1. Introduction to Computing Dr. Nadeem A Khan

2. Lecture 10

3. Sub Text1_KeyPress(KeyAscii as Integer) statements statements End Sub ► Text1_KeyPress event will occur when Text1 has the focus and a key is pressed Text1 has the focus and a key is pressed The Keypress Event Procedure

4. Sub Text1_KeyPress(KeyAscii as Integer) statements statements End Sub ► Keyascii  is a variable (of type Integer)  gets the ANSI value of the pressed key  value is used to display the corresponding character in the Text1 at the end of this procedure The Keypress Event Procedure (Contd.)

5. What will happen in these cases? Sub Text1_KeyPress(KeyAscii as Integer) Let KeyAscii =65 Let KeyAscii =65 End Sub Sub Text1_KeyPress(KeyAscii as Integer) Let KeyAscii =0 Let KeyAscii =0 End Sub The Keypress Event Procedure (Contd.)

6. ► Already known:  String  Single  Integer Data Types

7. ► Strings Storage: 10 bytes + string length Range: 0 to app. 2 billions chars. Declaration: Dim strVarLen As String (declares a string of a variable length) (declares a string of a variable length) Data Types (Contd.)

8. ► Fixed-Length Strings Storage: Length of string Range: 1 to app. 65, 400 chars. Declaration: Dim strFixLen As String * 2 (declares a string of a fix size of 2 chars.) Data Types (Contd.)

9. ► Fixed-Length Strings Usage:Example Dim strText As String * 5 Let strText = “Hello” Picture1.Print strText Let strText = “H” Picture1.Print strText Let strText = “HelloWorld” Picture1.Print strText Data Types

10. ► Fixed-Length Strings Usage:Result: Hello(Complete string) H…. (H followed 4 spaces) Hello(First 5 characters only) => The length is fix Data Types

11. ► Integers Storage: 2 bytes Range: -32,768 to 32,767 Declaration: Dim intExample As Integer (declares intExample as an Integer variable) (declares intExample as an Integer variable) Data Types (Contd.)

12. ► Integers Usage:Example Dim count As Integer Let count= 6 Picture1.Print count Let count= count+1 Picture1.Print count Let count= 6/5 Picture1.Print count Let count= 2.33333 * 2 Picture1.Print count Data Types

13. ► Integer Usage:Result67 1 (rounding to lower value) 5(rounding to higher value) => takes only whole number values => takes only whole number values Data Types

14. ► Long (Integer) Storage: 4 bytes Range: -2,147,483,648 to 2,147,483,647 Declaration: Dim lngExample As Long (declares lntExample as a long variable) (declares lntExample as a long variable) Data Types (Contd.)

15. ► Long (Integer) Usage: Same as Integer Type except the range is much larger much larger Data Types (Contd.)

16. ► Byte Storage: 1 byte Range: 0 to 255 Declaration: Dim bytExample As Byte (declares bytExample as a Byte type variable) (declares bytExample as a Byte type variable) Data Types (Contd.)

17. ► Byte Usage: Same as Integer Type except the range is positive and much smaller positive and much smaller Data Types (Contd.)

18. ► Boolean Storage: 2 bytes Range: TRUE(1) or FALSE(0) Declaration: Dim blnState As Boolean (declares a Boolean type variable blnState) Data Types (Contd.)

19. ► Boolean Usage:Example Dim blnExample As Boolean Let blnExample= FALSE Picture1.Print blnExample Let blnExample= 1 Picture1.Print blnExample Let blnExample= 6 Picture1.Print blnExample Let blnExample= -8*7+5.2 Let blnExample= -8*7+5.2 Picture1.Print blnExample Data Types (Contd.)

20. ► Boolean Usage:Example FALSE FALSE TRUE TRUE =>Values other than 0 are TRUE =>Values other than 0 are TRUE Data Types (Contd.)

21. ► Single (Precision Floating-Point) Storage: 4 bytes Range: -3.4…E38 to -1.4…E-45 (negative) 1.4…E-45 to 3.4…E38 (positive) Declaration: Dim sngAverage As Single (declares a Single type variable sngAverage) Data Types (Contd.)

22. ► Double (Precision Floating-Point) Storage: 8 bytes Range: -1.7…E308 to -4.9…E-324 (negative) 4.9…E-324 to 1.7…E308 (positive) 4.9…E-324 to 1.7…E308 (positive)Declaration: Dim dblAverage As Double (declares a Double type variable dblAverage) Data Types (Contd.)

23. ► Double Usage:Example Dim sngValue As Single, dblValue As Double Let sngValue= 1/3 Picture1.Print sngValue Let dblValue= 1/3 Picture1.Print dblValue Data Types (Contd.)

24. ► Double Usage:Result 0.3333333(Single precision) 0.333333333333333 (Double precision) => Value of 1/3 represented more accurately by double than by single by double than by single Data Types (Contd.)

25. ► Double Usage:Example Dim sngValue As Single, dblValue As Double Let sngValue= 1/3 Let sngValue= sngValue * 100000 Picture1.Print sngValue Let dblValue= 1/3 Let dblValue= dblValue * 100000 Picture1.Print dblValue Data Types (Contd.)

26. ► Double Usage:Result 33333.34(Single precision; rounding error) 33333.3333333333 (Double precision) => - The decimal point is floating; - Eventually both will be subjected to rounding errors value increases to large values value increases to large values - Still Double will remain more precise than Single Data Types (Contd.)

27. ► Currency Storage: 8 bytes Range: -922,337,203,685,477.5808 to 922,337,203,685,477.5807 922,337,203,685,477.5807Declaration: Dim curRevenue As Currency (declares a Currency type variable curRevenue) Data Types (Contd.)

28. ► Currency Usage:Example Dim curValue As Currency Let curValue= 1/3 Picture1.Print curValue Let curValue= 100*1/3 Picture1.Print curValue Data Types (Contd.)

29. ► Currency Usage:Result 0.3333 33333.3333 =>- The decimal point is NOT floating; - Could be used for currency and scientific research - No rounding problems for high values Data Types (Contd.)

30. ► Even more data types  Date Variable: for date and time  Object Variable  Variant Variable  ………. Read the book for more info. Data Types (Contd.)

31. Not only +,-,*,^ But also: But also: \ opeartor e.g: 5.1\2.04= 2 \ opeartor e.g: 5.1\2.04= 2 MOD operator e.g: 15.2 MOD 6=3 MOD operator e.g: 15.2 MOD 6=3 More on Operators

32. More on Operators (Contd.) ► Operator Precedence 1. ^ 2.- operator (indicating a negative value) 3. * and / operator 4. \ operator 5. MOD operator 6. + and - operator

33. Assignment 3 ► Assignment 3 has been posted to the CS101 Website. ► The Deadline is on 6:59 p.m. Wednesday Sept. 25.

34. Reading for today ► Scott Warner:  Chapter 2: Section 2.3(p60-p66)  Chpater5: Section 5.1