Cdbl() bug?

[segrobiur]

Hi there...

I've recently tried to convert a string into a double, using Cdbl(), however I think there is a bug in VB6 Pro...

Here's what I get:

dim A as string
dim B as double

A="123.45"

if I use it this way,

B= cdbl(A)

I get: B=12345 (Wrong!!!)

If I use this way,

B= cdbl(123.45)

I get: B=123.45 (Correct!!)

Does anybody else had this problem?

thanks in advance

[chemicalNova]

This works for me:

Dim A As String
Dim B As Double
 
A = "123.45"
 
B = CDbl(A)
MsgBox B

(I have Enterprise Edition)

See attachment..

Phreak

[segrobiur]

And is there a way to go around the bug??

I mean .. can I take the " " from the string variable and insert it in the cdbl() ?

[agmorgan]

Post a screenshot like phReAk did
I cant believe you are getting those answers when noone else is

[chemicalNova]

Does this work?

Dim A As String
Dim B As Double
 
A = "123.45"
 
B = CDbl(CInt(Left(A, 3)) + Right(A, 3))
MsgBox B

It kinda uses integers aswell, but works just the same.

Does it work for you?

Phreak

[segrobiur]

Well I don't have VB right now but I'll get the printscreen as soon as I can...

to tell you the truth that bug happend when I used an array of strings... when I called one of those variables inside the array I had the bug..

something like this:

dim A(10) as string
dim B as double

A(2)="123.45"
B=Cdbl(A(2))

as for phReAk's code... I don't think is the correct way to use since I chose double variable because I had very large (and no way to now their size) numbers, so I can't use integers...

[chemicalNova]

Will using the CDbl function fix this?

Dim A As String
Dim B As Double
 
A = "123.45"
 
B = CDbl(CDbl(Left(A, 3)) + Right(A, 3))
MsgBox B

?

Phreak

[segrobiur]

well... maybe, but the problem is I don't know how long will the number be (in the program) so I'll have to figure out a way to convert it ... (witch isn't that difficult - but it totally messes my code up)...
..
And since there is a function that SHOULD do this I don't see why I shoul develop new code....
..
but for what I'm seeing I'm the only one with this problem... so I'll soon get a print screen and see if there's more to it (??- I doubt)
..
Another thing ... I have no patches installed!! so it's VB6.0 Professional original..

[Buzby]

Believe me if there was a bug with CDBL it would've been noticed before version 6 sp 6...

I think you'll find there's a problem with your code and you're not passing the number to the CDBL function correctly.
Why not put a breakpoint on the CDBL command, and when the program hits it use the DEBUG window to discover what is being passed to the function...?

[manavo11]

Maybe instead of CDbl you can use Val? They both return a double...

[manavo11]

Or try using a comma (",") instead of a period (".").

[Buzby]

Stop blaming CDBL - there's nothing wrong with it - post the code you have in your project and we'll take a look.
Or but a DEBUG.PRINT [variable name] before the CDBL line to see what is actually being passed into the CDBL function.

[segrobiur]

...
... ok... here are the examples...
...
as you can see I kid you not... only the last worked... probably it has to do with assuming "." instead of ",".... but still it is a bug!!!

[Mc Brain]

As you can see, the msgbox tells you "123,45" and not "123.45". Try with a comma instead of a dot.

[Mc Brain]

This code:

Option Explicit
 
Private Sub Form_Load()
    Debug.Print CDbl("123.45")
    Debug.Print CDbl("123,45")
    
End Sub

returned:

12345
123,45

[segrobiur]

Well ... the thing is: in my project I import the data from a txt file and I can't control if the data inserted in the txt file is using "." or ","... so when I use the Cdbl(Variable) (being variable the data imported) I would expect it would recognise the data...

It may not be a problem unless I use a computer with a diferent language of windows installed... If I use the program in Win XP (English version) I would get the oposite... So I would have to know wich version of the windows is installed to be abble to convert right!!!....

Wich totally SUCKS ...

That's Why I hate Mcrosoft!!!

now that I got that out of my chest ... do you know of another function that converts strings to doubles (without bugs)???

By the way Buzby sometimes they also make mistakes!!

[Mc Brain]

It is not a bug!!

If you regional settings says that the "," is the decimal separator, then the "." would be interpreted as a thousand separator.

[segrobiur]

You may be right about that ... but still He didn't even recognised it as a thousand separator...

Is there a way to go around that?

[Mc Brain]

Who says it does not? It did recognize it as a thousand separator. When dealing with number the thousand separator "does not exist". So, "123.456" would be a string value, but its number would be 123456.

[Tec-Nico]

Oh come on... What is so difficult about doing this:

Public Function CDouble(ByVal strDouble As String) As Double
 Dim iPos As Integer
 Dim strAfterDot As String
  
 strDouble = Trim(strDouble)
 iPos = InStr(1, strDouble, ".")
 
 If iPos > 0 Then
  strAfterDot = Mid(strDouble, iPos + 1)
  CDouble = CInt(Mid(strDouble, 1, iPos)) + CInt(strAfterDot) * 0.1 ^ Len(strAfterDot)
 Else
  CDouble = CInt(strDouble)
 End If
End Function
 
Private Sub Form_Load()
 Debug.Print CDouble("13.22")
 Debug.Print CDouble("15467.09801")
 Debug.Print CDouble("12")
End Sub

[segrobiur]

thanks for the code... but I'll just use this one:

private sub solving_what_should_have_already_be_right( Mystring as string)

MyString=replace(MyString, ".", ",")

end sub

and hope that however runs my program has the same regional simbol for decimals...

[Mc Brain]

Originally posted by Tec-Nico
Oh come on... What is so difficult about doing this:

Public Function CDouble(ByVal strDouble As String) As Double
 Dim iPos As Integer
 Dim strAfterDot As String
  
 strDouble = Trim(strDouble)
 iPos = InStr(1, strDouble, ".")
 
 If iPos > 0 Then
  strAfterDot = Mid(strDouble, iPos + 1)
  CDouble = CInt(Mid(strDouble, 1, iPos)) + CInt(strAfterDot) * 0.1 ^ Len(strAfterDot)
 Else
  CDouble = CInt(strDouble)
 End If
End Function
 
Private Sub Form_Load()
 Debug.Print CDouble("13.22")
 Debug.Print CDouble("15467.09801")
 Debug.Print CDouble("12")
End Sub

As a matter of fact, it should be:

Public Function CDouble(ByVal strDouble As String) As Double
 Dim iPos As Integer
 Dim strAfterDot As String
  
 strDouble = Trim(strDouble)
 iPos = InStr(1, strDouble, ".")
 
 If iPos > 0 Then
  strAfterDot = Mid(strDouble, iPos + 1)
  CDouble = CInt(Mid(strDouble, 1, iPos)) + CInt(strAfterDot) * 0.1 ^ Len(strAfterDot)
 Else
  CDouble = [b]CDbl[/b](strDouble)
 End If
End Function