A processor has 512 distinct instructions and 80general purpose registers. A 24 bits instructionword has an opcode, one register operand anda memory operand. How many bits arereserved for memory operand field.
Question
A processor has 512 distinct instructions and 80general purpose registers. A 24 bits instructionword has an opcode, one register operand anda memory operand. How many bits arereserved for memory operand field.
Solution
To solve this problem, we need to first determine the number of bits needed for the opcode and the register operand.
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Opcode: Since there are 512 distinct instructions, we need log2(512) = 9 bits to represent the opcode.
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Register Operand: Since there are 80 general purpose registers, we need log2(80) = 7 bits to represent the register operand. However, since we can't have a fraction of a bit, we round this up to the nearest whole number, which is 8 bits.
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Now, we subtract the number of bits used for the opcode and the register operand from the total number of bits in the instruction word.
So, 24 bits (total bits in instruction word) - 9 bits (for opcode) - 8 bits (for register operand) = 7 bits.
Therefore, 7 bits are reserved for the memory operand field.
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