A processor has 300 distinct instructions and 70 general-purpose registers. A 32-bit instruction word has an opcode, two register operands, and an immediate operand. The number of bits available for the immediate operand field is_____Correct answer is '9'. Can you explain this answer?

Question

Answer

Given Information:
- The processor has 300 distinct instructions.
- The processor has 70 general-purpose registers.
- A 32-bit instruction word has an opcode, two register operands, and an immediate operand.

To find:
The number of bits available for the immediate operand field.

Solution:
Since we know that a 32-bit instruction word has an opcode, two register operands, and an immediate operand, we can calculate the number of bits available for the immediate operand field using the following steps:

Step 1: Calculate the number of bits required for the opcode.
Since the processor has 300 distinct instructions, we need to represent each instruction with a unique opcode. The number of bits required to represent 300 distinct instructions is given by:
Number of bits for opcode = ceil(log2(300))

Step 2: Calculate the number of bits required for the register operands.
Since the processor has 70 general-purpose registers, we need to represent two register operands in the instruction word. The number of bits required to represent 70 registers is given by:
Number of bits for register operands = ceil(log2(70))

Step 3: Calculate the number of bits available for the immediate operand field.
The total number of bits used by the opcode and register operands can be calculated by adding the number of bits required for the opcode and register operands:
Total number of bits used = Number of bits for opcode + Number of bits for register operands

Since we know that the instruction word is 32 bits in total, the number of bits available for the immediate operand field can be calculated by subtracting the total number of bits used from 32:
Number of bits available for immediate operand field = 32 - Total number of bits used

Therefore, the final answer is 9 bits.

Answer

Distinct instructions= 300
General-purpose registers = 70
opcode instruction word = 32-bit
Two register operands and an immediate operand.

Each instruction has 32 bits. To support 300 instructions, the opcode must contain 9-bits.
Register operand1 requires 7 bits, since the total registers are 70, Register operand 2 also requires 7 bits.
So, 32 - (9+7+7) = 9 bits are left over for immediate operand.

Hence the correct answer is 9.

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