Cdbl() bug?

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

This works for me:

VB Code:

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

(I have Enterprise Edition)

See attachment..

Phreak

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() ?

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

Does this work?

VB Code:

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

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...

Will using the CDbl function fix this?

VB Code:

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

?

Phreak

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..

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...?

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

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

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.

...
... 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!!!

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

This code:

VB Code:

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

returned:

Quote:

---

12345
123,45

---

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:mad: ...

That's Why I hate Mcrosoft!!!

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

By the way Buzby sometimes they also make mistakes!!

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.

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

Is there a way to go around that?

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.

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

VB Code:

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

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...

Quote:

---

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

VB Code:

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:

VB Code:

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