morris mano

Question

• A. If a computer has 128 operation codes and 512 k address, how many bits would be required for

(1) Single address instruction

(2) Two address instruction

Answer 1

No of bits required for opcode are: 7 bits (2^7=128)

No of bits required for address :19 bits as 512K =2^9*2^10

For single address instruction we just have the opcode and single address .So 7+19=26 bits are required.

For double address instructions we have the opcode and two addresses so 7+19+19=45 bits are required.

Correct me if I am wrong