09/30/2006

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A student wrote, asking about problem 2(b) on the last problem set.
BUT, I was heading out the door to catch a flight, so I asked Veynu
to respond.  As we always do, if the response carries information
that could be useful to others, we prefer to share it with you all.

Yale Patt

> From narasima@ece.utexas.edu Thu Sep 28 03:05:36 2006
> Subject: Re: Problem Set
> Date: Thu, 28 Sep 2006 03:05:37 -0500
>
> Hello ********, I am responding on behalf of Dr. Patt
> since he is out of town for a few days.
>
> > Prof. Patt,
> >
> > On question #2 letter 'b', should it read from 12 to 60
> > instead of 4 to 60?
>
> No, it should stay as it is, 4 to 60 is correct.
>
> > 4 is the number of address lines, but there are
> > 12 locations since it has an addressability of 3 bits.
>
> No, 4 is NOT the number of address lines.  The number of
> address lines refers to how many bits wide the address is.
> In the example memory from figure 3.21 in the book, the number
> of address lines is 2 which is represented by A[1:0].  So,
> with 2 bits of address we can uniquely identify 2^2 = 4 memory
> locations.  Another way of saying this is that the "address
> space" of the memory is 4 locations.  So in this example memory,
> there are 4 memory locations, not 12 as you said above.  Within
> each of the 4 memory locations, 3 bits are stored, so we would
> say the memory has an "addressability" of 3 bits.
>
> I hope this clears things up, if it is still unclear please
> see a TA in office hours.
>
> > Thanks for yout time,
> > <<name withheld ...>>
>
> You are welcome.
>
> --Veynu

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