The instruction set architecture for a simple computer must support access to 64 KB of byte-addressable memory

Question:

The instruction set architecture for a simple computer must support access to 64 KB of byte-addressable memory space and eight 16-bit general-purpose CPU registers.

a. If the computer has three-operand machine language instructions that operate on the contents of two different CPU registers to produce a result that is stored in a third register, how many bits are required in the instruction format for addressing registers?

b. If all instructions are to be 16 bits long, how many op codes are available for the three-operand, register operation instructions described above (neglecting, for the moment, any other types of instructions that might be required)?

c. Now assume (given the same 16-bit instruction size limitation) that, in addition to the instructions described in (a), there are a number of additional two-operand instructions to be implemented, for which one operand must be in a CPU register while the second operand may reside in a main memory location or a register. If possible, detail a scheme that allows for at least 50 register-only instructions of the type described in (a) plus at least 10 of these two-operand instructions. (Show how you would lay out the bit fields for each of the machine language instruction formats.) If this is not possible, explain in detail why not and describe what would have to be done to make it possible to implement the required number and types of machine language instructions.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Question Posted: