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?

This is sometimes referred to as parsing

Answer 2

The process of changing one data type into another.

Question 3

What is a character set?

What are some examples?

Answer 3

• A list of 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

Left shift the following 8-bit number 3 places: 00111010

What mathematical operation is this equivalent to?

Answer 6

Remove the required number of bits from the left

Add the same number of zeros to the right

Answer = 11010000

Equivalent to multiplying the number

Question 7

Right shift the following 8-bit number 3 places: 00111010

What mathematical operation is this equivalent to?

Answer 7

Remove the required number of bits from the right

Add the same number of zeros to the left

Answer = 00000111

Equivalent to dividing the number

Question 8

Convert 177 to an unsigned 8-bit binary number

Answer 8

• Use the binary number line (128 | 64 | 32 | 16 | 8 | 4 | 2 | 1)
• 128+32+16+1 = 177

Answer = 10110001

Question 9

Convert the unsigned 8-bit binary number 10110010 to denary

Answer 9

Use the binary number line (128 | 64 | 32 | 16 | 8 | 4 | 2 | 1)

128+32+16+2 = 178

Question 10

Convert the denary number 188 to Hex

Answer 10

• Binary = 010111100
• Split from right to left into nibbles = 1011 1100
• Convert to denary using 8421 number line for each nibble
• 1011 = 11 | 1100 = 12
• Remember A = 10 | F = 15

Answer = BC

Question 11

Convert the hex FE to a denary number

Answer 11

• Remember A = 10 | F = 15
• F = 15 | E = 14
• Write each character in a 4-bit nibble = 1111 1110
• Combine into one binary number = 11111110
• Convert to denary = 128+64+32+16+8+4+2

Answer = 254

Question 12

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

Answer 12

• Write out the positive version of 49 in binary = 00110001
• Flip and add one to the bits (2’s Complement) = 11001111

Answer = 11001111

Alternative methods exist using (-128) + 64 + 8 + 4 + 2 + 1 = -49

Question 13

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

Answer 13

• No flipping required as it is a positive number
• You only need to flip when it requires a negative number

Answer = 00110001

Question 14

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

Answer 14

• Your left-most bit (most significant) is now only available to store a 1 for negative OR 0 for positive.
• Use the remaining bits to make the number (-) 0110001

Answer = 10110001

Question 15

Add the following two binary numbers

01101010 + 00111111

Answer 15

Stack the numbers on top of each other and use the rules of binary addition going from right to left

• 0 + 0 = 0
• 1 + 0 = 1
• 1 + 1 = 0 (carry 1)
• 1 + 1 + 1 = 1 (carry 1)

01101010
00111111 +

Answer = 10101001

Carries - 11111100

Question 16

Subtract the following two binary numbers

00101111 - 00010111

Answer 16

• Convert the second number to a negative version (flip+add 1)
• Stack the original first number on top of the new negative second number
• Follow the rules of binary addition

Answer = 00011000

An overflow should always occur using this method

An alternative borrowing method is also suitable

Question 17

Mask 11010101 using the AND mask 10101010

Answer 17

• Stack the two binary numbers on top of each other
• Going from right to left - apply the rules of an AND gate

Answer = 10000000

Question 18

Mask 11010101 using the OR mask 10101010

Answer 18

• Stack the two binary numbers on top of each other
• Going from right to left - apply the rules of an OR gate

Answer = 11111111

Question 19

Mask 11010101 using the XOR mask 10101010

Answer 19

• Stack the two binary numbers on top of each other
• Going from right to left - apply the rules of an XOR gate

Answer = 01111111

Question 20

Normalise 0001011000 001100

Answer 20

• A normalised number can only start with 01 if positive, or 10 if it is negative
• Remove the excess bits from the front of the number - be careful not to change the sign of the number. (01011000)
• For each digit removed - pad the right of the mantissa with 0s (0101100000)
• Convert the original exponent to denary and subtract the number of number bits that you removed from the front of the mantissa (12-2)
• Convert the exponent back to binary (it could be a negative number)

Answer = 0101100000 001010

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

Alternative methods are available

Question 21

Normalise
1110011000 001100

Answer 21

• A normalised number can only start with 01 if positive, or 10 if it is negative
• Remove the excess bits from the front of the number - be careful not to change the sign of the number (10011000)
• For each digit removed - pad the right of the mantissa with 0s (1001100000)
• Convert the original exponent to denary and subtract the number of number bits that you removed from the front of the mantissa (12-2)
• Convert the exponent back to binary (it could be a negative number)

Answer = 1001100000 001010

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

Alternative methods are available

Question 22

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

Answer 22

• Write out the number in fixed point binary (0101.01)
• Float the binary point to be to the left of the first 1 (0.10101)
• Write out the number without the binary point (010101) - this is your mantissa. If it is not long enough then pad on the right with zeros.
• In binary, write out how many places you floated the binary point (011) this is your exponent. If it is not long enough then pad on the left with zeros.

