Computer organisation

Question

A machine support 16 bit instruction format size of address field is 4 bit the computer uses expanding opcode technique and 34 two address instructions and 100 one address instructions. The number of zero address instruction it can support is......

Answer 1

Instruction size = 16 bits, hence total number of encodings will be = 2¹⁶

Size of address field is 4 bit , so encodings will be = 2⁴

This machine uses 0,1,2 address instructions.

34 2-address instructions will take = 34 * 2⁴ * 2⁴ encodings.

100 1-address instructions will take = 100 * 2⁴  encodings

N 0-address instructions will take = (2¹⁶ - (34 * 2⁴ * 2⁴  + 100 * 2⁴  ))encodings = 55232 = which will also be the total 0 - address instructions.

Answer 2

Total Two address instruction can be  = 28 = 256

Two address instruction used = 34

Total left two address instruction = 256-34 = 222.

One address instruction possible from 222 two address instruction = 222*24 = 222*16

One address instruction used = 100

Total left one address instruction = 222*16 - 100 = 3452

Zero address instruction possible from 3452 one address instruction = 3452*24 = 3452*16 = 55232

Comments

I am not able to understand. Can someone elaborate?

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how total 2-address instruction are 2^8?

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what is meant by expanding opcode technique?

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See in a simple manner 2 address instructions means instructions will be having two addresses :

one for source/destination  and

other for destination

So if we have 4 bits then we can generate 2^4 different addresses for 1 address so here we are having 2 addresses here so we can geberate 2^2  × 2 ^2 addresses