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Defining The Problem & Planning Software Solutions ,[object Object]
Syllabus ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Why are they needed? 3 2 1
Hi I am going to tell you about why computers require data types and data structures.
INPUT Firstly, when we use a computer that is classified as INPUT.
This raw rata is converted into binary – machine code via the electronic signals on the computers keyboard.
As you know this information needs to be processed so information can be generated and OUTPUTTED.
However, this data must be stored in similar categories for processing.
A B C 1 55 7 8 90 22/3/80 $22.00 All the data is filtered into categories for storage and subsequent processing. Integer Character String Date
So therefore we need data types to organise data for processing.
Examples of Data DATA
How is data stored? ,[object Object],[object Object],[object Object]
How is data stored? 1001 1002 1003 1004 1005 Computer -  Store 10 in 1001 Programmer –  Num1 = 10 10 Num1
How is data stored? 1001 1002 1003 1004 1005 Computer -  Store 12 in 1002 Programmer –  Num2 = 12 10 12 Num2
How is data stored? 1001 1002 1003 1004 1005 Computer –  ADD 1001 to 1002 and STORE in 1003 Programmer –  Total = Num1 + Num2  10 12 22 Total
Number Representation ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Binary Number System ,[object Object],[object Object],[object Object],[object Object]
Binary Number System Number 1 0 1 1 1 Binary Place 2  4 2  3 2  2 2  1 2  0 Binary Place Value 16 8 4 2 1 Binary Calculation 1 x 2  4 0 x 2  3 1 x 2  2 1 x 2  1 1 x 2  0 Binary Value 16 0 4 2 1 Binary Total = 16 + 4 + 2 + 1 = 23
Octal Number System ,[object Object],[object Object]
Octal Number System Number 3 4 5 2 1 Octal Place 8  4 8  3 8  2 8  1 8  0 Octal Place Value 4 096 512 64 8 1 Octal Calculation 3 x 8  4 4 x 8  3 5 x 8  2 2 x 8  1 1 x 8  0 Octal Value 12 288 2 048 320 16 1 Octal Total = 12 288 + 2 048 + 320 + 16 + 1 = 14 673
Hexadecimal Number System ,[object Object],[object Object]
Hexadecimal Number System Number 1 2 4 A C Hex Place 16  4 16  3 16  2 16  1 16  0 Hex Place Value 65 536 4 096 256 16 1 Hex Calculation 1 x 16  4 2 x 16  3 4 x 16  2 10 x 16  1 12 x 16  0 Hex Value 65 536 8 192 1 024 160 12 Hex Total = 65 536 + 8 192 + 1 024 + 160 + 12 = 74 924
Activity 1 ,[object Object],[object Object]
Activity 2 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Activity 3 Decimal Binary Octal Hexadecimal 45 10 1010111 2 75 8 3FC 16
Data Types ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Activity 4 DATA TYPE DESCRIPTION EXAMPLES Integer 5 0 -4 87 Floating Point 2.764 -67665.3 2.3 x 10 8 String Cat Daggy2 Eggplant Boolean True  False Yes No Date 16/1/2006 4 th  June 2001 Currency $5.84 -$58.97 $1005.46
Variables and Constants ,[object Object],[object Object],[object Object],[object Object],[object Object]
Naming Variables ,[object Object],[object Object],[object Object],[object Object]
Activity 5 ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Homework ,[object Object],[object Object]

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Data types

  • 1.
  • 2.
  • 3. Why are they needed? 3 2 1
  • 4. Hi I am going to tell you about why computers require data types and data structures.
  • 5. INPUT Firstly, when we use a computer that is classified as INPUT.
  • 6. This raw rata is converted into binary – machine code via the electronic signals on the computers keyboard.
  • 7. As you know this information needs to be processed so information can be generated and OUTPUTTED.
  • 8. However, this data must be stored in similar categories for processing.
  • 9. A B C 1 55 7 8 90 22/3/80 $22.00 All the data is filtered into categories for storage and subsequent processing. Integer Character String Date
  • 10. So therefore we need data types to organise data for processing.
  • 12.
  • 13. How is data stored? 1001 1002 1003 1004 1005 Computer - Store 10 in 1001 Programmer – Num1 = 10 10 Num1
  • 14. How is data stored? 1001 1002 1003 1004 1005 Computer - Store 12 in 1002 Programmer – Num2 = 12 10 12 Num2
  • 15. How is data stored? 1001 1002 1003 1004 1005 Computer – ADD 1001 to 1002 and STORE in 1003 Programmer – Total = Num1 + Num2 10 12 22 Total
  • 16.
  • 17.
  • 18. Binary Number System Number 1 0 1 1 1 Binary Place 2 4 2 3 2 2 2 1 2 0 Binary Place Value 16 8 4 2 1 Binary Calculation 1 x 2 4 0 x 2 3 1 x 2 2 1 x 2 1 1 x 2 0 Binary Value 16 0 4 2 1 Binary Total = 16 + 4 + 2 + 1 = 23
  • 19.
  • 20. Octal Number System Number 3 4 5 2 1 Octal Place 8 4 8 3 8 2 8 1 8 0 Octal Place Value 4 096 512 64 8 1 Octal Calculation 3 x 8 4 4 x 8 3 5 x 8 2 2 x 8 1 1 x 8 0 Octal Value 12 288 2 048 320 16 1 Octal Total = 12 288 + 2 048 + 320 + 16 + 1 = 14 673
  • 21.
  • 22. Hexadecimal Number System Number 1 2 4 A C Hex Place 16 4 16 3 16 2 16 1 16 0 Hex Place Value 65 536 4 096 256 16 1 Hex Calculation 1 x 16 4 2 x 16 3 4 x 16 2 10 x 16 1 12 x 16 0 Hex Value 65 536 8 192 1 024 160 12 Hex Total = 65 536 + 8 192 + 1 024 + 160 + 12 = 74 924
  • 23.
  • 24.
  • 25. Activity 3 Decimal Binary Octal Hexadecimal 45 10 1010111 2 75 8 3FC 16
  • 26.
  • 27. Activity 4 DATA TYPE DESCRIPTION EXAMPLES Integer 5 0 -4 87 Floating Point 2.764 -67665.3 2.3 x 10 8 String Cat Daggy2 Eggplant Boolean True False Yes No Date 16/1/2006 4 th June 2001 Currency $5.84 -$58.97 $1005.46
  • 28.
  • 29.
  • 30.
  • 31.