1.4.1 Data Types

Question 1

What are the 5 primitive data types?

Answer 1

• Real / Floating Point – Stores decimal numbers (3.141)
• Character - A single letter, number or special character (‘H’)
• String – A collection of characters (“Hello World”)
• Boolean – TRUE or FALSE
• Integer – A positive or negative whole number (24, -34)

Question 2

What is casting?

Answer 2

The process of changing one data type into another.

Question 3

What is a character set?

What are some examples?

Answer 3

• List of all the characters the computer can represent.
• Each character is represented by a unique binary value.
• Used to map binary values to characters.
• Examples: UNICODE and ASCII

Question 4

Describe the ASCII character set

Answer 4

• ASCII is a character set which is a subset of UNICODE
• Uses 7 bits, or 8 bits for extended ASCII
• Fewer characters can be represented than UNICODE
• Characters from different languages cannot be represented in ASCII

Question 5

Describe the UNICODE character set

Answer 5

• Each character is represented by 1-4 bytes.
• It supports a very large number of characters
• It is backwards compatible with ASCII
• Text using UNICODE rather than ASCII would take up more storage (roughly 4 times more)

Question 6

Convert 177 to an unsigned 8-bit binary number

Answer 6

10110001

128+32+16+1 = 177

Question 7

Convert the unsigned 8-bit binary number 10110010 to denary

Answer 7

128+32+16+2 = 178

Question 8

Convert 188 to Hex

Answer 8

BC

Question 9

Convert the hex FE to a denary number

Answer 9

11111111 = 254

Question 10

Convery -49 to an 8-bit binary number using two’s complement

Answer 10

11001111

Question 11

Convert 49 to an 8-bit binary number using two’s complement

Answer 11

00110001

Question 12

Convert -49 to an 8-bit binary number using sign and magnitude

Answer 12

10110001

Question 13

Add the following two binary numbers

01101010 + 00111111

Answer 13

10101001

Carries - 11111100

Question 14

Subtract the following two binary numbers

00101111 - 00010111

Answer 14

00011000

An overflow should have occurred

Question 15

Normalise
0001011000 001100

Answer 15

0101100000 001010

(Remove the two extra bits from the front of the mantissa and reduce the exponent value by 2)

Question 16

Normalise
1110011000 001100

Answer 16

1001100000 001010

(Remove the two extra bits from the front of the mantissa and reduce the exponent value by 2)

Question 17

Convert 5.25 to an 6-bit mantissa and 3-bit exponent

Answer 17

010101 011

Question 18

Convert -5.25 to an 6-bit mantissa and 3-bit exponent

Answer 18

101011 011

Question 19

Convert 01110 0001 to floating point denary

Answer 19

1.75

Question 20

Convert 10010 0001 to floating point denary

Answer 20

-1.75

Question 21

Convert 0.125 to an 6-bit mantissa and 3-bit exponent

Answer 21

010000 110

Question 22

Convert -0.125 to an 6-bit mantissa and 3-bit exponent

Answer 22

110000 110

Remember to normalise

Question 23

Why do we normalise floating point numbers?

Answer 23

Allows for more accuracy/precision from the given number of bits

So that the representation of each binary value is unique

Question 24

What impact does increasing the size of the exponent have?

Answer 24

Increasing the number of bits used for the exponent increases the size of the number that can be stored.

Question 25

What impact does increasing the size of the mantissa have?

Answer 25

Increasing the number of bits used for the mantissa increases the precision of the number that can be stored.