Lab 6b

We need more time to absorb some concepts. Therefore today we will review. A form file (*.frm) contains the specification for what happens when the project executes that form. It has all the information about the controls that appear on the form (size, names, positions, contents, etc.) AND it has code to say what happens when the user interacts with a control. List below the controls that we know how to place on a form and interact with:

Command Button, Label, Textbox, Scroll Bar (Horizontal and Vertical), and list box
We know how to use the command button, a text box, a label, an option button, a list box, and we know how to change them properties on these to be named and look like we want them to.
Buttons, Text Boxes, Labels, Scroll Bar, Option Button, Check Box, List Boxes.
Label, text box, command button, list box, check box and option button
Buttons, textboxes, labels, text fields, and captions or file names.

TextBox, CommandButton, OptionButton, ListBox, VscrollBar, Label, CheckBox, HScrollBar, PictureBox, Pointer
LABEL, TEXTBOX, COMBO BOX, LIST BOX, VSCROLLBAR, HSCROLLBAR, FRAME, CHECKBOX, OPTION BOX, DRIVESCROLL BAR, COMMANDBUTTON, LINE, DATA, FILELISTBOX, IMAGE, OLE, SHAPE, TIMER. ßnot so, these controls are in the toolbox, but we don’t know how to use them all yet.

The code section of the form begins with the statement Option Explicit. What does this tell the compiler?

Forces variables to be explicitly declared.
The option explicit tells the computer that the statement is forced to be explicitly declared.
Forces variables to be explicity declared.
The option explicit forces the variables to be explicitly declared. It does not let you name the same variable more then once.

Forces explicit declarations or variable declarations
To force variable declarations if none are made.
THAT A VARIBLE WILL BE DECLARED EXPLICTLY.

The rest of the code area has subroutines in it unless it has a Dim statement. A Dim statement outside a subroutine means what?

           Declares that these data areas apply to all sub routines
It sets a variable as something, such as Dim sum as Integer, sets the variable (any valid identifier) as an integer.
Declares a variable and type, such as integer.   ßmore than this, what’s significant about position
Each variable that has been assigned, has that value throughout the program, not just the subroutine.
It means that it has variables outside of the main subroutine. ßthere is no such thing as a main subroutine as far as we know. This was explained on page 145.
    That it is in the general declaration (a global variable) and available to every event procedure.
         THAT A VARIBLE IS BEING DECLARED.ßmore than this, where

The subroutines that we have observed so far are a block of code statements, the first of which begins PRIVATE SUB… and ends with END SUB. (look at page 144 for an example). What comes after and on the same line as the PRIVATE SUB?

The name of the command button or other control
After Private Sub comes the name of the item (i.e command button, list box, etc) that you are programming.
Name of the designated field and the action that activates it.ßaction is for event procedures only
. The name of the control

Either a command1 which is unnamed button, click or form1 which is also another unnamed form1ßI know of no occasion so far in the text where form1 occurred on the Private Sub line, please point it out to me. There must be some justification for making this assertion!
The name of the control / command (i.e., cmdEnter_Click())
THE OBJECT OF FOCUS, SUCH AS LABEL, TEXT, COMMAND BUTTONßabsolutely not—these are controls, not subroutine names.

Inside the subroutine block we find three kinds of statements: statements which declare local names (data items), statements that change values of data items, and statements that change which statement will be executed next (flow control statements). Give an example of each kind of statement:

X=1

DIM x as integer

CmdDisplay_Clickß this does none of the above and is not a statement.
Dim message as String (data items)

mAccessCode = txtDisplay.Text (change values)

Select Case mAccessCode (flow control statment)

Local names: Text1.Text = “Hello”ßno different than the next statement—not a declaration

Changing values: Text1.Text = Text1.Text + 1

Flow Control: If x > 5 Then, Print x, Else Print y, End If
Declaration Dim A As Integer

Change value x = a + 1

Flow control If a > 2 Then

Print a

                              End If
: In the first item or the data items, an example would be a product of <=1000, for a statement that changes the value an example would be the product =5 * product, and for a flow control statement it is what makes it go in the order that it does. An example might be a loop and how many times that it loops or a do until. ß I specifically asked for you to write three example VB statements. You have written no statements.   A statement starts with a keyword.
Declaration Statement:                     Dim message as String

Change Values: x = x + 1

Flow Control Statements: If num1 <= num2 Then

LblDisplay4.Caption = num1

ElseIf num1 > num2 Then

LblDisplay4.Caption = num2

IF/THEN, IF/THEN/ELSE, DO UNTIL/LOOP ßthese are keywords. NOT example statements

Pages 163 to 165 summarize almost everything we have studied so far. You need to know the forms of all the Basic statements mentioned. But you also have to understand how to use the different statements. Most statements have parts (phrases) and all statements are identified by keywords.
a. many Basic statements may contain an expression. Define this term and give as examples three different Basic statements that contain an expression.

