I don't understand why 2 of my codes have overflow error
Sub varr3_1()
Dim x As Single
Dim y As Single
For x = 1 To 2 Step 0.2
y = Sqr((x - 1) / (x + 1))
Debug.Print x, y
Next x
End Sub
Sub varr3_3()
Dim x As Single
Dim z As Double
For x = 3 To 8 Step 0.9
z = 2
While (z > 1)
z = Log(x) + Tan(2 * x)
Debug.Print z
Wend
Next x
End Sub
I tried to change Single to Integer and so on but there is still a problem
The first sub Var3_1() works fine for me.
The second sub gets stuck in an endless loop on the second iteration of x. The value of z = 19.86... which will always be greater than 1 so the While/Wend loop never exits thus resulting in an eventual overflow.
Related
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'm trying to find and remove outliers from many columns of data, but it doesn't clear the cells that contain the outliers when I run the code. I tried just printing "colLength" within the first For loop, and that did nothing either. Advice on where I went wrong or how I might be able to fix it?
Sub Outliers()
Dim calc As Worksheet
Set calc = ThisWorkbook.Sheets("Sheet2")
Dim num As Double
Dim x As Integer
Dim y As Integer
Dim colLength As Integer
'Variables for upper fence, lower fence, first quartile, third quartile, and interquartile range
Dim upper As Double
Dim lower As Double
Dim q1 As Double
Dim q3 As Double
Dim interquartRange As Double
For y = 1 To y = 49
'Find last row of the column
colLength = calc.Cells(Rows.count, y).End(xlUp).Row
'Calculate first and third quartiles
q1 = Application.WorksheetFunction.Quartile(Range(Cells(2, y), Cells(colLength, y)), 1)
q3 = Application.WorksheetFunction.Quartile(Range(Cells(2, y), Cells(colLength, y)), 3)
interquartRange = q3 - q1
'Calculate upper and lower fences
upper = q3 + (1.5 * interquartRange)
lower = q1 - (1.5 * interquartRange)
For x = 2 To x = colLength
num = calc.Cells(x, y)
'Remove outliers
If num > upper Or num < lower Then
Range(calc.Cells(x, y)).ClearContents
End If
Next x
Next y
End Sub
For y = 1 To y = 49 should be For y = 1 To 49. Similarly For x = 2 To x = colLength should be For x = 2 To colLength
Try this in a new module and you will see and understand the difference ;)
Dim Y As Long
Sub SampleA()
For Y = 1 To Y = 49
Debug.Print Y
Next Y
End Sub
Sub SampleB()
For Y = 1 To 49
Debug.Print Y
Next Y
End Sub
I have this code below, and I'm getting an overflow error at the line:
s = s + (x Mod 10) [first line in the Do Loop]
Why? I declared x and s to be of type Double. Adding two doubles, why is this not working?
Thanks for your help.
Public Sub bidon1()
Dim i As Double, x As Double, s As Double, k As Byte, h As Byte
Dim y(1 To 6) As Double
For i = 1 To 1000000
x = i ^ 3
Do
s = s + (x Mod 10)
x = x \ 10
Loop Until x = 0
If s = x Then
k = k + 1
y(k) = x
If y(6) > 0 Then
For h = 1 To 6
Debug.Print y(h)
Next
Exit Sub
End If
End If
Next
End Sub
The problem is that the VBA mod operator coerces its arguments to be integers (if they are not already so). It is this implicit coercion which is causing the overflow. See this question: Mod with Doubles
On Edit:
Based on your comments, you want to be able to add together the digits in a largish integer. The following function might help:
Function DigitSum(num As Variant) As Long
'Takes a variant which represents an integer type
'such as Integer, Long or Decimal
'and returns the sum of its digits
Dim sum As Long, i As Long, s As String
s = CStr(num)
For i = 1 To Len(s)
sum = sum + Val(Mid(s, i, 1))
Next i
DigitSum = sum
End Function
The following test sub shows how it can be used to correctly get the sum of the digits in 999999^3:
Sub test()
Dim x As Variant, y As Variant
Debug.Print "Naive approach: " & DigitSum(999999 ^ 3)
y = CDec(999999)
x = y * y * y
Debug.Print "CDec approach: " & DigitSum(x)
End Sub
Output:
Naive approach: 63
CDec approach: 108
Since 999999^3 = 999997000002999999, only the second result is accurate. The first result is only the sum of the digits in the string representation of the double 999999^3 = 9.99997000003E+17
Trying to write a macro to insert a hyphen at specific points in a text string depending on how long the string is or delete all text after said point.
i.e
- if 6 characters, insert a hyphen between char 4+5 or delete all text after char 4
- if 7 characters, insert a hyphen between char 5+6 or delete all text after char 5
Ideally i would love to be able to truncate the string at that point rather than hyphenate the text but i couldn't get my head around how to make it work so i decided to hyphen and then just run a find and replace '-*' to remove the unwanted characters. Can get this working on small sample sets 100-300 cells but i need the code to be able to go through workbooks with 70,000+ cells. I've tried tweaking the code to stop the memory issue but now i can't seem to get it to work.
