-15 in the description of the I/O cycle and labels this "I/O address", which is, at least, somewhat misleading.

An older (1979) MOSTEK Z80 manual I found, shows only A0-A7 in the I/O cycle diagrams as port address, but does state the fact that register B is put to the upper half of the address bus on an I/O cycle.

I would neither suspect a design bug nor a documentation problem: The behavior is very clearly documented, apparently even in early manuals. It is also implied you simply shouldn't evaluate the upper 8 address lines on an I/O cycle. You still can, but if you do so, you will not be able to make sensible use of some of the I/O instructions.

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    "the fact that OUT <PORT>,A puts the contents of the accumulator on A8-A15 is not very useful, as A would hold the byte to output" - this is build upon an assumption that we want to output all 8 bits. I know several examples (16-bit ports on ZX Spectrum 128k), where several bits of A select the correct port and other bits contain actual output data. This may be a little too clever to envisage during the design stage, I agree. At the same time, I'd tone down the either/or emphasis in your answer; one can use both block I/O commands and some of the bits of the I/O addresses simultaneously.
    – introspec
    Sep 30, 2018 at 17:57
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Well, the Z-80 CPU documentation that I have, a Zilog Z-80 product spec, PS017801-0602, on page 20, clearly shows the behavior you noted in your question.

In addition, the traditional I/O instructions, such as:

IN  A,(n)

show the address bus as n to A0 ~ A7 and A to A8 ~ A15

Clearly this "feature" is documented. I suspect it was just not used much.

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    Both the Sinclair ZX spectrum and the Amstrad CPC used 16 bit port addresses. So the feature was used...
    – Martin Rosenau
    Sep 30, 2018 at 10:51
  • Like I said, not used much. From the question, it seems that the ZX 81 and Spectrum machines used A8 ~ A15 to transfer extra data rather than to create a large I/O space.
    – Peter Camilleri
    Sep 30, 2018 at 13:34
  • The Spectrum was really only using one bit of the address space for its single I/O port. If they were stuck with only 8-bit I/O they probably could have used a 7x6 layout at the cost of more complicated keyboard membrane wiring.
    – user3570736
    Sep 30, 2018 at 15:35
  • @user3570736, I think they both meant ZX Spectrum 128K, where the PSG chip ports as well as the memory control port are all essentially 16-bit.
    – introspec
    Sep 30, 2018 at 17:59
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16bit I/O addressing capability of Z80 is clearly an improvement over 8080 8bit I/O address space.

This capability is rarely used

It was actually extensively used even in 80ies era computers. For example, ZX Spectrum 48k used addresses A15..A8 to select keyboard row to scan when CPU read #xxFE port. ZX Spectrum 128k, in addition, used 16bit I/O ports with addresses like #7FFD, #BFFD and #FFFD.

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  • Indeed, this answer is more correct in a way than the chosen one. The 16 bit capability was not much touted to sell the Z80 as a 8080 Intel drop in chip, however the ZX Spectrum use those instructions.
    – Rui F Ribeiro
    May 28, 2019 at 15:30

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