ISRO2020-8

Question

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. $16$

Comments

option b) is correct 6-bit for each register indentifier. So 2^6 = 64 registers.

Answer 1

$\underline{\textbf{Answer:}\Rightarrow}$

$\underline{\textbf{Explanation:}\Rightarrow}$

Opcode

Reg1

Reg2

Immediate Operand

 

Number of instructions which are supported $=70$

$13-\text{bits}$ support the maximum value of immediate operand $=8191$

Now,

Number of bits needed for opcode $=7\;\text{bits}$

Number of bits left to represent register bits $=12$

which means $6-\text{bits}$ per register.

Hence, at max. $\mathbf{64}$ registers.

$\therefore \;\mathbf{(b)}$ is the correct option.

 

Comments

Sir How many registers the processor has?

why we will not add r1 + r2 i.e is 64+ 64 = 128

we can do this when asked How many total registers are there?

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Good question but see, the $12$ bits are already equally divided between two registers. Therefore, we get $6$ bits per register. So, the maximum possible value of registers are $\mathbf{2^{6}}$.

Since you have already divided $12$ bits equally in the beginning, you don’t have to make it $64 +64$.

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the maximum possible value of registers are 2^64 ?

2^6 (typo)

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Right.

Thanks for the correction.

Answer 2

$32$ bit instruction contains opcode, two registers say $R1, R2$ and one unsigned immediate operand. 

Since 70 instructions are supported, each instructions will take $log70$= 7 bit for each opcode.

Maximum value of opcode is $8191$ that means it takes 13 bit ranging $0-8191$

Opcode(7 bit)

R1

R2

Immediate Op.(13bit)

Bits needed for two registers = $32-7-13=12$

Each register takes 6 bits, i.e maximum 64 registers are possible.

Hence Option B) is correct

Comments

32-7-13 = 12 and each register will take 6 bit.

number of register will be 64