An expression is a function that interprets with operators like AND and OR.ß Boolean expressions use the operators AND and OR, String and numeric expressions may not.

Ex: If not 5<6 then falseßafter the then comes a statement, this is an error.

Do: x=5:Loop while 5<6ß this is invalid, needs a LOOP

If x=5 then trueßerror else x<5
If/Then expression - if a certain variable is true then a certain an expression is displayed if it is false then another expression is displayed.

While/Wend expressionßexpression not defined or illustrated!
Expression: combination of data names with operators.

If x > 5 Then

ElseIf Grade >= 80 Thenßtwo Ifs are not different (even if elseis), use For, Case

Do Until product > 1000
An expression is a sequence of operations and characters where the characters are data names and logical.

Example:

DIM A As integer

X = A + 1

If X > 5 Then

Print “Yipee”

End If

Do: LoopßLoop ends the loop(this is an empty loop—repeats nothing but never ends (no way to get out) I hope the IDE won’t let you write this.

Print “Yipee”

While X > 5ßmissing Wend
A string that sets up a command for a specified function. Ex. ßnot an adequate definition. Please identify what the specified function is below in each statement, and explain what string is involved, and define what “sets up a command” means.
Yellow = command1.caption , ßan assignment statement moving text from caption to Yellow. A variable name does not make a very explanatory expression

Print false, ßprint statements often have expressions, but the constant False is not a very interesting expression

end, ßincomplete (End Sub or End Function) No expression in this statement

text1.text= TextßLet statement, Text, perhaps a variable name, is not a very interesting expression
Expressions are a sequence of operators and character strings, which is a combination of data names with operators. The program evaluates these expressions.

Examples: 1. Dim X as Integer

X = 1 to 100 Step 2ßtwo statements, where is the expression?

2. If grade >= 50 Then

LblState.Caption = “Passed”

3. Do While Product <= 1000

Product = Product * 2
EXPRESSION- STATES THE PURPOSE OF THE KEYWORDS OR THE STATEMENT. ß has little to do with expressions, which are formulas that signify values. IF/THEN, IF/THEN/ELSE, FOR/NEXT.ßnot example statements.

Expressions result in a value once they have been evaluated. This value is reached in a systematic way by the compiler, but sometimes the programmer thinks a different value will be computed. What causes this kind of error?

If you true to combine two different type of values and the compiler converts them

differently than what you think. Ex: if you want to combine a + 6 the compiler will

convert both to numeric expressions and if you expect a letter it will not be what

you expect.
A wrong equation or sequence that either calculates the wrong value or has the wrong value in memory. Either of these mistakes will produce an errored value.
The programmer has the order of operations out of sequence according to the precedence rules.
The programmer may have used fixed values which cannot be change by the programmer ßdo you mean he wrote 5 when he meant 3? What else has “fixed values” and what does this say about the difference between how the programmer wrote the expression and the computer evaluated it?
The programmer forgets the precedence in which the compiler handles the data. For instance, incorrect use of parenthesis or forgetting the order in which arithmetic operations are performed.

THE PROGRAMMER GETS CONFUSED ABOUT WHAT IS GOING ON IN THE PROGRAM DUE TO THE PRECEDNCE OF THE CODE.

