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A processor has 128 distinct instructions. A 24-bit instruction word has an opcode, register, and operand. The number of bits available for the operand field is 7. The maximum possible value of the general-purpose register is ______.
Correct answer is '1023'. Can you explain this answer?
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A processor has 128 distinct instructions. A 24-bit instruction word h...
No. of bits required for 128 instructions = log2 128 = 7
No. of bits required for operand field = 7
No. of bits required for register field = 24 – 7 – 7 = 10
Maximum no. of registers = 210 = 1024
i.e. from 0 to 1023. Hence, maximum value = 1023
Most Upvoted Answer
A processor has 128 distinct instructions. A 24-bit instruction word h...
To determine the maximum possible value of the general-purpose register, we need to consider the number of bits available for the register field in the instruction word.
Given that the instruction word is 24 bits long and the operand field has 7 bits, we can calculate the number of bits available for the register field.
- The opcode field and the operand field together occupy 7 + 7 = 14 bits.
- Therefore, the register field must occupy the remaining bits, which is 24 - 14 = 10 bits.
Now, let's calculate the maximum possible value of a 10-bit register field.
- In binary representation, a 10-bit field can represent 2^10 = 1024 different values.
- However, since we start counting from 0, the actual maximum value is 1024 - 1 = 1023.
Therefore, the maximum possible value of the general-purpose register is 1023.
In summary:
- The instruction word is 24 bits long.
- The opcode and operand fields together occupy 14 bits.
- Therefore, the register field occupies the remaining 24 - 14 = 10 bits.
- A 10-bit field can represent 1024 different values.
- The maximum value of the register is therefore 1024 - 1 = 1023.
Given that the instruction word is 24 bits long and the operand field has 7 bits, we can calculate the number of bits available for the register field.
- The opcode field and the operand field together occupy 7 + 7 = 14 bits.
- Therefore, the register field must occupy the remaining bits, which is 24 - 14 = 10 bits.
Now, let's calculate the maximum possible value of a 10-bit register field.
- In binary representation, a 10-bit field can represent 2^10 = 1024 different values.
- However, since we start counting from 0, the actual maximum value is 1024 - 1 = 1023.
Therefore, the maximum possible value of the general-purpose register is 1023.
In summary:
- The instruction word is 24 bits long.
- The opcode and operand fields together occupy 14 bits.
- Therefore, the register field occupies the remaining 24 - 14 = 10 bits.
- A 10-bit field can represent 1024 different values.
- The maximum value of the register is therefore 1024 - 1 = 1023.
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A processor has 128 distinct instructions. A 24-bit instruction word has an opcode, register, and operand. The number of bits available for the operand field is 7. The maximum possible value of the general-purpose register is ______.Correct answer is '1023'. Can you explain this answer?
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