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I'm studying for my final and don't understand this question. Here is the full question (from Stallings 8th edition):

Consider a hypothetical microprocessor generating a 16-bit address (e.g., assume the program counter and the address registers are 16 bits wide) and having a 16-bit data bus.

a. What is the maximum memory address space that the processor can access directly if it is connected to a “16-bit memory”?

b. What is the maximum memory address space that the processor can access directly if it is connected to an “8-bit memory”?

c. What architectural features will allow this microprocessor to access a separate “I/O space”?

d. If an input and an output instruction can specify an 8-bit I/O port number, how many 8-bit I/O ports can the microprocessor support? How many 16-bit I/O ports? Explain.

If I'm following, the answer to "a" is the processor is generating 2^16 = 64 Kbytes and the bus is 16-bit so the answer it 64 Kbytes.

However, for "b", the answer is 64 Kbytes are being generated but since the bus is only 8-bit, it requires twice as many cycles.

For the answer to "c", I am clueless and I think that carries over into me being clueless on "d". :-\

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  • $\begingroup$ for part c I think the answer could be the use of an expansion bus $\endgroup$
    – Abu
    Commented Dec 7, 2019 at 20:35
  • $\begingroup$ An 8086 processor with 16 bit PC and address registers could access 2^20 + 2^16 bytes of memory. $\endgroup$
    – gnasher729
    Commented Dec 8, 2019 at 13:18

2 Answers 2

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The CPU uses the same address/data bus to access both the memory space and the I/O space. They are however distinguished by an extra signal that tells whether the address/data is in the memory space or the I/O space. For instance, in x86 this signal is called $M/\overline{IO}$ (pin 28).

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  • $\begingroup$ Thanks @ran-g If the pin is "high" (meaning "set to 1"), it will know the data is in the I/O space? If it is "low" (meaning "set to 0"), it will know the data is in the memory space? (Are my responses to "a" and "b" correct?) $\endgroup$
    – harperville
    Commented Apr 21, 2019 at 13:19
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    $\begingroup$ the notation $M/\overline{IO}$ means $=1$ for memory and $=0$ for IO. $\endgroup$
    – Ran G.
    Commented Apr 21, 2019 at 14:05
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(For a) & b), my take is $2^{16}$ of whatever are addressable units - the bus size not being an indication (see e.g. Intel 8086↔8088). It does not read 16 address lines/signals, nor does it mention address multiplexing.
Implementations not supplying a signal for every architectural address bit are somewhat common: physical address space may be smaller (or, historically, larger) than the architectural address space.)

c) In the programming model, the pivotal feature is existence of I/O instructions mentioned in d). Which does not specify whether there are separate instructions for 8-bit and 16-bit ports.
d) assuming no separate instructions for accessing 8-bit and 16-bit ports, both share "an 8-bit address space".

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