The value of an expression occupies an anonymous location in memory, but it has a type determined by the operators and operands. We say that the type of the value is the type of the expression. Give the types of each of the following expressions (assume x and y are single with values 3.1416 and 2.7x10¹⁰):
a. –2 + 7.4 Single because of the decimal point
b. 7.4 \ 5 Integer because of the backslash
c. 4 mod 2 + y Single because y was declared single
d. 3 ^ -2 Single
e. 5>x Single because x was declared singleßrelational expressions are Boolean valued
f. “+” & x singleßimpossible with a String operator
g.   Text1.Text + 1 Objectßwe’ve done this often, always with an Integer result
h.   (5>x) OR (y<x^10) Boolean
–2 + 7.4 = IntegerßNo!
b. 7.4 \ 5 = Integer
c. 4 mod 2 + y = IntegerßNo!
d. 3 ^ -2 = Integer
e. 5>x = Boolean
f. “+” & x = BooleanßNo!
g.   Text1.Text + 1 = Integer
h.   (5>x) OR (y<x^10) = Boolean
. –2 + 7.4                                single
b. 7.4 \ 5                                 integer
c. 4 mod 2 + y             single
d. 3 ^ -2                                  single
e. 5>x                                     Boolean
f. “+” & x                                string
g.   Text1.Text + 1                    integer (numeric value)
h.   (5>x) OR (y<x^10 )           Boolean
2 + 7.4 single
b. 7.4 \ 5 integer
c. 4 mod 2 + y single
d. 3 ^ -2 single
e. 5>x Boolean
f. “+” & x LongßString operator, string value
g.   Text1.Text + 1 variantßwe have not used this data type but we have used this expression!
h.   (5>x) OR (y<x^10) Boolean
a. –2 + 7.4                              Single
b. 7.4 \ 5                                 Integer
c. 4 mod 2 + y                         Single
d. 3 ^ -2                                  Single
e. 5>x                                     Boolean
f. “+” & x                                String
g.   Text1.Text + 1                    Integer (numeric value)
h.   (5>x) OR (y<x^10)            Boolean

a. –2 + 7.4 INTEGER
b. 7.4 \ 5 INTEGER
c. 4 mod 2 + y SINGLE
d. 3 ^ -2 SINGLE
e. 5>x BOOLEAN
f. “+” & x STRING
g. Text1.Text + 1 INTEGER
h. (5>x) OR (y<x^10) BOOLEAN

Some Basic statements contain phrases that denote ranges of values. Give examples of two such statements.

The statement IF an For can contain ranges.ßthe If cannot have a range
String$(17, “@”) and Format$(amount, “currency”)ßneither illustrate a range
For x = 1 to 100 Step 5 ……..

If x >= 5000 Then ……ßthis is NOT a range expression
The for and the case are statements that denote ranges of values, For row = 1 to 10 , Case 165 to 245
For x = 1 to 100 Step 2

If x >= 1000 Then …
SELECT CASE AND FOR/NEXT

Knowing the valid forms of statements is less than half the battle. You can easily write a statement that will be accepted by the compiler but not produce the effect you wanted. This is the difference between a logic error and a syntax error. On page 123, the first four parts of exercise 4.6 illustrate both kinds of errors. Classify which is which.

a) Syntax error

b) Syntax error

c) Logic error

Syntax error
) syntax b) syntax c) logic d) logic e) logic f) syntax
a) Syntax error because you didn’t use Then with the If statement

b) Syntax error because Loop is missing to complete the statement

c) Logic error because y = y + 1 should be in the loop (before Wend)

d) Logic error because it would loop continuously (it should be z = z –1)

A. is a syntax error which refers to errors usually caught by the compiler, B,C, and D is a logic error because it did not perform the task thought out to run.
(a) syntax (b) syntax (c) logic (d) logic (it would loop continuously)
A-SYNTAX B- SYNTAX C- SYNTAX D-LOGIC E-SYNTAX

for exercise 4.12a (p.125) if x contains 573 before the loop, write below the FOR/NEXT loop that will produce the identical result.

For x = 573 To 500 Step x - 5

Print x;

Next x
Dim x As Integer

For x = 573 to 500 Step –5

Print x

Next x
Dim x As Integer

For x = 573 To 500 Step -5

Print x

Next x
For x = 573 To 500 Step x - 5

Print x;

Next x
Dim x as Integer

For x = 573 to 500 Step -5

Print x

Next x
For X = 573 To 500 Step -5

Print X;

Rem ENDßdoesn’t hurt but then it doesn’t help either!

Next X

for exercise 4.12b tell me how many time the loop will execute (there are three answers, and you must state what conditions determine which answer)

(1) If pcm starts out not equal to q it will not enter the loop loops 0 times

(2)If pcm starts out equal to q it enters the loop is changed in the loop to a different value, we have an infinite loop

(3) If the txtInput is changed in the program to q it loops once
If pcm does not = q, then the loop will not be executed.

