Computer Science, asked by Kavyashree8079, 1 year ago

A computer uses memory unit of 512 words of 16 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 to specify one of 8registers and an address part. How many bits are there in each part?

Answers

Answered by adikhan
3
well your question is not clear enough but i will try to get the meaning.
..
wait
indirect bit = 1
adress part = 2^9(512)=9 bits
register = 2^3 (8) = 3 bits
operation code =3 bits
Answered by payalchatterje
0

Answer:

Required number of bits are 4 bits.

Explanation:

Given,

Memory unit = 512 K words

1 word = 32 bits = 4 B

1  \: word = 32  \: bits = 4  \: B

Binary instruction code stored in 1 word of memory

Addressing mode = 1 bit for direct or indirect

Register code = ln (256) = 8 bits

Since, registers are already addressed, only memory needs to be addressed Address  \:  \: part = ln  \: (512 K) = 19  \: bits

Operation mode = = 32 – 1 – 8 – 19 = 4  \: bits

Therefore, bits for addressing mode part, opcode part, register code part and the address part = (1, 4, 8, 19)

Required number of bits are 4 bits.

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