A computer uses a memory unit of 512 K words of 32 bits each. A binary instruction code is stored in one word of the memory. The instruction has four parts: an addressing mode field to specify one of the two-addressing mode (direct and indirect), an operation code, a register code part to specify one of the 256 registers and an address part. How many bits are there in addressing mode part, opcode part, register code part and the address part?
(A)
1, 3, 9, 19
(B)
1, 4, 9, 18
(C)
1, 4, 8, 19
(D)
1, 3, 8, 20
Answer: (C)
Explanation:
=> Indirect addressing bit = 1
=> 512 = 19 bits
=> There are 256 registers.So we need ⌈log 256⌉ = 8 bits for identifying a register. i.e register code part needs 8 bits.
=> There are 512K words.i.e 512∗210 = 2 19
=> So we need ⌈log219⌉=19
=> ⌈log2219⌉=19 bits for identifying a word. i.e,address part needs 19 bits
=> 19+8+1-32 = 4
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