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UGC-NET | UGC NET CS 2018 Dec – II | Question 14

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A Computer uses a memory unit with 256K word of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code and a register code part to specify one of 64 registers and an address part. How many bits are there in operation code, the register code part and the address part ?
(A) 7, 7, 18
(B) 18, 7, 7
(C) 7, 6, 18
(D) 6, 7, 18


Answer: (C)

Explanation: Indirect 1 bit
Address 28 (256kB) * 210 (1024 bytes/kB) = 218 ==> 18 bits

Reg 64 registers = 26 => 6 bits

OP-code 32 – 1 – 18 – 6 bits = 7 bits.

Option (C) is correct.

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Last Updated : 02 Nov, 2021
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