prime number in vba excel 2003 - excel

I'm analyzing a code from the website and I tried it on my side as well but seems it doesn't work. Could you please tell me why? would greatly appreciate your help.
Thanks
Private Sub CommandButton1_Click()
Dim N, D As Single
Dim tag As String
N = Cells(2, 2)
Select Case N
Case Is < 2
MsgBox "It is not a prime number"
Case Is = 2
MsgBox "It is a prime number"
Case Is > 2
D = 2
Do
If N / D = Int(N / D) Then
MsgBox "It is not a prime number"
tag = "Not Prime"
Exit Do
End If
D = D + 1
Loop While D <= N - 1
If tag <> "Not Prime" Then
MsgBox "It is a prime number"
End If
End Select
End Sub

The single biggest problem I see is using Single instead of Integer or Long. Primes are positive integers and are not thought of in the context of decimal values (as far as I know). Thus by using a singles and comparing them to divided ints, you're opening yourself up to nasty edge-case rouding errors.
The line If N / D = Int(N / D) Then is using a poor method to see whether or not the numbers are prime. It's assuming that every time you divide a floating point number (in this case, the single) by the divisor, if it has a decimal remainder, then the integer conversion of that remainder will not be equal. However, I've run into rounding errors sometimes with floating point numbers when trying to compare answers, and in general, I've learned to avoid using floating point to int conversions as a way of comparing numbers.
Here's some code you might try instead. Some things to note:
I've changed the types of N and D so that they are Longs and not Singles. This means they are not floating point and subject to possible rounding errors.
I've also explicitly converted the cell value to a long. This way you know in your code that you are not working with a floating point type.
For the comparison, I've used Mod, which returns the remainder of the N divided by D. If the remainder is 0, it returns true and we know we don't have a prime. (Note: Remainder is often used with \, which only returns the integer value of the result of the division. Mod and \ are commonly used in precise division of integer types, which in this case is very appropriate.
Lastly, I changed your message box to show the actual number being compared. Since the number in the cell is converted, if the user enters a floating point value, it will be good for them to see what it was converted to.
You'll probably also note that this code runs a lot faster than your code when you get to high numbers in the hundreds of millions.
'
Sub GetPrime()
Dim N As Long
Dim D As Long
Dim tag As String
N = CLng(Cells(2, 2))
Select Case N
Case Is < 2
MsgBox N & " is not a prime number"
Case Is = 2
MsgBox N & " is a prime number"
Case Is > 2
D = 2
Do
If N Mod D = 0 Then
MsgBox N & " is not a prime number"
tag = "Not Prime"
Exit Do
End If
D = D + 1
Loop While D <= N - 1
If tag <> "Not Prime" Then
MsgBox N & " is a prime number"
End If
End Select
End Sub
NOTE: I changed the name of the procedure to be GetPrime. In your code, you had:
Private Sub CommandButton1_Click()
In the line above, you are defining a procedure (also called a method or sometimes just referred to as a sub). The word Sub indicates you are defining a procedure in code that returns no value. (Sometimes you might see the word Function instead of Sub. This means the procedure returns a value, such as Private Function ReturnANumber() As Long.) A procedure (Sub) is a body of code that will execute when called. Also worth noting, an excel macro is stored in VBA as a Sub procedure.
In your line of code, CommandButton1_Click() is the name of the procedure. Most likely, this was created automatically by adding a button to an Excel spreadsheet. If the button is tied to the Excel spreadsheet, CommandButton1_Click() will execute each time the button is pressed.
In your code, Private indicates the scope of the procedure. Private generally means that the procedure cannot be called outside of the module or class in which it resides. In my code, I left out Private because you may want to call GetPrime from a different module of code.
You mentioned in your comments that you had to change the name of my procedure from GetPrime() to CommandButton1_Click(). That certainly works. However, you could also have simply called GetPrime from within CommandButton1_Click(), like below:
Private Sub CommandButton1_Click()
'The following line of code will execute GetPrime() '
'Since GetPrime does not have parameters and does not return a value, '
'all you need to do is put the name of the procedure without the () '
GetPrime
End Sub
'Below is the entire code for the Sub GetPrime() '
Sub GetPrime()
'The body of the code goes below: '
' ... '
End Sub
Hopefully this helped to explain a little bit about VBA to further your understanding!

