Chapter 5, Problem 5E

a.

Explanation of Solution

Explanation:

The following table describes about the 2’s complement representation of an integer for the given value

...

Value of an integer	2’s complement
2	0010
A	1010
C	1100
2	0010

b.

Explanation of Solution

Explanation:

• 2A of big endian represents the first bit as 0 which is a positive number and the most significant bit for the 2’s complement representation is SIGN...

c.

Explanation of Solution

Explanation:

The following table describes about the representation of Institute of Electrical and Electronic Engineers (IEEE) single precision floating point

Sign	Exponent	Mantissa
-1	-8	-23

The following table describes about the decimal equivalent for the above values

Sign	Exponent	Mantissa
-1	-8	-23
0	010101011	10000100000100000011011

The decimal equivalent of the address stored at address 100 is $1$

d.

Explanation of Solution

Explanation:

The size of the data stored by the address 100 is 8 Bytes

The little endian is stored at last and the most significant bit is 0 and the absolute value of the integer is n-1 bits.

Therefore, the little endian of 32-bit integer number stored at address 100 is $1$

e.

Explanation of Solution

Explanation:

• 1B of little endian represents the first bit as 0 which is a positive number and the most significant bit for the 2’s complement representation is SIGN...

f.

Explanation of Solution

Explanation:

The following table describes about the representation of Institute of Electrical and Electronic Engineers (IEEE) single precision floating point

Sign	Exponent	Mantissa
-1	-8	-23

The following table describes about the decimal equivalent for the above values

Sign	Exponent	Mantissa
-1	-8	-23
0	010101011	10000100000100000011011

Little endian is reverse of big endian and hence the decimal equivalent is calculated as follows

$\begin{array}{l}\Rightarrow {(-1)}^{\text{sign}}\times {(1-(\text{fraction}))}_2\times 2^{(\text{e}-12}\end{array}$