Question

(a) How many locations of memory can you address with 12-bit memory address? (b) How many bits are required to address a 2-Mega-location memory, i.e, what should the length of the memory address be in order to access a 2M memory? (c) Assume intArray is an integer array with 20 elements stored in memory and its starting memory address is in $s2. What is the memory address for element intArray [10]? (d) Assume charArray is a character array with 20 elements stored in memory and its starting memory address is in $t5. What is the memory address for element charArray[5]?

Answer 1

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Let,

The address of 1-bit  memory  to add in 2 location:

$\to \frac{0}{1} =2^1 \ (\frac{m}{m} \ location)$

The address of 2-bit  memory to add in 4 location:

$\to \frac{\frac{00}{01}}{\frac{10}{11}} =2^2 \ (\frac{m}{m} \ location)$

similarly,

Complete 'n'-bit memory address' location number is = $2^n.$Here, 12-bit memory address, i.e. n = 12, hence the numeral. of the addressable locations of the memory:

$= 2^n \\\\ = 2^{12} \\\\ = 4096$

In point b:

$\to Let \ Mega= 10^6$

              $=10^3\times 10^3\\\\= 2^{10} \times 2^{10}$

So,

$\to 2 \ Mega =2 \times 2^{20}$

                 $= 2^1 \times 2^{20}\\\\= 2^{21}$

The memory position for '$2^n$' could be 'n' m bits'  

It can use $2^{21}$ bits to address the memory location of 21.  

That is to say, the 2-mega-location memory needs '21' bits.  

Memory Length = 21 bit Address

In point c:

$i^{th}$ element array addresses are given by:

$\to address [i] = B+w \times (i-LB)$

$_{where}, \\\\B = \text {Base address}\\w= \text{size of the element}\\L B = \text{lower array bound}$

$\to B=\ 52\\\to w= 4 byte\\ \to L B= 0\\\to address = 10$

$\to address [10] = \ 52 + 4 \times (10-0)$

                       $= \ 52 + 40 \ bytes$

1 term is 4 bytes in 'MIPS,' that is:

$= \ 52 + 10 \ words\\\\ = \ 512$

In point d:

$\to base \ address = \ t 5$

When MIPS is 1 word which equals to 32 bit :

In Unicode, its value is = 2 byte

In ASCII code its value is = 1 byte

both sizes are  < 4 byte

Calculating address:

$\to address [5] = \ t5 + 4 \times (5-0)$

                     $= \ t5 + 4 \times 5\\ \\ = \ t5 + 20 \\\\= \ t5 + 20 \ bytes \\\\= \ t5 + 5 \ words \\\\= \ t 10 \ words$