Sat, 7 Mar 2009, 23:35

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A student writes (and I apologize for the delay since I am on the east coast
for a few days):

Hi Prof. Patt,

Problem Set 2 had a question:

6 Answer the following short questions:

a. A memory's addressability is 64 bits. What does that
tell you about the sizes of the MAR and the MDR?

The posted solution states:

6 a. MDR is 64 bits (same as addressability).

Hmmn.

We cannot tell the size of the MAR, since it depends on the
number of memory locations and does not depend
on the addressability (the number of bits in each location).

Well, at least one of the solutions is unambiguous.

We have confusion here, regarding the size of MDR.
Should not the size of MDR be dependent on the size of data
bus (architecture dependent), rather than addressability of memory?

Apparently one of my TAs is busy challenging you.  I did not realize this
had slipped in.  The real answer is that you are correct - MDR is not
dependent on addrssability, ...BUT also NOT on the size of the data bus as
you suggest.  Rather, the size of MDR is generally the same size as the size
of the most common data type.

So, what then?  Since the addressibility is 64 bits, my TA must have expected
you to think: Each 64 bits has its own unique address.  Hmmn, sounds like
floating point to me and 64 bits must be the default size of a data element
(64 bit double floating point).  Taking that one step further (the MDR
accomodates the normal data size), we have MDR containing 64 bits.

Truth is, I would have preferred my TA was not so tricky, but (a) giving him
some rope encourages him to challenge you and that is a good thing, and (b)
it is only one item on a problem set and won't really affect your grade,
except that if you chew on it (like you are clearly doing), it could result
in a better grade on the exam, and that WILL affect your grade.

Good luck on the midterm.

Yale Patt

Thanks

<<name withheld to protect the student who was tricked by my TA.

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