Question
Given a 16-bit instruction format with 4 bits for the
opcode and 12 bits for the address, calculate the maximum number of different opcodes that can be accommodated.Solution
The number of different opcodes that can be accommodated is given by 2^n, where n is the number of bits dedicated to the opcode. In this case, n = 4 (4 bits for opcode) Maximum number of opcodes = 2^4 = 16
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