Test by Bikram | Mock GATE | Test 1 | Question: 53

Question

A computer uses a memory unit with $512K$ words of $64$ 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 $256$ registers), and an address part. The number of bits in the operation code are _______.

Comments

why answer is given in range

i think it is only 36

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Is it ideal to assume word-addressable or byte-addressable memory here?

Answer 1

64-(indirect bit(1)+num of register(8)+adressing bit(19)

=64-28

=36

Comments

I also assumed byte addressable and got 33

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@giggz07 and @pratyush

yes by default we use Byte Addressable , as nothing is mention there in the question .. 

i will consider both value which come as Byte and Word addressable ..

now add that range in the answer.

If interested , read:

http://stackoverflow.com/questions/2724449/difference-between-word-addressable-and-byte-addressable

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why indirect bit is 1?