1.4.1 Data Types Flashcards

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1
Q

What are the 5 primitive data types?

A
  • 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)
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2
Q

What is casting?

A

The process of changing one data type into another.

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3
Q

What is a character set?

What are some examples?

A
  • 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
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4
Q

Describe the ASCII character set

A
  • 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
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5
Q

Describe the UNICODE character set

A
  • 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)
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6
Q

Convert 177 to an unsigned 8-bit binary number

A

10110001

128+32+16+1 = 177

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7
Q

Convert the unsigned 8-bit binary number 10110010 to denary

A

128+32+16+2 = 178

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8
Q

Convert 188 to Hex

A

BC

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9
Q

Convert the hex FE to a denary number

A

11111111 = 254

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10
Q

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

A

11001111

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11
Q

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

A

00110001

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12
Q

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

A

10110001

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13
Q

Add the following two binary numbers

01101010 + 00111111

A

10101001

Carries - 11111100

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14
Q

Subtract the following two binary numbers

00101111 - 00010111

A

00011000

An overflow should have occurred

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15
Q

Normalise
0001011000 001100

A

0101100000 001010

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

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16
Q

Normalise
1110011000 001100

A

1001100000 001010

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

17
Q

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

A

010101 011

18
Q

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

A

101011 011

19
Q

Convert 01110 0001 to floating point denary

A

1.75

20
Q

Convert 10010 0001 to floating point denary

A

-1.75

21
Q

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

A

010000 110

22
Q

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

A

110000 110

23
Q

Why do we normalise floating point numbers?

A

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

So that the representation of each binary value is unique

24
Q

What impact does increasing the size of the exponent have?

A

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

25
Q

What impact does increasing the size of the mantissa have?

A

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