I'm not sure where you copied this code from, but it's terribly inefficient. If I may:
Dim N, D As Long will cause D to be a Long, and N to be a variant. As you may know, variants are one of the slowest data types available. This line should be: Dim N As Long, D As Long
You only need to test every other number as an even number will always be divisible by two. (Therefore can not possibly be prime).
You don't need to test all the way up to N. You only need to test up to the Square Root of N. This is because after the square root the factors just switch sides, so you are just retesting values.
For Loops only evaluate the For-Line once for the life of the loop, but Do and While loops evaluate their conditional on every loop, so N-1 is being evaluated many, many times. Store this value in a variable if you want to use a Do Loop.
Ok, so now that we have dispensed with the blah, blah, blah, here is the code. I structured it so you can use it as a UDF from Excel as well (Ex: =ISPRIME(A2)):
Option Explicit
Sub GetPrime()
Dim varValue As Variant
varValue = Excel.ActiveSheet.Cells(2&, 2&).Value
If IsNumeric(varValue) Then
If CLng(varValue) = varValue Then
If IsPrime(varValue) Then
MsgBox varValue & " is prime", vbInformation, "Prime Test"
Else
MsgBox varValue & " is not prime", vbExclamation, "Prime Test"
End If
Exit Sub
End If
End If
MsgBox "This operation may only be performed on an integer value.", vbCritical, "Tip"
End Sub
Public Function IsPrime(ByVal num As Long) As Boolean
Dim lngNumDiv As Long
Dim lngNumSqr As Long
Dim blnRtnVal As Boolean
''//If structure is to optimize logical evaluation as AND/OR operators do not
''//use short-circuit evaluation in VB.'
If num = 2& Then
blnRtnVal = True
ElseIf num < 2& Then 'Do nothing, false by default.
ElseIf num Mod 2& = 0& Then 'Do nothing, false by default.
Else
lngNumSqr = Sqr(num)
For lngNumDiv = 3& To lngNumSqr Step 2&
If num Mod lngNumDiv = 0& Then Exit For
Next
blnRtnVal = lngNumDiv > lngNumSqr
End If
IsPrime = blnRtnVal
End Function

You can optimise it further (and make it more readable, in my opinion) by making the following changes. First performance:
Use longs, not floats. This will result in a huge speed increase.
You don't need to check up to n-1, only the square root of n. That's because if a factor d greater than sqrt(n) exists, its counterpart n/d would have already been found under sqrt(n). We use a special variable for this so that we don't get overflow by calculating divisor2. It also speeds it up by calculating that once rather than calculating the square every time through the loop (even though getting the square root is undoubtedly slower than squaring, it only happens once).
Do a special check first for multiples of two then you need only check that your number is a multiple of an odd number, effectively doubling the speed (not checking if you're a factor of a multiple of two).
Use the modulo operator rather than division/multiplication.
Now readability:
Use descriptive variable names.
Use a boolean for boolean values (not a string like tag).
Move the message box logic down to the bottom, based on the isPrime boolean, rather than scattering the messages amongst your code.
With all those changes, the following code can detect a 9-digit prime number (795,028,841) in well under a second. In fact, we can detect the largest 31-bit prime (2,147,483,647) in the same time.
Based on benchmarks (putting a 10,000-iteration for loop around the select), it takes 35 seconds on my box to detect that 31-bit prime. That's about 285 times per second - hopefully that'll be fast enough for you :-)
Option Explicit
Public Sub Go()
Dim number As Long
Dim divisor As Long
Dim maxdivisor As Long
Dim isPrime As Boolean
number = CLng(Cells(2, 2))
Select Case number
Case Is < 2
isPrime = False
Case Is = 2
isPrime = True
Case Is > 2
isPrime = True
If number mod 2 = 0 Then
isPrime = False
Else
maxdivisor = CLng(Sqr(number)) + 1
divisor = 3
Do
If number mod divisor = 0 Then
isPrime = False
Exit Do
End If
divisor = divisor + 2
Loop While divisor <= maxdivisor
End If
End Select
If isPrime Then
MsgBox "Number (" & number & ") is prime"
Else
MsgBox "Number (" & number & ") is not prime"
End If
End Sub

Related

Why is this IF statement not working correctly? The values of the variables in the immediate window show that it should be working [duplicate]

