I have the following code:
Sub Main()
Dim a As Integer = 8 * 60
Dim b As Integer
Dim c As Integer
If a < (6 * 60) Then
b = 0 And c = 0
ElseIf a >= 6 * 60 And a < 9 * 60 Then
b = 30 And c = 1
Else
b = 45 And
c = 1
End If
MsgBox(b)
End Sub
Thinks i dont understand and where i need someones help:
"c=0" and "c=1" are underlined with the error: Strict on doesnt allow implicit convertation from boolean to integer. WHY? I declared c as integer!
Variable "b" and "c" are always "0" even though in the case above they should be b=30 and c = 1.
can anyone please explain me this behaviour.
You are using the And keyword where it is not allowed. And is a logical operator (along with Or, AndAlso, OrElse.)
The following should work.
Sub Main()
Dim a As Integer = 8 * 60
Dim b As Integer
Dim c As Integer
If a < (6 * 60) Then
b = 0
c = 0
ElseIf a >= 6 * 60 And a < 9 * 60 Then
b = 30
c = 1
Else
b = 45
c = 1
End If
MsgBox(b)
End Sub
Related
I'm writing VB Code and I see a question below
There are three positive integers A, B, C
If A is greater than B, C is equal to A+B
If A is less than or equal to B, then C is equal to A-B.
Please use IF...Then and Select/Switch Case to write a program, both of which are used in this program, and additional variables can be added by yourself.
I would like to ask how to write this question, as far as I know, the answer just need only IF Then or Select Case can be realized?
Dim A As Double = 3
Dim B As Double = 2
Dim C As Double = 1
Dim D As Double = 0
D = A - B
Select Case D
Case D > 0
C = A + B
Case D < 0
C = A - B
End Select
If D > 0 Then
C = A + B
ElseIf D < 0 Then
C = A - B
End If
In the real world you wouldn't need to use both an If/Then statement and a Select/Case statement. Since this appears to be homework, I believe the exercise is to see if you can use both conditional check.
I would setup two functions that accept two arguments (a and b) that return the value of c. In your first function use an If/Then and in your second function use a Select/Case.
E.g.
Private Function IfThenOption(a As Integer, b As Integer) As Integer
Dim c As Integer
If (a > b) Then
c = a + b
ElseIf (a < b) Then
c = a - b
Else
c = a
End If
Return c
End Function
Private Function SelectCaseOption(a As Integer, b As Integer) As Integer
Dim c As Integer
Select Case True
Case a > b
c = a + b
Case a < b
c = a - b
Case Else
c = a
End Select
Return c
End Function
Example: https://dotnetfiddle.net/kwcyWc
I am getting the "next without for" error. I checked other questions on this and looked for any open if statements or loops in my code, but could find none. I'm need an extra set of eyes to catch my error here.
I am trying to loop through this code and advance the torque value 3 times each times it gets to the 30th i.
'This is Holzer's method for finding the torsional natural frequency
Option Explicit
Sub TorsionalVibrationAnalysis_()
Dim n As Integer 'position along stucture
Dim m As Integer
Dim i As Long 'frequency to be used
Dim j As Variant 'moment of inertia
Dim k As Variant 'stiffness
Dim theta As Long 'angular displacement
Dim torque As ListRow 'torque
Dim lambda As Long 'ListRow 'omega^2
Dim w As Variant
Dim s As Long
'equations relating the displacement and torque
n = 1
Set j = Range("d2:f2").Value 'Range("d2:f2").Value
Set k = Range("d3:f3").Value
'initial value
Set w = Range("B1:B30").Value
For i = 1 To 30
'start at 40 and increment frequency by 20
w = 40 + (i - 1) * 20
lambda = w ^ 2
theta = 1
s = 1
Do While i = 30 & s <= 3
torque = lambda * j(1, s)
s = s + 1
End
m = n + 1
theta = theta - torque(i, n) / k(n)
torque(i, m) = torque(i, n) + lambda * j(m) * theta
If m = 4 & i < 30 Then
w(i) = 40 + (i - 1) * 20
lambda = w(i) ^ 2
ElseIf m = 4 & i >= 30 Then
Cells([d], [5+i]).display (i)
Cells([e], [5+i]).display (theta)
Cells([f], [5+i]).display (torque)
Else
End If
If m <> 4 Then
n = n + 1
End If
Next i
End Sub
You are trying to terminate your While with an End instead of Loop
Try changing your End to Loop in your Do While loop. I think you are terming the loop when you hit that End
Proper indentation makes the problem rather apparent.
You have:
For i = 1 To 30
'...
Do While i = 30 & s <= 3
'...
End
'...
If m = 4 & i < 30 Then
'...
ElseIf m = 4 & i >= 30 Then
'...
Else
End If
If m <> 4 Then
'...
End If
Next i
But run it through Rubberduck's Smart Indenter and you get:
For i = 1 To 30
'...
Do While i = 30 & s <= 3
'...
End
'...
If m = 4 & i < 30 Then
'...
ElseIf m = 4 & i >= 30 Then
'...
Else
End If
If m <> 4 Then
'...
End If
Next i
End Sub
Notice how the End other answers are pointing out, is clearly not delimiting the Do While loop.
The Next i is inside the Do While block, which isn't terminated - when the VBA compiler encounters that Next i, it doesn't know how it could possibly relate to any previously encountered For statement, and thus issues a "Next without For" compile error.
Use an indenter.
I am trying to convert some old VB code to .Net but am having a problem with the Rnd function.
