Ace Test Series: CO & Architecture - Cache Memory

Question

 

Comments

I am getting 19 bits as the answer anybody verify

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Yes it should be 16+3 = 19.

Assuming memory is byte addressable

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@Kunal Kadian 3 came from??

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@manisha11 tag bits will obviously be needed and for block we will take as byte addressable and so 64 bits  = 64/8 = 8 bytes = 2^3 , so min value of x = 16+3 = 19 

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oh yes 64/8 thanks

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I am also getting 19 as answer Any one please Verifiy….

Answer 1

To get minimun value of X , you should consider that it is block addressable where block size should be 64 bits.

So minimun number of bits will be only for tag bits.

So x=16.

Comments

@Rishabh Agrawal

does this mean here one word is of 64bits ?

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Actually somewhere it is Block addressable while somewhere it is word addressable.Just different terminology.

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@kman30

Yes