A computer uses a memory unit with 256K words of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code, a register code part to specify one of 64 registers, and an address part. How many bits are there in the operation code, the register code part and the address part?

1. A7,6,18
2. B6,7,18
3. C7,7,18
4. D18,7,7 x86

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Correct answer: A. 7,6,18

Explanation

256K = 2¹⁸ words, so the address needs 18 bits; 64 registers need 6 bits; the indirect bit takes 1. Opcode = 32 − 1 − 6 − 18 = 7.