I have trouble comparing 2 double in Excel VBA
suppose that I have the following code
Dim a as double
Dim b as double
a = 0.15
b = 0.01
After a few manipulations on b, b is now equal to 0.6
however the imprecision related to the double data type gives me headache because
if a = b then
//this will never trigger
end if
Do you know how I can remove the trailing imprecision on the double type?
You can't compare floating point values for equality. See this article on "Comparing floating point numbers" for a discussion of how to handle the intrinsic error.
It isn't as simple as comparing to a constant error margin unless you know for sure what the absolute range of the floats is beforehand.
if you are going to do this....
Dim a as double
Dim b as double
a = 0.15
b = 0.01
you need to add the round function in your IF statement like this...
If Round(a,2) = Round(b,2) Then
//code inside block will now trigger.
End If
See also here for additional Microsoft reference.
It is never wise to compare doubles on equality.
Some decimal values map to several floating point representations. So one 0.6 is not always equal to the other 0.6.
If we subtract one from the other, we probably get something like 0.00000000051.
We can now define equality as having a difference smaller that a certain error margin.
Here is a simple function I wrote:
Function dblCheckTheSame(number1 As Double, number2 As Double, Optional Digits As Integer = 12) As Boolean
If (number1 - number2) ^ 2 < (10 ^ -Digits) ^ 2 Then
dblCheckTheSame = True
Else
dblCheckTheSame = False
End If
End Function
Call it with:
MsgBox dblCheckTheSame(1.2345, 1.23456789)
MsgBox dblCheckTheSame(1.2345, 1.23456789, 4)
MsgBox dblCheckTheSame(1.2345678900001, 1.2345678900002)
MsgBox dblCheckTheSame(1.2345678900001, 1.2345678900002, 14)
As has been pointed out, many decimal numbers cannot be represented precisely as traditional floating-point types. Depending on the nature of your problem space, you may be better off using the Decimal VBA type which can represent decimal numbers (base 10) with perfect precision up to a certain decimal point. This is often done for representing money for example where 2-digit decimal precision is often desired.
Dim a as Decimal
Dim b as Decimal
a = 0.15
b = 0.01
Late answer but I'm surprised a solution hasn't been posted that addresses the concerns outlined in the article linked in the (currently) accepted answer, namely that:
Rounding checks equality with absolute tolerance (e.g. 0.0001 units if rounded to 4d.p.) which is rubbish when comparing different values on multiple orders of magnitude (so not just comparing to 0)
Relative tolerance that scales with one of the numbers being compared meanwhile is not mentioned in the current answers, but performs well on non-zero comparisons (however will be bad at comparing to zero as the scaling blows up around then).
To solve this, I've taken inspiration from Python: PEP 485 -- A Function for testing approximate equality to implement the following (in a standard module):
Code
'#NoIndent: Don't want to lose our description annotations
'#Folder("Tests.Utils")
Option Explicit
Option Private Module
'Based on Python's math.isclose https://github.com/python/cpython/blob/17f94e28882e1e2b331ace93f42e8615383dee59/Modules/mathmodule.c#L2962-L3003
'math.isclose -> boolean
' a: double
' b: double
' relTol: double = 1e-09
' maximum difference for being considered "close", relative to the
' magnitude of the input values
' absTol: double = 0.0
' maximum difference for being considered "close", regardless of the
' magnitude of the input values
'Determine whether two floating point numbers are close in value.
'Return True if a is close in value to b, and False otherwise.
'For the values to be considered close, the difference between them
'must be smaller than at least one of the tolerances.
'-inf, inf and NaN behave similarly to the IEEE 754 Standard. That
'is, NaN is not close to anything, even itself. inf and -inf are
'only close to themselves.
'#Description("Determine whether two floating point numbers are close in value, accounting for special values in IEEE 754")
Public Function IsClose(ByVal a As Double, ByVal b As Double, _
Optional ByVal relTol As Double = 0.000000001, _
Optional ByVal absTol As Double = 0 _
) As Boolean
If relTol < 0# Or absTol < 0# Then
Err.Raise 5, Description:="tolerances must be non-negative"
ElseIf a = b Then
'Short circuit exact equality -- needed to catch two infinities of
' the same sign. And perhaps speeds things up a bit sometimes.
IsClose = True
ElseIf IsInfinity(a) Or IsInfinity(b) Then
'This catches the case of two infinities of opposite sign, or
' one infinity and one finite number. Two infinities of opposite
' sign would otherwise have an infinite relative tolerance.
'Two infinities of the same sign are caught by the equality check
' above.
IsClose = False
Else
'Now do the regular computation on finite arguments. Here an
' infinite tolerance will always result in the function returning True,
' since an infinite difference will be <= to the infinite tolerance.
'This is to supress overflow errors as we deal with infinity.
'NaN has already been filtered out in the equality checks earlier.
On Error Resume Next
Dim diff As Double: diff = Abs(b - a)
If diff <= absTol Then
IsClose = True
ElseIf diff <= CDbl(Abs(relTol * b)) Then
IsClose = True
ElseIf diff <= CDbl(Abs(relTol * a)) Then
IsClose = True
End If
On Error GoTo 0
End If
End Function
'#Description "Checks if Number is IEEE754 +/- inf, won't raise an error"
Public IsInfinity(ByVal Number As Double) As Boolean
On Error Resume Next 'in case of NaN
IsInfinity = Abs(Number) = PosInf
On Error GoTo 0
End Function
'#Description "IEEE754 -inf"
Public Property Get NegInf() As Double
On Error Resume Next
NegInf = -1 / 0
On Error GoTo 0
End Property
'#Description "IEEE754 +inf"
Public Property Get PosInf() As Double
On Error Resume Next
PosInf = 1 / 0
On Error GoTo 0
End Property
'#Description "IEEE754 signaling NaN (sNaN)"
Public Property Get NaN() As Double
On Error Resume Next
NaN = 0 / 0
On Error GoTo 0
End Property
'#Description "IEEE754 quiet NaN (qNaN)"
Public Property Get QNaN() As Double
QNaN = -NaN
End Property
Updated to incorporate great feedback from Cristian Buse
Examples
The IsClose function can be used to check for absolute difference:
assert(IsClose(0, 0.0001233, absTol:= 0.001)) 'same to 3 d.p.?
... or relative difference:
assert(IsClose(1234.5, 1234.6, relTol:= 0.0001)) '0.01% relative difference?
... but generally you specify both and if either tolerance is met then the numbers are considered close. It has special handling of +-infinity which are only close to themselves, and NaN which is close to nothing (see the PEP for full justification, or my Code Review post where I'd love feedback on this code :)
The Currency data type may be a good alternative. It handles relatively large numbers with fixed four digit precision.
Work-a-round??
Not sure if this will answer all scenarios, but I ran into a problem comparing rounded double values in VBA. When I compared to numbers that appeared to be identical after rounding, VBA would trigger false in an if-then compare statement.
My fix was to run two conversions, first double to string, then string to double, and then do the compare.
Simulated Example
I did not record the exact numbers that caused the error mentioned in this post, and the amounts in my example do not trigger the problem currently and are intended to represent the type of issue.
Sub Test_Rounded_Numbers()
Dim Num1 As Double
Dim Num2 As Double
Let Num1 = 123.123456789
Let Num2 = 123.123467891
Let Num1 = Round(Num1, 4) '123.1235
Let Num2 = Round(Num2, 4) '123.1235
If Num1 = Num2 Then
MsgBox "Correct Match, " & Num1 & " does equal " & Num2
Else
MsgBox "Inccorrect Match, " & Num1 & " does not equal " & Num2
End If
'Here it would say that "Inccorrect Match, 123.1235 does not equal 123.1235."
End Sub
Sub Fixed_Double_Value_Type_Compare_Issue()
Dim Num1 As Double
Dim Num2 As Double
Let Num1 = 123.123456789
Let Num2 = 123.123467891
Let Num1 = Round(Num1, 4) '123.1235
Let Num2 = Round(Num2, 4) '123.1235
'Add CDbl(CStr(Double_Value))
'By doing this step the numbers
'would trigger if they matched
'100% of the time
If CDbl(CStr(Num1)) = CDbl(CStr(Num2)) Then
MsgBox "Correct Match"
Else
MsgBox "Inccorrect Match"
End If
'Now it says Here it would say that "Correct Match, 123.1235 does equal 123.1235."
End Sub
Depending on your situation and your data, and if you're happy with the level of precision shown by default, you can try comparing the string conversions of the numbers as a very simple coding solution:
if cstr(a) = cstr(b)
This will include as much precision as would be displayed by default, which is generally sufficient to consider the numbers equal.
This would be inefficient for very large data sets, but for me was useful when reconciling imported data which was identical but was not matching after storing the data in VBA Arrays.
Try to use Single values if possible.
Conversion to Double values generates random errors.
Public Sub Test()
Dim D01 As Double
Dim D02 As Double
Dim S01 As Single
Dim S02 As Single
S01 = 45.678 / 12
S02 = 45.678
D01 = S01
D02 = S02
Debug.Print S01 * 12
Debug.Print S02
Debug.Print D01 * 12
Debug.Print D02
End Sub
45,678
45,678
45,67799949646
45,6780014038086

