A byte-addressable processor has a 20-bit address space. If 16 KiB of that address space is reserved for memory-mapped I/O registers, how much address space remains for other uses?

Options

  • A. 1008 KiB
  • B. 1000 KiB
  • C. 1016 KiB
  • D. More than one of the above
  • E. None of the above

Correct Answer (Detailed Explanation is Below)

A. 1008 KiB

Detailed Explanation

Correct Answer: Option A (1008 KiB).

Explanation: A 20-bit address space contains 2^20 bytes = 1 MiB = 1024 KiB. The remaining address space is 1024 - 16 = 1008 KiB, assuming the reserved region is part of the total address space and does not overlap with other allocations.
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BPSC TRE 4 operating system practice set 3

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