If pcm = q, the loop will execute an infinite number of times.UNLESS case below!

If TextInput.Text does not = q, the loop will execute only once.
pcm is not equal to q –> zero loops

pcm is equal to q & txtInput.Text = q –> infinite loop

pcm is equal to q & txtInput.Text not = q –> loops 1 time
It will loop three times Once through the pcm, then through the qfactor?? and finally into the txtinput.text.ßNOT!
pcm not equal to q, do the loop 0 times

If pcm starts out equal to q, it gets changed in the loop to q, and you have an infinite loop.

If pcm starts out equal to q, and textInput is not q, then it loops once.

So, enter the loop 0 times, once, or infinite.
THE THREE CASES ARE 0 TO 1, INDEFINTE, 0ßno conditions stated!

for exercise 4.12c assume that the loop begins and ends. How many times did it add J to largeJ?

Once One or infiniteßthen it would not end! 1 time
Once, because any value that was less than huge J would make it an infinite loop and any value greater than Huge J would make it only cycle once.

Once

THE CASES WOULD BE GO THROUGH ONCE AND INDEFINTELY AFTER GOING THROUGH LOOP ONCE.ßsounds like two different answers and the computer will only do one of them.

for exercise 4.12d: assume display has the value “135”.   Give an initial value for n such that
a. nothing will be printed    136
b. the loop will print exactly 8 times   .000164ßtoo big, prints only 7 times
c. the loop will (try to) print an infinite number of times. 0
a) 136 b) .0001   c) 0
n > 136

(b) n = 0.00003

nothing will be printed 200
b. the loop will print exactly 8 times .000164
c. the loop will (try to) print an infinite number of times. 0

anything greater than 135

b. 0.00003

c. If n = 0
nothing will be printed A NUMBER GREATER THAN DISPLAY(135)
b. the loop will print exactly 8 times N=N^8ß I don’t know what value this is.
c. the loop will (try to) print an infinite number of times. N=0

for exercise 4.7 the last statement of the subroutine is to change the Caption property of what?

The caption bar on the GUI interface ßno such thing (do you mean window?)
The caption of the form

GUI interface window (title bar of run window) (form 1 changes to Value is 385)
Value isßwhat object is this (objects have properties, Caption is one such property)
The GUI interface window (title bar of run window)
COMMANDBUTTON1ßcommon misunderstanding!

for exercise 4.7,
a. convert the While loop into the equivalent FOR/NEXT loop and write it here.
b. the value of y that is printed each time through the loop depends on x. What will the value be (note that it prints in a column)?
c. a statement of the form <data item> = <data item> + y has been used many times so far such as Text1.Text=Text1.Text + 1. Each time the statement is executed we say that the value in <data item> is incremented by y. When we execute the statement repeatedly in a loop, what is the result?

a) Dim x As Integer, y As Integer, Mysteryvalue as Integer

For x = 1 to 10 Step 1

y = x ^ 2

Print y

Mysteryvalue = Mysteryvalue + y

Next x

Caption = “Value is “ & Mysteryvalue

b) The square of x. Which is 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

The <data item> is repeatedly incremented by y as the loop continues.
mysteryValue = 0

For x = 1 To 10 Step 1

y = x ^ 2

Print y

mysteryValue = mysteryValue + y

Next x

Caption = "value is " & mysteryValue

For x = 1 To 10 Step x ^ 2

y = x ^ 2

Print y

Mysteryvalue = Mysteryvalue + y
Next x

b. The value of Y is dependant on the value x which is the result of the square of x
c. It continues to increase by one unless otherwise specified.ßhow can it be otherwise specified? it is what it is, the question is what does it accomplish!

For x = 1 to 10 Step 1

y = x ^ 2

Print y

MysteryValue = MysteryValue + y

Next x

Caption = “value is “ & MysteryValue

(b) 1 It assigns the value of x-squared to y, and prints y.

100

( c ) The value will be getting larger or smaller based on “y,” because the Text1.Text is being incremented each time through the loop by “y.”

