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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?
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Most Upvoted Answer
A processor has 300 distinct instructions and 70 general-purpose regis...
Concept:
The given data,
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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Community Answer
A processor has 300 distinct instructions and 70 general-purpose regis...
Calculation:
- Total number of instructions = 300
- Total number of registers = 70
- Size of instruction word = 32 bits
Breakdown of instruction word:
- Opcode field: Used to specify the operation to be performed (variable length)
- Register operand 1: Used to specify the first register (log2(70) = 7 bits)
- Register operand 2: Used to specify the second register (log2(70) = 7 bits)
- Immediate operand: Used to specify immediate data (remaining bits)
Calculation for immediate operand:
- Total bits available in instruction word = 32 bits
- Bits used by opcode = variable
- Bits used by register operand 1 = 7 bits
- Bits used by register operand 2 = 7 bits
- Total bits used = 7 + 7 = 14 bits
- Remaining bits for immediate operand = 32 - 14 = 18 bits
Conclusion:
- The number of bits available for the immediate operand field is 18 bits.
- However, since the correct answer is given as '9', it is possible that there may be a mistake in the options provided or in the calculation.
- Total number of instructions = 300
- Total number of registers = 70
- Size of instruction word = 32 bits
Breakdown of instruction word:
- Opcode field: Used to specify the operation to be performed (variable length)
- Register operand 1: Used to specify the first register (log2(70) = 7 bits)
- Register operand 2: Used to specify the second register (log2(70) = 7 bits)
- Immediate operand: Used to specify immediate data (remaining bits)
Calculation for immediate operand:
- Total bits available in instruction word = 32 bits
- Bits used by opcode = variable
- Bits used by register operand 1 = 7 bits
- Bits used by register operand 2 = 7 bits
- Total bits used = 7 + 7 = 14 bits
- Remaining bits for immediate operand = 32 - 14 = 18 bits
Conclusion:
- The number of bits available for the immediate operand field is 18 bits.
- However, since the correct answer is given as '9', it is possible that there may be a mistake in the options provided or in the calculation.
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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?
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