A computer has an instruction format with 4 fixed fields: an opcode field; a mod

Question

A computer has an instruction format with 4 fixed fields: an opcode field; a mode field; a register address field and a memory address field. The ISA of this computer must support 4 addressing modes, 16 registers and a linear space of 220 bytes of byte-addressable memory. Assume an instruction is 32 bits long.

Answer the following:

a. How many bits must the mode field have?

b. How many bits must the register address field have?

c. How many bits must the memory address field have?

d. How many bits does the opcode field have?

e. How many operations can be defined?

Explanation / Answer

bits for mode field-2bits(2^2=4. 4 modes are possible i.e using 2 bits we can achieve 4 combinations of modes)

bits for registers-4bits(2^4=16)

bits for 220byte byte addressable memory - 8bits. (As this is byte addressable You need log2(n) bits to address n bytes. For example, you can store 256 different values in an 8 bit number, so 8 bits can address 256 bytes. Then again if it was for than 256 say for example 512 so by calculation it would require 9bits but as it is byte addressable so 8bits has to be represented at a time hence we will need another 8bits instead of 1 xtra bit i.e for 512byte of memory in byte addressable we need 16bits.)

bits for opcode-18bits(32-(2+4+8)=18) ( 32 bit address - (2 bit mode + 4 bit register + 8 bit byte addressable memory)

operations can be defined-2^18

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