Old Code
Private Function Decode() As String
Dim r As Integer
Dim x As Integer
Dim c As Integer
Dom Code As String = "m[n-Msr0Xn*ca8qiGeIL""7'&;,_*EV{M;[{2bEmg8u!^s*+O37!692{-Y4IS"
x = Int(Rnd(-7))
For r = 1 To Len(Code)
x = Int(Rnd() * 96)
c = Asc(Mid(Code, r, 1))
c = c + x
If c >= 126 Then c = c - 126 + 32
Decode = Decode & Chr$(c)
Next
End Function
The decoded text is "Bet you needed more than a pencil and paper to get this one!"
This is what I have done:
Private Function Decode() As String
Dim r As Integer
Dim x As Integer
Dim c As Integer
Dim Answer As String
Dim Code As String = "m[n-Msr0Xn*ca8qiGeIL""7'&;,_*EV{M;[{2bEmg8u!^s*+O37!692{-Y4IS"
x = CType(Microsoft.VisualBasic.VBMath.Rnd(-7), Integer)
For r = 0 To sList.Length - 1
x = CType(Microsoft.VisualBasic.VBMath.Rnd() * 96 - 0.5, Integer)
c = Asc(sList.Substring(r, 1))
c = c + x
If c >= 126 Then c = c - 126 + 32
Answer &= Chr(c)
Next
Return Answer
End Function
but this is what I get "Bet you needed morB th(n a pencil and paper to get this one!"
I suspect its how I am castng to an int but I can't figure out how.
When I run your "Old Code" (after correcting the typos) under Office VBA (which should have the same Rnd implementation as VB6), I get the same result as you say you get from VB.Net. Therefore, there must be an error in the string assigned to Code.
Public Function Decode() As String
Dim r As Integer
Dim x As Integer
Dim c As Integer
Dim Code As String
Code = "m[n-Msr0Xn*ca8qiGeIL""7'&;,_*EV{M;[{2bEmg8u!^s*+O37!692{-Y4IS"
x = Int(Rnd(-7))
For r = 1 To Len(Code)
x = Int(Rnd() * 96)
c = Asc(Mid(Code, r, 1))
c = c + x
If c >= 126 Then c = c - 126 + 32
Decode = Decode & Chr$(c)
Next
End Function
Thanks to TnTinMan for leading me to the solution. The problem was created when I copied the string. I used a I instead of an l. The font that was used has these two characters looking identical. I also mistook ' for `.
so basically all I had to do was subtract .5
I'm trying to create a code that uses the false position method to find the roots of an equation. The equation is as follows:
y = x^(1.5sin(x)) * e^(-x/7) + e^(x/10) - 4
I used a calculator to find the roots, and they are 6.9025, 8.8719, and 12.8079.
My VBA code is as follows:
Option Explicit
Function Func(x)
Func = (x ^ (1.5 * Sin(x))) * Exp(-x / 7) + Exp(x / 10) - 4
End Function
Function FalsePos(Guess1, Guess2)
Dim a, b, c As Single
Dim i As Integer
a = Guess1
b = Guess2
For i = 0 To 1000
c = a - Func(a) * (b - a) / (Func(b) - Func(a))
If (Func(c) < 0.00001) Then
i = 1001
Else
If Func(a) * Func(c) < 0 Then
b = c
Else
a = c
End If
End If
Next
FalsePos = c
End Function
My problem is that when I call the function and use for example 4 and 8 as my two guesses, the number it returns is 5.29 instead of the root between 4 and 8 which is 6.9025.
Is there something wrong with my code or am I just not understanding the false position method correctly?
You should use Double for precision with Maths problems. Three other notes about coding that you may not be aware of:
dim a, b, c as Single
will dim a and b as Variants, and c as a Single, and you can use Exit For to escape from a for loop, rather than setting the control variable out of the bounds. Finally, you should define the outputs of a Function by specifying As ... after the closing parenthesis.
You should use breakpoints (press F9 with the carrot in a line of code to breakpoint that line), then step through the code by pressing F8 to advance line-by-line to see what is happening, and keep your eye on the Locals window (Go to View > Locals)
This is the code with the above changes:
Function Func(x As Double) As Double
Func = (x ^ (1.5 * Sin(x))) * Exp(-x / 7) + Exp(x / 10) - 4
End Function
Function FalsePos(Guess1 As Double, Guess2 As Double) As Double
Dim a As Double, b As Double, c As Double
Dim i As Integer
a = Guess1
b = Guess2
For i = 0 To 1000
c = a - Func(a) * (b - a) / (Func(b) - Func(a))
If (Func(c) < 0.00001) Then
Exit For
Else
If Func(a) * Func(c) < 0 Then
b = c
Else
a = c
End If
End If
Next
FalsePos = c
End Function
I'm trying to figure out how I can use the same comparison operator that Excel uses in sorting a mix of alphanumeric values like the following:
0
9
34
51
123abc
15
a
a1b23c
i
z
34ui
10
d
1
12
When sorting this, this is the result:
0
1
9
10
12
15
34
51
123abc
34ui
a
a1b23c
d
i
z
Is it possible to use the comparison operator that Excel uses to get this result? Or is it necessary to create my own function for this?
I just went ahead and created a comparison function with the same return values as StrComp() since it seems there isn't one already.
Function ExcelCompare(ByVal str1 As String, ByVal str2) As Integer
Dim isnum1 As Boolean
Dim isnum2 As Boolean
isnum1 = IsNumeric(str1)
isnum2 = IsNumeric(str2)
ExcelCompare = StrComp(str1, str2)
If isnum1 And Not isnum2 Then
ExcelCompare = -1
ElseIf Not isnum1 And isnum2 Then
ExcelCompare = 1
ElseIf isnum1 And isnum2 Then
Dim num1 As Double
Dim num2 As Double
num1 = CDbl(str1)
num2 = CDbl(str2)
If num1 = num2 Then
ExcelCompare = 0
ElseIf num1 < num2 Then
ExcelCompare = -1
Else
ExcelCompare = 1
End If
End If
End Function