The flow of control statements in Basic are ALL block statements, that is, they are comprised, like subroutines, of a header statement, an END statement, and have body statements in between. List all of the different forms of the header statements and for each the form of the END statement.
If/Then: End If, If/Then/Else: End If, Select Case: End Select, While: Wend, Do While: Loop, For: Next.
If ……Then……<Else If……Else……>End If < > is not part of header or end

Do Until …… Loop

Do While …… Loop

Do …… Loop While (Until)

While … … Wend

For ……Next

. Select case-caseßnot end statement, <else-elseif>, <sub-endsub>, if-endif, for-next, while-wend, do-while-loop-until??, <let-until>ß< > indicates invalid pairs
SELECT CASE …< CASE …> END SELECT

IF ….THEN <…. ELSE IF … ELSE …> END IF

OR can be just IF…THEN<…ELSE….>END IF

OR can be just IF…THEN if a single statementßMUST have End If

DO UNTIL… LOOP

DO WHILE … LOOP

PRIVATE SUB … END SUBßNot flow control

WHILE … WEND

FOR…NEXT

DO … LOOP UNTIL

DO … LOOP WHILE

If you know how to write valid statements that achieve your purpose, you are able to read valid statements and predict what they will accomplish. Do that for exercises 5.10, 5.11, 5.12, and 5.13, showing the output of each below.

5.10:

Prints & in as many rows and columns as indicated by input windows

5.11:

Prints 6 rows of “*” that is 5 columns long

5.12:

3, 5, 8, 11, 11

5.13:

5, 11, 18, 26, 30, 35, 41, 44, 48, 50, 50*

5.10 Print “&” x number of times across and y number of times down

If you entered 3 for x and 2 for y you would get: &&&

&&&

5.11 i produces 2 loops

j produces 3 loops

k prints 4 *

This produces 4 *’s in 6 rows

i j k Print

1 1 1 à *

2 à *

3 à *

4 à *

2 1 à *

2 à *

3 à *

4 à *

3 1 à *

2 à *

3 à *

4 à *

2 1 1 à *

2 à *

3 à *

4 à *

2 1 à *

2 à *

3 à *

4 à *

3 1 à *

2 à *

3 à *

4 à *

5.12 Outside For Loop starts at 0 increments to 4 (step 1)

Inside loop says while c < p, go through select sequence.

First time through, p = 0, c = 0, so it does not do while loop

So, it increment p to 1, gets to while statement and c = 0, so it enters the while loop.

Select Case (p + c – 1) will equal 1 + 0 – 1 = 0, so it uses the first case, and m = 1 now m = 1, and

P = 1, so case 1,2,3 m = m + 2 is selected and m = 3; then you fall out of select print “ m” which is “3”, then increments c by 1, making it 1, and back up to top for loop, make it 2. This continues until the final printout is:

3 5 8 11

The last 11 appears on its own line, because of the last Print and Print m statements, which are outside the outside For loop

5.13 e b t w list box

4 0 5 5 à

1 6 11 à

2 7 18 à

3 8 26 à

3 0 4 30 à

1 5 35 à

2 6 41 à

2 0 3 44 à

1 4 48 à

1 0 2 50 à

0 - - - à 50*

5.10It makes the same number of columns and rows that you have requested 5.11 it does not run because of a syntax error 5.12 it runs the amount in 2 then 3 then 3 again or in other words a pattern of 3 5 8 11 then repeats 5.13 it won’t run because it has a syntax error for the Next variable, it won’t be able to use it as a cyclic variable.ßthey all run without syntax errors.
5.10 This is a table format (rows/columns). The inside loop prints the rows, and the inside loop sets the values for the number of & on a column (x), and the outer loops set the number of rows (y). For example, if you entered 2 for x and 3 for y, it would print:

5.11 3 loops. The program goes through the first 2 loops, and then runs the k loop 4 times, producing “****”

Once it goes through the k loop 4 times, it goes back up to the j loop, which increments by 1, making the value 2, and it goes back through the k loops another 4 times creating another

“****”

Then it falls out of the k loop again and increments the j loop to 3, goes back through k producing another “****.”

Then, it falls out of the k loop, falls out of the j loop and increments the “i” loop to 2, and then goes back through the j loop sequence, creating 3 more rows of “****.”

The final output is:

* * * *

5.12 Outside For loop starts at 0 increments to 4 (step 1)

Inside loop says while c < p, go through Select sequence.

First time through, p = 0, c = 0, so it does not do while loop.

So, it increments p to 1, gets to while statement and c=0, so it enters the while loop.