How do you correct incorrect multiplication in Excel VBA (automatic round down/missing decimals)? [duplicate]

I have trouble comparing 2 double in Excel VBA
suppose that I have the following code
Dim a as double
Dim b as double
a = 0.15
b = 0.01
After a few manipulations on b, b is now equal to 0.6
however the imprecision related to the double data type gives me headache because
if a = b then
//this will never trigger
end if
Do you know how I can remove the trailing imprecision on the double type?
You can't compare floating point values for equality. See this article on "Comparing floating point numbers" for a discussion of how to handle the intrinsic error.
It isn't as simple as comparing to a constant error margin unless you know for sure what the absolute range of the floats is beforehand.
if you are going to do this....
Dim a as double
Dim b as double
a = 0.15
b = 0.01
you need to add the round function in your IF statement like this...
If Round(a,2) = Round(b,2) Then
//code inside block will now trigger.
End If
See also here for additional Microsoft reference.
It is never wise to compare doubles on equality.
Some decimal values map to several floating point representations. So one 0.6 is not always equal to the other 0.6.
If we subtract one from the other, we probably get something like 0.00000000051.
We can now define equality as having a difference smaller that a certain error margin.
Here is a simple function I wrote:
Function dblCheckTheSame(number1 As Double, number2 As Double, Optional Digits As Integer = 12) As Boolean
If (number1 - number2) ^ 2 < (10 ^ -Digits) ^ 2 Then
dblCheckTheSame = True
Else
dblCheckTheSame = False
End If
End Function
Call it with:
MsgBox dblCheckTheSame(1.2345, 1.23456789)
MsgBox dblCheckTheSame(1.2345, 1.23456789, 4)
MsgBox dblCheckTheSame(1.2345678900001, 1.2345678900002)
MsgBox dblCheckTheSame(1.2345678900001, 1.2345678900002, 14)
As has been pointed out, many decimal numbers cannot be represented precisely as traditional floating-point types. Depending on the nature of your problem space, you may be better off using the Decimal VBA type which can represent decimal numbers (base 10) with perfect precision up to a certain decimal point. This is often done for representing money for example where 2-digit decimal precision is often desired.
Dim a as Decimal
Dim b as Decimal
a = 0.15
b = 0.01
Late answer but I'm surprised a solution hasn't been posted that addresses the concerns outlined in the article linked in the (currently) accepted answer, namely that:
Rounding checks equality with absolute tolerance (e.g. 0.0001 units if rounded to 4d.p.) which is rubbish when comparing different values on multiple orders of magnitude (so not just comparing to 0)
Relative tolerance that scales with one of the numbers being compared meanwhile is not mentioned in the current answers, but performs well on non-zero comparisons (however will be bad at comparing to zero as the scaling blows up around then).
To solve this, I've taken inspiration from Python: PEP 485 -- A Function for testing approximate equality to implement the following (in a standard module):
Code
'#NoIndent: Don't want to lose our description annotations
'#Folder("Tests.Utils")
Option Explicit
Option Private Module
'Based on Python's math.isclose https://github.com/python/cpython/blob/17f94e28882e1e2b331ace93f42e8615383dee59/Modules/mathmodule.c#L2962-L3003
'math.isclose -> boolean
' a: double
' b: double
' relTol: double = 1e-09
' maximum difference for being considered "close", relative to the
' magnitude of the input values
' absTol: double = 0.0
' maximum difference for being considered "close", regardless of the
' magnitude of the input values
'Determine whether two floating point numbers are close in value.
'Return True if a is close in value to b, and False otherwise.
'For the values to be considered close, the difference between them
'must be smaller than at least one of the tolerances.
'-inf, inf and NaN behave similarly to the IEEE 754 Standard. That
'is, NaN is not close to anything, even itself. inf and -inf are
'only close to themselves.
'#Description("Determine whether two floating point numbers are close in value, accounting for special values in IEEE 754")
Public Function IsClose(ByVal a As Double, ByVal b As Double, _
Optional ByVal relTol As Double = 0.000000001, _
Optional ByVal absTol As Double = 0 _
) As Boolean
If relTol < 0# Or absTol < 0# Then
Err.Raise 5, Description:="tolerances must be non-negative"
ElseIf a = b Then
'Short circuit exact equality -- needed to catch two infinities of
' the same sign. And perhaps speeds things up a bit sometimes.
IsClose = True
ElseIf IsInfinity(a) Or IsInfinity(b) Then
