Dim ArrayName(UpperBound) As Datatype An array is declared using the following syntax: An array’s...

Dim ArrayName(UpperBound) As DatatypeDim ArrayName(UpperBound) As Datatype

An array is declared using the following syntax:

An array’s name obeys the same rules as when declaring a variable: it begins with a letter and may be followed by any number of letters, underscores, or digits. An array name can be as small as one letter or as large as 255 letters, underscores, and digits.

Individual values within an array are selected by using an index.

The lowest index of an array is 0, while the upper bound of the array is the highest index in the array.

An index must be a Short, an Integer, or a Long data type.

The data type is any valid variable like a Short, an Integer, or a String.

ArraysArrays

Computer Memory

shtGrades (short)Index 0 1 2 3

Values 0 0 0 0

Dim shtGrades(3) As ShortDim shtGrades(3) As Short

Computer Memory

shtGrades (short)Index 0 1 2 3

Values 21 0 0 0

shtGrades(0) = 21 shtGrades(0) = 21

We can assign an individual variable by using its name, and index.

Index and subscript are synonymous

Dim shtAges() As Short = {21, 23, 45, 12, 23, 65, 12}

Dim intLoopCount As Integer

Dim intMaxVal As Integer = 0

For intLoopCount = 0 To UBound(shtAges)

If intMaxVal < shtAges(intLoopCount) Then

intMaxVal = shtAges(intLoopCount)

End If

Next intLoopCount

bntRedDemo.Text = intMaxVal.ToString

Dim intLoopCount As Integer

End If

Next intLoopCount

bntRedDemo.Text = intMaxVal.ToString

Finding the largest value in an array

Dim intLoopCount, intLocation As Integer

intLocation = intLoopCount

End If

Next intLoopCount

bntRedDemo.Text = “Value ” & str(intMaxVal) & “, at ” & str(intLocation)

End If

Next intLoopCount

We often want to know the location of the max value..

Dim shtA() As Short = {-21, -23, -45, -12, -23, -65, -12}

For intLoopCount = 0 To UBound(shtA)

If intMaxVal < shtA(intLoopCount) Then

intMaxVal = shtA(intLoopCount)

End If

Next intLoopCount

Dim shtA() As Short = {-21, -23, -45, -12, -23, -65, -12}

End If

Next intLoopCount

Our code can fail because of an incorrect assumption…

Dim shtA() As Short = {-21, -23, -45, -12, -23, -65, -12}

Dim intMaxVal As Integer = - 2147483648

End If

Next intLoopCount

Dim shtA() As Short = {-21, -23, -45, -12, -23, -65, -12}

Dim intMaxVal As Integer = - 2147483648

End If

Next intLoopCount

Here the incorrect assumption is fixed

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

Dim intMinVal As Integer = 2147483647

If intMinVal > shtA(intLoopCount) Then

intMinVal = shtA(intLoopCount)

End If

Next intLoopCount

bntRedDemo.Text = “Value ” & str(intMinVal ) & “, at ” & str(intLocation)

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

If intMinVal > shtA(intLoopCount) Then

End If

Next intLoopCount

Consider finding the minimum value, again a pessimistic assumption

Note relational test

Note pessimistic assumption

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

If intMinVal >= shtA(intLoopCount) Then

End If

Next intLoopCount

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

End If

Next intLoopCount

The previous slide found the first occurrence of the lowest value, this code finds the last occurrence, do you see the difference?

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

End If

Next intLoopCount

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

End If

Next intLoopCount

From now on I won’t bother showing this stuff, you can assume it from context

Dim intFindMe As Short = 45

Dim blnWasFound As Boolean = False

If intFindMe = shtA(intLoopCount) Then

blnWasFound = True

End If

Next intLoopCount

If blnWasFound Then bntRedDemo.Text = "Found at " & Str(intLocation)Else bntRedDemo.Text = "Not found!"End If

Dim intFindMe As Short = 45

Dim blnWasFound As Boolean = False

If intFindMe = shtA(intLoopCount) Then

blnWasFound = True

End If

Next intLoopCount

If blnWasFound Then bntRedDemo.Text = "Found at " & Str(intLocation)Else bntRedDemo.Text = "Not found!"End If

If IntFindMe was say, 1111

{21, 23, 45, 12, 23, 65, 12}{21, 23, 45, 12, 23, 65, 12}

Note the pessimistic assumption. Question: Does it find the first or last occurrence?

Dim shtA() As Short = {25,25,50,51,52,53,53,56,57,58}

Dim intMaxChange As Integer = 0

If intMaxChange < (shtA(intLoopCount + 1) - shtA(intLoopCount)) Then

intMaxChange = shtA (intLoopCount)

End If

Next intLoopCount

bntRedDemo.Text = “Max change at” & Str(intLocation)

Dim shtA() As Short = {25,25,50,51,52,53,53,56,57,58}

End If

Next intLoopCount

Lets try to find the location of the greatest (positive change)

Looks good, but there is a bug!

Dim shtA() As Short = {25,25,50,51,52,53,53,56,57,58}

For intLoopCount = 0 To UBound(shtA) - 1

End If

Next intLoopCount

Dim shtA() As Short = {25,25,50,51,52,53,53,56,57,58}

For intLoopCount = 0 To UBound(shtA) - 1

End If

Next intLoopCount

This fixes our bug…This fixes our bug…

Let us try to find a repeated item

Dim intInnerLoop, intOuterLoop As Integer

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

Dim blnWasRepeat As Boolean = False

For intInnerLoop = 0 To UBound(shtA)

bntRedDemo.Text = Str(shtA(intInnerLoop))

Next intInnerLoop

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

bntRedDemo.Text = Str(shtA(intInnerLoop))

Next intInnerLoop

21 23 45 12…

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

For intOuterLoop = 0 To UBound(shtA)

bntRedDemo.Text = Str(shtA(intOuterLoop)) & " " & Str(shtA(intInnerLoop))

Next intInnerLoop

Next intOuterLoop

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

Next intInnerLoop

Next intOuterLoop

21 21 21 23 21 45 12 12…

intOuterLoop = 0 {21, 23, 45, 12, 23, 65, 12}

intInnerLoop = 0

intInnerLoop = 1

intInnerLoop = 2

Str(shtA(intOuterLoop)) & " " & Str(shtA(intInnerLoop))

…12 21

intOuterLoop = 1

intOuterLoop = 6

{21, 23, 45, 12, 23, 65, 12}

intOuterLoop = 0

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

If intOuterLoop <> intInnerLoop

Next intInnerLoop

End If

Next intOuterLoop

Dim shtA() As Short = {21, 23, 45, 12, 23, 65, 12}

Next intInnerLoop

End If

Next intOuterLoop

…12 21

intOuterLoop = 1

intOuterLoop = 6

{21, 23, 45, 12, 23, 65, 12}

intOuterLoop = 0

If shtA(intOuterLoop) = shtA(intOuterLoop)

blnWasRepeat = True

End If

Next intInnerLoop

Next intOuterLoop

If shtA(intOuterLoop) = shtA(intOuterLoop)

blnWasRepeat = True

End If

Next intInnerLoop

Next intOuterLoop

Dim ArrayName(UpperBound) As Datatype An array is declared using the following syntax: An array’s...

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