Sub Postcodesplitter()
Dim b As Range, w As Long, c As Range, x As Long, d As Range, y As Long
For Each b In Selection
w = Len(b)
If w = 8 And InStr(b, "-") = 0 Then b = Application.WorksheetFunction.Replace(b, 15 - w, 0, "-")
For Each c In Selection
x = Len(c)
If x = 7 And InStr(c, "-") = 0 Then c = Application.WorksheetFunction.Replace(c, 13 - x, 0, "-")
For Each d In Selection
y = Len(d)
If y = 6 And InStr(d, "-") = 0 Then d = Application.WorksheetFunction.Replace(d, 11 - y, 0, "-")
Next
Next
Next
End Sub
That's the original code i put together, but it caused memory issues over 300 target cells. I'm a pretty bad coder even at the best of times but with some advice from a friend i tried this instead.
Sub Postcodesplitter()
Dim b As Range, x As Long
If (Len(x) = 6) Then
b = Application.WorksheetFunction.Replace(b, 11 - x, 0, "-")
Else
If (Len(x) = 7) Then
b = Application.WorksheetFunction.Replace(b, 13 - x, 0, "-")
Else
If (Len(x) = 8) Then b = Application.WorksheetFunction.Replace(b, 15 - x, 0, "-")
End Sub
But this just throws out errors when compiling. I feel like im missing something really simple.
Any tips?
It looks as though you want to truncate to two less than the existing number of characters, if that number is 6-8? If so, something like this:
Sub Postcodesplitter()
Dim data
Dim x as Long
Dim y as Long
data = Selection.Value
For x = 1 to ubound(data,1)
for y = 1 to ubound(data, 2)
Select Case Len(data(x, y))
Case 6 to 8
data(x, y) = left(data(x, y), len(data(x, y)) - 2)
end select
next y
next x
selection.value = data
End Sub
First off, here is what I have so far:
Option Explicit
Dim y As Variant
Dim yforx As Variant
Dim yfork As Variant
Dim ynew As Variant
Dim ymin As Variant
Dim x As Variant
Dim xmin As Variant
Dim k As Variant
Dim kmin As Variant
Dim s As Variant
Dim Z As Variant
Dim Track As Variant
Sub PracticeProgram()
'Selects the right sheet
Sheets("PracticeProgram").Select
'y = k ^ 2 * (x ^ 2 + 2 * x * k - 6) / (x + k) ^ 2
'these are the bounds we are stepping through
Track = 0
x = 1
xmin = 1
k = 1
kmin = 1
y = 100000000
yforx = 100000
yfork = 1000000000
Do
y = 100000000
For x = 0 To 1000 Step 0.1
ynew = kmin ^ 2 * (x ^ 2 + 2 * x * kmin - 6) / (x + kmin) ^ 2
'This checks the new y-value against an absurdly high y-value we know is wrong. if it is less than this y-value, we keep the x-value that corresponds with it.
If ynew < y Then
xmin = x
y = ynew
yforx = y
xmin = Application.Evaluate("=Round(" & xmin & ", 3)")
Else
End If
Next
MsgBox (yforx)
For k = 0 To 1000 Step 0.1
y = k ^ 2 * (xmin ^ 2 + 2 * xmin * k - 6) / (xmin + k) ^ 2
If ynew < y Then
kmin = k
y = ynew
yfork = y
kmin = Application.Evaluate("=Round(" & kmin & ",3)")
Else
End If
Next
MsgBox (yfork)
Loop Until (Abs(yforx - yfork) < 10)
End Sub
This program is supposed to find the values of x and k in order to minimize the value of y. This is a practice for a much more complicated program that will use this same concept. In my actual program y, k, and x will all be greater than zero no matter what, but since it was hard to think of a simple equation whose results would be in the shape of a parabola opening up, I decided to allow negative answers for this practice program.
Basically, it should bounce back and forth between the equations finding the ideal values for x and k until finally it has a minimal answer for y using ideal answers for both x and k. I'm not sure what the actual answer is, so I'm letting it stop within a range of 10. If it works, I'll make it smaller, but I don't want the program going for forever, just in case.
MY PROBLEM: I keep getting overflow errors! I'm trying to round the values for xmin and kmin to three figures after the decimal, but it doesn't seem to be helping. Am I using them wrong? Can someone help me get this program working?
You're doing a division by zero. xmin = 0, k = 0, (xmin + k) ^ 2 = 0. (I'm not sure why it isn't reporting division by zero.)
A suggestion: use the Locals pane to see the value of local variables. You can also use the Watch pane to see the value of expressions you want to monitor.