'This catches the case of two infinities of opposite sign, or
' one infinity and one finite number. Two infinities of opposite
' sign would otherwise have an infinite relative tolerance.
'Two infinities of the same sign are caught by the equality check
' above.
IsClose = False
Else
'Now do the regular computation on finite arguments. Here an
' infinite tolerance will always result in the function returning True,
' since an infinite difference will be <= to the infinite tolerance.
'This is to supress overflow errors as we deal with infinity.
'NaN has already been filtered out in the equality checks earlier.
On Error Resume Next
Dim diff As Double: diff = Abs(b - a)
If diff <= absTol Then
IsClose = True
ElseIf diff <= CDbl(Abs(relTol * b)) Then
IsClose = True
ElseIf diff <= CDbl(Abs(relTol * a)) Then
IsClose = True
End If
On Error GoTo 0
End If
End Function
'#Description "Checks if Number is IEEE754 +/- inf, won't raise an error"
Public IsInfinity(ByVal Number As Double) As Boolean
On Error Resume Next 'in case of NaN
IsInfinity = Abs(Number) = PosInf
On Error GoTo 0
End Function
'#Description "IEEE754 -inf"
Public Property Get NegInf() As Double
On Error Resume Next
NegInf = -1 / 0
On Error GoTo 0
End Property
'#Description "IEEE754 +inf"
Public Property Get PosInf() As Double
On Error Resume Next
PosInf = 1 / 0
On Error GoTo 0
End Property
'#Description "IEEE754 signaling NaN (sNaN)"
Public Property Get NaN() As Double
On Error Resume Next
NaN = 0 / 0
On Error GoTo 0
End Property
'#Description "IEEE754 quiet NaN (qNaN)"
Public Property Get QNaN() As Double
QNaN = -NaN
End Property
Updated to incorporate great feedback from Cristian Buse
Examples
The IsClose function can be used to check for absolute difference:
assert(IsClose(0, 0.0001233, absTol:= 0.001)) 'same to 3 d.p.?
... or relative difference:
assert(IsClose(1234.5, 1234.6, relTol:= 0.0001)) '0.01% relative difference?
... but generally you specify both and if either tolerance is met then the numbers are considered close. It has special handling of +-infinity which are only close to themselves, and NaN which is close to nothing (see the PEP for full justification, or my Code Review post where I'd love feedback on this code :)
The Currency data type may be a good alternative. It handles relatively large numbers with fixed four digit precision.
Work-a-round??
Not sure if this will answer all scenarios, but I ran into a problem comparing rounded double values in VBA. When I compared to numbers that appeared to be identical after rounding, VBA would trigger false in an if-then compare statement.
My fix was to run two conversions, first double to string, then string to double, and then do the compare.
Simulated Example
I did not record the exact numbers that caused the error mentioned in this post, and the amounts in my example do not trigger the problem currently and are intended to represent the type of issue.
Sub Test_Rounded_Numbers()
Dim Num1 As Double
Dim Num2 As Double
Let Num1 = 123.123456789
Let Num2 = 123.123467891
Let Num1 = Round(Num1, 4) '123.1235
Let Num2 = Round(Num2, 4) '123.1235
If Num1 = Num2 Then
MsgBox "Correct Match, " & Num1 & " does equal " & Num2
Else
MsgBox "Inccorrect Match, " & Num1 & " does not equal " & Num2
End If
'Here it would say that "Inccorrect Match, 123.1235 does not equal 123.1235."
End Sub
Sub Fixed_Double_Value_Type_Compare_Issue()
Dim Num1 As Double
Dim Num2 As Double
Let Num1 = 123.123456789
Let Num2 = 123.123467891
Let Num1 = Round(Num1, 4) '123.1235
Let Num2 = Round(Num2, 4) '123.1235
'Add CDbl(CStr(Double_Value))
'By doing this step the numbers
'would trigger if they matched
'100% of the time
If CDbl(CStr(Num1)) = CDbl(CStr(Num2)) Then
MsgBox "Correct Match"
Else
MsgBox "Inccorrect Match"
End If
'Now it says Here it would say that "Correct Match, 123.1235 does equal 123.1235."
End Sub
Depending on your situation and your data, and if you're happy with the level of precision shown by default, you can try comparing the string conversions of the numbers as a very simple coding solution:
if cstr(a) = cstr(b)
This will include as much precision as would be displayed by default, which is generally sufficient to consider the numbers equal.
This would be inefficient for very large data sets, but for me was useful when reconciling imported data which was identical but was not matching after storing the data in VBA Arrays.
Try to use Single values if possible.
Conversion to Double values generates random errors.
Public Sub Test()
Dim D01 As Double
Dim D02 As Double
Dim S01 As Single
Dim S02 As Single
S01 = 45.678 / 12
S02 = 45.678
D01 = S01
D02 = S02
Debug.Print S01 * 12
Debug.Print S02
Debug.Print D01 * 12
Debug.Print D02
End Sub
45,678
45,678
45,67799949646
45,6780014038086

