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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?

Respuesta :

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Answer:

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

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