Open In App
Related Articles

ISRO | ISRO CS 2020 | Question 7

Improve Article
Save Article
Like Article

Consider a 32-bit processor which supports 70 instructions. Each instruction is 32 bit long and has 4 fields namely opcode, two register identifiers and an immediate operand of unsigned integer type. Maximum value of the immediate operand that can be supported by the processor is 8191. How many registers the processor has ?
(A) 32
(B) 64
(C) 128
(D) 16

Answer: (B)

Explanation: Given, 32-bit processor,

Number of bits required for Opcode = ceiling(log2(70)) = 7

Number of bits required for Immediate operands = ceiling(log2(8191)) = ceiling (12.99) = 13

Now, Number of bits left to represents register bits = 32 – (7+13) = 12.

Since there are two register operands (i.e., Reg1 and Reg2), so 6 bits for each register.

Hence, number of registers = 26 = 64 registers.

Option (B) is correct.

Quiz of this Question

Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!

Last Updated : 10 May, 2020
Like Article
Save Article
Similar Reads