Answer = 010101 011

Alternative methods are available

Question 23

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

Answer 23

• Write out the number in fixed point binary (0101.01)
• Float the binary point to be to the left of the first 1 (0.10101)
• Write out the number without the binary point (010101) - this is your mantissa. If it is not long enough then pad on the right with zeros.
• In binary, write out how many places you floated the binary point (011) this is your exponent. If it is not long enough then pad on the left with zeros.
• We need the mantissa to be negative - so flip and add one to the bits.

Answer = 101011 011

Alternative methods are available

Question 24

Convert 01110 0001 to floating point denary

Answer 24

• Convert the exponent (0001) to denary (Exp = 1)
• If the mantissa is negative then flip and add one to it.
• Add the binary point to the left of the first 1 in the mantissa (0.1110)
• Float the binary point as dictated by the exponent - Right if negative exponent OR left if positive exponent (01.110 in fixed point).
• Calculate the denary value - normal binary to the left of the binary point and fractions to the right of it (1/2 | 1/4 | 1/8 | 1/16)

Answer = 1.75

Alternative methods are available

Question 25

Convert **10010 0001** to floating point denary

Answer 25

* Convert the exponent (0001) to denary (Exp = 1) * If the mantissa is negative then flip and add one to it (01110) * Add the binary point to the left of the first 1 in the mantissa (0.1110) * Float the binary point as dictated by the exponent - Right if negative exponent OR left if positive exponent (01.110 in fixed point). * Calculate the denary value - normal binary to the left of the binary point and fractions to the right of it (1/2 | 1/4 | 1/8 | 1/16) **Answer = -1.75** ## Footnote Alternative methods are available

Question 26

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

Answer 26

* Write out the number in fixed point (0.0010) - 1/8 * Calculate how many places the binary point needs to float to be to the left of the left-most one (2 places to the right, this is a negative movement) * Remove the additional zeros from the front of the mantissa to get it to be 01 for positive or 10 for negative. * Pad the mantissa on the right hand side to make it to the required number of bits (010000) * Write out your exponent 010 - flip to be negative 110 **Answer = 010000 110** ## Footnote **Any number less than 0.5 would have a negative exponent**

Question 27

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

Answer 27

* Ignore the negative - write out the number in fixed point (0.0010) - 1/8 * Calculate how many places the binary point needs to float to be to the left of the left-most one (2 places to the right, this is a negative movement) * Remove the additional zeros from the front of the mantissa to get it to be 01 for positive or 10 for negative. * Pad the mantissa on the right hand side to make it to the required number of bits (010000) * Flip and add one to the mantissa to make it negative. * Write out your exponent 010 - flip to be negative 110 **Answer = 110000 110** ## Footnote Alternative methods are available

Question 28

**Add** the following floating point binary numbers stored in the two's complement format. **0110 0010 and 0100 0011**

Answer 28

* Calculate the exponent of the first number (0010 = 2) * Write out the first number with the binary point - Initially it will be before the left-most one - float it the required number of places dictated by the exponent (011.0) * Calculate the exponent of the second number (0011 = 3) * Calculate the second number - float it the required places (0100.0) * Stack the two numbers on top of each other and align the binary points * Perform binary addition from right to left (0111.00) * Using the set number of bits, write out the mantissa without its point (0111) * In binary write out how far you moved the binary point. Pad to the left to get it to the correct number of bits (3 = 0011) **Answer = 0111 0011** ## Footnote Always check your answer by converting both numbers to denary and doing the normal addition (3 + 4 = 7)

Question 29

Why do we normalise floating point numbers?

Answer 29

* Allows for more **accuracy/precision** from the given number of bits * So that the representation of each binary value is unique

Question 30

**Subtract** the following floating point binary numbers stored in the two's complement format. 010010 0100 - 010010 0010

Answer 30

* Calculate the exponent of the first number (0100 = 4) * Write out the first number with the binary point - Initially it will be before the left-most one - float it the required number of places dictated by the exponent (01001.0) * Calculate the exponent of the second number (0010 = 2) * Calculate the second number - float it the required places (010.010) * Flip and add one to the second number (101.110) * Stack the two numbers on top of each other and align the binary points - pad the negative number on the left with 1's if required. * Perform binary addition from right to left (0110.110 - you will get an overflow) * Using the set number of bits, write out the mantissa without its point (0111) * In binary write out how far you moved the binary point. Pad to the left to get it to the correct number of bits (3 = 0011) **Answer = 0110110 0011** ## Footnote Always check your answer by converting both numbers to denary and doing the normal subtraction (9 - 2.25 = 6.75)

Question 31

What impact does **increasing** the size of the **exponent** have?

Answer 31

Increases the size of the number that can be stored.

Question 32

What impact does **increasing** the size of the **mantissa** have?

Answer 32

Increases the precision of the number that can be stored.