Select Case (p + c – 1) will equal 1 + 0 – 1 = 0, so it uses the first case, and m = 1, now m = 1, and

P = 1, so Case 1, 2, 3 m = m + 2 is selected and m = 3; then you fall out of select print “m” which is “3,” then increments c by 1, making it 1, and back up to top for loop, making it 2. This continues until the final printout is:

3 5 8 11

The last 11 appears on its own line, because of the last Print and Print m statements, which are outside the outside For loop.

5.13 It prints the numbers “5, 11, 18, 26, 30, 35, 41, 44, 48, 50, and 50*

E values start at 4 and increment by –1 until they reach 0.

For loop inside starts with b=0 and goes to (e-1)

The first value of t is 5.

e b t w list box

4 0 5 5 à 5

1 6 11 à 11

2 7 18 à 18

3 8 26 à 26

3 0 4 30 à 30

1 5 35 à 35

2 6 41 à 41

2 0 3 48 à 44

1 4 48 à 48

1 0 2 50 à 50

0 - - - à 50*

Give the code below for 5.16 ( C ) and (D) (use the Space$ function)

VERSION 5.00

Begin VB.Form Form1

Caption = "Form1"

ClientHeight = 3195

ClientLeft = 60

ClientTop = 345

ClientWidth = 4680

BeginProperty Font

Name = "Courier"

Size = 9.75

Charset = 0

Weight = 700

Underline = 0 'False

Italic = 0 'False

Strikethrough = 0 'False

EndProperty

LinkTopic = "Form1"

ScaleHeight = 3195

ScaleWidth = 4680

StartUpPosition = 3 'Windows Default

Begin VB.CommandButton Command2

Caption = "D"

BeginProperty Font

Name = "MS Sans Serif"

Size = 13.5

Charset = 0

Weight = 700

Underline = 0 'False

Italic = 0 'False

Strikethrough = 0 'False

EndProperty

Height = 615

Left = 2760

TabIndex = 1

Top = 2040

Width = 735

End

Begin VB.CommandButton Command1

Caption = "C"

BeginProperty Font

Name = "MS Sans Serif"

Size = 13.5

Charset = 0

Weight = 700

Underline = 0 'False

Italic = 0 'False

Strikethrough = 0 'False

EndProperty

Height = 615

Left = 720

TabIndex = 0

Top = 2040

Width = 735

End

Attribute VB_Name = "Form1"

Attribute VB_GlobalNameSpace = False

Attribute VB_Creatable = False

Attribute VB_PredeclaredId = True

Attribute VB_Exposed = False

Option Explicit

Private Sub Command1_Click()

Dim row As Integer

Dim col As Integer

Cls

For row = 1 To 10

Print Space$(row - 1);

For col = 10 To row Step -1

Print "*";

Next col

Next row

End Sub

Private Sub Command2_Click()

Dim row As Integer

Dim col As Integer

Cls

For row = 1 To 10 Step 1

Print Space$(10 - row);

For col = 1 To row

Print "*";

Next col

Next row

End Sub

5.16

( C )

STARZ2.FRM:

VERSION 5.00

Begin VB.Form Starz2

Caption = "Starz2"

ClientHeight = 3195

ClientLeft = 60

ClientTop = 345

ClientWidth = 4680

BeginProperty Font

Name = "Courier New"

Size = 12

Charset = 0

Weight = 400

Underline = 0 'False

Italic = 0 'False

Strikethrough = 0 'False

EndProperty

LinkTopic = "Form2"

ScaleHeight = 3195

ScaleWidth = 4680

StartUpPosition = 3 'Windows Default

Begin VB.CommandButton Starz2

Caption = "Starz2"

BeginProperty Font

Name = "Courier New"

Size = 12

Charset = 0

Weight = 700

Underline = 0 'False

Italic = 0 'False

Strikethrough = 0 'False

EndProperty

Height = 495

Left = 1800

TabIndex = 0

Top = 1320

Width = 1215

End

Attribute VB_Name = "Starz2"

Attribute VB_GlobalNameSpace = False

Attribute VB_Creatable = False

Attribute VB_PredeclaredId = True

Attribute VB_Exposed = False

Private Sub Starz2_Click()

Dim nStar As Integer

Dim row As Integer

For row = 10 To 1 Step -1

Print Space$(10 - row);

For nStar = 1 To row Step 1

Print "*";

Next nStar

Next row

End Sub

( D )