VBA: How to a edit a code to eliminate the out of stack space error?

This code used to work, but recently I am getting an error (out of stack space).
I think the code is failing because I am calling a function too many times without exiting/ending.
If that is the case, how many times can you call a function and is there something I can do to fix this?
I am not the original author of this code.
I included the sub where the error occurs.
Sub CalculatePct(e As Variant)
Dim G As Integer
Dim pct As Double
Dim Owned100Pct As Boolean
If entities(e) < 0 Then
pct = 0
Owned100Pct = True ' Keeps track if the entity exists in the table other than as a parent
For G = 1 To UBound(MainArray, 1)
If MainArray(G, colEntity) = e Then
Owned100Pct = False
If entities(MainArray(G, colParent)) = -1 Then
'If we don't know the parent's ownership percentage, go and calculate it
CalculatePct MainArray(G, colParent)
End If
pct = pct + CDbl(MainArray(G, colPct)) / 100 * entities(MainArray(G, colParent))
End If
Next
If Owned100Pct Then
'Assume 100% owned if we don't know the parentage
'("Outside" entities won't go through here as they are already set to 0%)
entities(e) = 1
Else
'Store the entity's percentage
entities(e) = pct
End If
End If
End Sub
A #TimWilliams noted in the comments - you have an endless recursion loop.
Highlighting the problem area:
Sub CalculatePct(e As Variant)
[...]
If entities(MainArray(G, colParent)) = -1 Then
CalculatePct MainArray(G, colParent)
End If
[...]
End Sub
e is the parameter, and entities(e) is checked. In the code, MainParent(G, colParent) is used in place of e, so the next call to the routine gives e = MainParent(G, colParent).
Up to the point in the code, you do not change the value of G, colParent. entities or MainArray. So if entities(MainArray(G, colParent)) = -1 it will be forever calling itself.
Without knowing anything else about the code (including if recursion is necessary) I cannot suggest any definitive solutions. However, some things to consider:
Rewriting to be a loop instead of a recursive call
Making the recursive call to a subset of MainArray
Doing any amendments to G or colParent prior to the recursive
call
You've offered no indication of what line the error occurs on nor what MainArray represents but I'm guessing that MainArray has grown to a size greater than what can be accessed with a signed short integer.
Change the declaration of your iteration variable to a Signed Long Integer. This raises the functional limit of the variable from 32,767 iterations to 2,147,483,647.
Dim G As Long

VBA Greater Than Function Not Working

I have an issue where I am trying to compare a values that can be alphanumeric, only numeric, or only alphabetic.
The code originally worked fine for comparing anything within the same 100s group (IE 1-99 with alphabetic components). However when I included 100+ into it, it malfunctioned.
The current part of the code reads:
For j = 1 To thislength
If lennew < j Then
enteredval = Left("100A", lennew)
ElseIf lennew >= j Then
enteredval = Left("100A", j)
End If
If lenold < j Then
cellval = Left("67", lenold)
ElseIf lenold >= j Then
cellval = Left("67", j)
End If
'issue occurs here
If enteredval >= cellval Then
newrow = newrow+1
End If
Next j
The issue occurs in the last if statement.
When cycling through the 100 is greater than the 67 but still skips over. I tried to declare them both as strings (above this part of code) to see if that would help but it didn't.
What I am trying to accomplish is to sort through a bunch of rows and find where it should go. IE the 100A should go between 100 and 100B.
Sorry lennew=len("100A") and lennold=len("67"). And thislength=4or whatever is larger of the two lengths.
The problem is that you're trying to solve the comparison problem by attacking specific values, and that's going to be a problem to maintain. I'd make the problem more generic by creating a function that supplies takes two values returns -1 if the first operand is "before" the second, 0 if they are the same, and 1 if the first operand is "after" the second per your rules.
You could then restructure your code to eliminate the specific hardcoded prefix testing and then just call the comparison function directly, eg (and this is COMPLETELY untested, off-the-cuff, and my VBA is VERRRRRY stale :) but the idea is there: (it also assumes the existence of a simple string function called StripPrefix that just takes a string and strips off any leading digits, which I suspect you can spin up fairly readily yourself)
Function CompareCell(Cell1 as String, Cell2 as String) as Integer
Dim result as integer
Dim suffix1 as string
Dim suffix2 as string
if val(cell1)< val(cell2) Then
result = -1
else if val(cell1)>val(cell2) then
result = 1
else if val(cell1)=val(cell2) then
if len(cell1)=len(cell2) then
result =0
else
' write code to strip leading numeric prefixes
' You must supply StripPrefix, but it's pretty simple
' I just omitted it here for clarity
suffix1=StripPrefix(cell1) ' eg returns "ABC" for "1000ABC"
suffix2=StripPrefix(cell2)
if suffix1 < suffix2 then
result = -1
else if suffix1 > suffix2 then
result = 1
else
result = 0
end if
end if
return result
end function
A function like this then allows you to take any two cell references and compare them directly to make whatever decision you need:
if CompareCell(enteredval,newval)>=0 then
newrow=newrow+1
end if

