He memory unit of a computer has 2 20 words. The computer has instruction format with four fields; an operation code field, a mode field to specify one of 4 addressing modes, a register address field to specify one of 6 5 processor registers, and a memory address.

Assume an instruction is 32 bits long. Answer the following:

A: How large must the mode field be? B: How large must the register field be? C: How large must the address field be? D: How large is the opcode field?

Question

Jj

Computers & Technology

17 January, 22:38

He memory unit of a computer has 2 20 words. The computer has instruction format with four fields; an operation code field, a mode field to specify one of 4 addressing modes, a register address field to specify one of 6 5 processor registers, and a memory address.

Assume an instruction is 32 bits long. Answer the following:

A: How large must the mode field be? B: How large must the register field be? C: How large must the address field be? D: How large is the opcode field?

+5

Answers (1)

Know the Answer?

Giovanny Bowman · Answer 1

Opcode = 3

Mode = 2

RegisterRegister = 7

AR = 20

Explanation:

a) Number of addressing modes = 4 = 22, So it needs 2 bits for 4 values

Number of registers = 65 = 1000001 in binary, So it needs 7 bits

AR = 20

Bits left for opcode = 32 - (2+7+20) = 3