STARZ.FRM:

VERSION 5.00

Begin VB.Form Starz

Caption = "Form1"

ClientHeight = 3195

ClientLeft = 60

ClientTop = 345

ClientWidth = 4680

BeginProperty Font

Name = "Courier New"

Size = 12

Charset = 0

Weight = 400

Underline = 0 'False

Italic = 0 'False

Strikethrough = 0 'False

EndProperty

LinkTopic = "Form1"

ScaleHeight = 3195

ScaleWidth = 4680

StartUpPosition = 3 'Windows Default

Begin VB.CommandButton Starz

Caption = "Starz"

BeginProperty Font

Name = "Courier New"

Size = 12

Charset = 0

Weight = 700

Underline = 0 'False

Italic = 0 'False

Strikethrough = 0 'False

EndProperty

Height = 495

Left = 1800

TabIndex = 0

Top = 1320

Width = 1215

End

Attribute VB_Name = "Starz"

Attribute VB_GlobalNameSpace = False

Attribute VB_Creatable = False

Attribute VB_PredeclaredId = True

Attribute VB_Exposed = False

Private Sub Starz_Click()

Dim nStar As Integer

Dim row As Integer

For row = 1 To 10 Step 1

Print Space$(10 - row);

For nStar = 1 To row Step 1

Print "*";

Next nStar

Next row

End Sub

Insert the frm file below that solves exercise 5.11 where an InputBox is used to get the maximum number of stars and, with a different caption, to notify if an invalid number is input. The first time the box should request “Input a positive odd number up to 25”. Depending on the error error detected in the user’s input, the box will change to “Input a odd number between 1 and 25”, “Input a positive odd number less than 26”, “Input a positive number up to 25 but not divisible by 2”, or “Input a whole odd number up to 25”

Option Explicit

Private Sub Command1_Click()

Dim nstar As Integer

Dim row As Integer

Dim sp As Integer

Dim a As String

Dim b As String

Dim c As String

Dim d As String

Dim e As String

Dim maxNum As Integer

Dim valid As Boolean

Cls

a = "Enter a positive odd number up to 25:"

b = "Enter an odd number between 1 and 25:"

c = "Enter a positive odd number less than 26:"

d = "Enter a positive number up to 25 but not divisible by 2:"

e = "Enter a whole odd number up to 25:"

valid = False

maxNum = InputBox(a)

If maxNum >= 25 Then

InputBox (b)

Else

If maxNum <= 0 Then

InputBox (c)

Else

If maxNum Mod 2 = 0 Then

InputBox (c)

Else

Rem Data Okay

valid = True

End If

Loop Until valid

sp = maxNum \ 2

For row = 1 To maxNum Step 2

Print Space$(sp);

sp = sp - 1

For nstar = 1 To row

Print "*";

Next nstar

Next row

sp = 1

For row = maxNum - 2 To 1 Step -2

Print Space$(sp);

sp = sp + 1

For nstar = 1 To row Step 1

Print "*";

Next nstar

Next row

End Sub

Private Sub Command2_Click()

End Sub

Private Sub Command1_Click()

Dim nstar As Integer

Dim row As Integer

Dim sp As Integer

Dim a As String

Dim valid As Boolean

Dim maxNum As Single

a = "Enter a positive odd number up to 25:"

valid = False

maxNum = InputBox(a)

If maxNum >= 25 Then

a = "Enter an odd number between 1 and 25:"

Else

If maxNum <= 0 Then

a = "Enter a positive odd number greater then 0 and less than 26:"

Else

If CInt(maxNum) Mod 2 = 0 Then

a = "Enter a positive number up to 25 but not divisible by 2:"

Else

If maxNum - CInt(maxNum) <> 0 Then

a = "Enter a whole odd number up to 25:"

Else

Rem Data Okay

valid = True

End If

Loop Until valid

sp = maxNum \ 2

For row = 1 To maxNum Step 2

Print Space$(sp);

sp = sp - 1

For nstar = 1 To row

Print "*";

Next nstar

Next row

sp = 1

For row = maxNum - 2 To 1 Step -2

Print Space$(sp);

sp = sp + 1

For nstar = 1 To row Step 1

Print "*";

Next nstar

Next row

End Sub

Private Sub Command2_Click()

End

End Sub

Private Sub Command3_Click()

Cls

End Sub