Compare double in VBA precision problem

I have trouble comparing 2 double in Excel VBA
suppose that I have the following code
Dim a as double
Dim b as double
a = 0.15
b = 0.01
After a few manipulations on b, b is now equal to 0.6
however the imprecision related to the double data type gives me headache because
if a = b then
//this will never trigger
end if
Do you know how I can remove the trailing imprecision on the double type?
You can't compare floating point values for equality. See this article on "Comparing floating point numbers" for a discussion of how to handle the intrinsic error.
It isn't as simple as comparing to a constant error margin unless you know for sure what the absolute range of the floats is beforehand.
if you are going to do this....
Dim a as double
Dim b as double
a = 0.15
b = 0.01
you need to add the round function in your IF statement like this...
If Round(a,2) = Round(b,2) Then
//code inside block will now trigger.
End If
See also here for additional Microsoft reference.
It is never wise to compare doubles on equality.
Some decimal values map to several floating point representations. So one 0.6 is not always equal to the other 0.6.
If we subtract one from the other, we probably get something like 0.00000000051.
We can now define equality as having a difference smaller that a certain error margin.
Here is a simple function I wrote:
Function dblCheckTheSame(number1 As Double, number2 As Double, Optional Digits As Integer = 12) As Boolean
If (number1 - number2) ^ 2 < (10 ^ -Digits) ^ 2 Then
dblCheckTheSame = True
Else
dblCheckTheSame = False
End If
End Function
Call it with:
MsgBox dblCheckTheSame(1.2345, 1.23456789)
MsgBox dblCheckTheSame(1.2345, 1.23456789, 4)
MsgBox dblCheckTheSame(1.2345678900001, 1.2345678900002)
MsgBox dblCheckTheSame(1.2345678900001, 1.2345678900002, 14)
As has been pointed out, many decimal numbers cannot be represented precisely as traditional floating-point types. Depending on the nature of your problem space, you may be better off using the Decimal VBA type which can represent decimal numbers (base 10) with perfect precision up to a certain decimal point. This is often done for representing money for example where 2-digit decimal precision is often desired.
Dim a as Decimal
Dim b as Decimal
a = 0.15
b = 0.01
Late answer but I'm surprised a solution hasn't been posted that addresses the concerns outlined in the article linked in the (currently) accepted answer, namely that:
Rounding checks equality with absolute tolerance (e.g. 0.0001 units if rounded to 4d.p.) which is rubbish when comparing different values on multiple orders of magnitude (so not just comparing to 0)
Relative tolerance that scales with one of the numbers being compared meanwhile is not mentioned in the current answers, but performs well on non-zero comparisons (however will be bad at comparing to zero as the scaling blows up around then).
To solve this, I've taken inspiration from Python: PEP 485 -- A Function for testing approximate equality to implement the following (in a standard module):
Code
'#NoIndent: Don't want to lose our description annotations
'#Folder("Tests.Utils")
Option Explicit
Option Private Module
'Based on Python's math.isclose https://github.com/python/cpython/blob/17f94e28882e1e2b331ace93f42e8615383dee59/Modules/mathmodule.c#L2962-L3003
'math.isclose -> boolean
' a: double
' b: double
' relTol: double = 1e-09
' maximum difference for being considered "close", relative to the
' magnitude of the input values
' absTol: double = 0.0
' maximum difference for being considered "close", regardless of the
' magnitude of the input values
'Determine whether two floating point numbers are close in value.
'Return True if a is close in value to b, and False otherwise.
'For the values to be considered close, the difference between them
'must be smaller than at least one of the tolerances.
'-inf, inf and NaN behave similarly to the IEEE 754 Standard. That
'is, NaN is not close to anything, even itself. inf and -inf are
'only close to themselves.
'#Description("Determine whether two floating point numbers are close in value, accounting for special values in IEEE 754")
Public Function IsClose(ByVal a As Double, ByVal b As Double, _
Optional ByVal relTol As Double = 0.000000001, _
Optional ByVal absTol As Double = 0 _
) As Boolean
If relTol < 0# Or absTol < 0# Then
Err.Raise 5, Description:="tolerances must be non-negative"
ElseIf a = b Then
'Short circuit exact equality -- needed to catch two infinities of
' the same sign. And perhaps speeds things up a bit sometimes.
IsClose = True
ElseIf IsInfinity(a) Or IsInfinity(b) Then
'This catches the case of two infinities of opposite sign, or
' one infinity and one finite number. Two infinities of opposite
' sign would otherwise have an infinite relative tolerance.
'Two infinities of the same sign are caught by the equality check
' above.
IsClose = False
Else
'Now do the regular computation on finite arguments. Here an
' infinite tolerance will always result in the function returning True,
' since an infinite difference will be <= to the infinite tolerance.
'This is to supress overflow errors as we deal with infinity.
'NaN has already been filtered out in the equality checks earlier.
On Error Resume Next
Dim diff As Double: diff = Abs(b - a)
If diff <= absTol Then
IsClose = True
ElseIf diff <= CDbl(Abs(relTol * b)) Then
IsClose = True
ElseIf diff <= CDbl(Abs(relTol * a)) Then
IsClose = True
End If
On Error GoTo 0
End If
End Function
'#Description "Checks if Number is IEEE754 +/- inf, won't raise an error"
Public IsInfinity(ByVal Number As Double) As Boolean
On Error Resume Next 'in case of NaN
IsInfinity = Abs(Number) = PosInf
On Error GoTo 0
End Function
'#Description "IEEE754 -inf"
Public Property Get NegInf() As Double
On Error Resume Next
NegInf = -1 / 0
On Error GoTo 0
End Property
'#Description "IEEE754 +inf"
Public Property Get PosInf() As Double
On Error Resume Next
PosInf = 1 / 0
On Error GoTo 0
End Property
'#Description "IEEE754 signaling NaN (sNaN)"
Public Property Get NaN() As Double
On Error Resume Next
NaN = 0 / 0
On Error GoTo 0
End Property
'#Description "IEEE754 quiet NaN (qNaN)"
Public Property Get QNaN() As Double
QNaN = -NaN
End Property
Updated to incorporate great feedback from Cristian Buse
Examples
The IsClose function can be used to check for absolute difference:
assert(IsClose(0, 0.0001233, absTol:= 0.001)) 'same to 3 d.p.?
... or relative difference:
assert(IsClose(1234.5, 1234.6, relTol:= 0.0001)) '0.01% relative difference?
... but generally you specify both and if either tolerance is met then the numbers are considered close. It has special handling of +-infinity which are only close to themselves, and NaN which is close to nothing (see the PEP for full justification, or my Code Review post where I'd love feedback on this code :)
The Currency data type may be a good alternative. It handles relatively large numbers with fixed four digit precision.
Work-a-round??
Not sure if this will answer all scenarios, but I ran into a problem comparing rounded double values in VBA. When I compared to numbers that appeared to be identical after rounding, VBA would trigger false in an if-then compare statement.
My fix was to run two conversions, first double to string, then string to double, and then do the compare.
Simulated Example
I did not record the exact numbers that caused the error mentioned in this post, and the amounts in my example do not trigger the problem currently and are intended to represent the type of issue.
Sub Test_Rounded_Numbers()
Dim Num1 As Double
Dim Num2 As Double
Let Num1 = 123.123456789
Let Num2 = 123.123467891
Let Num1 = Round(Num1, 4) '123.1235
Let Num2 = Round(Num2, 4) '123.1235
If Num1 = Num2 Then
MsgBox "Correct Match, " & Num1 & " does equal " & Num2
Else
MsgBox "Inccorrect Match, " & Num1 & " does not equal " & Num2
End If
'Here it would say that "Inccorrect Match, 123.1235 does not equal 123.1235."
End Sub
Sub Fixed_Double_Value_Type_Compare_Issue()
Dim Num1 As Double
Dim Num2 As Double
Let Num1 = 123.123456789
Let Num2 = 123.123467891
Let Num1 = Round(Num1, 4) '123.1235
Let Num2 = Round(Num2, 4) '123.1235
'Add CDbl(CStr(Double_Value))
'By doing this step the numbers
'would trigger if they matched
'100% of the time
If CDbl(CStr(Num1)) = CDbl(CStr(Num2)) Then
MsgBox "Correct Match"
Else
MsgBox "Inccorrect Match"
End If
'Now it says Here it would say that "Correct Match, 123.1235 does equal 123.1235."
End Sub
Depending on your situation and your data, and if you're happy with the level of precision shown by default, you can try comparing the string conversions of the numbers as a very simple coding solution:
if cstr(a) = cstr(b)
This will include as much precision as would be displayed by default, which is generally sufficient to consider the numbers equal.
This would be inefficient for very large data sets, but for me was useful when reconciling imported data which was identical but was not matching after storing the data in VBA Arrays.
Try to use Single values if possible.
Conversion to Double values generates random errors.
Public Sub Test()
Dim D01 As Double
Dim D02 As Double
Dim S01 As Single
Dim S02 As Single
S01 = 45.678 / 12
S02 = 45.678
D01 = S01
D02 = S02
Debug.Print S01 * 12
Debug.Print S02
Debug.Print D01 * 12
Debug.Print D02
End Sub
45,678
45,678
45,67799949646
45,6780014038086

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