Re: gettingstarted Digest, Vol 2, Issue 24

[Aubrey Todd <atodd at copper dot net>]

>
Yes, I am stepping through the Debugger watching the numbers change.

> >   If Pos>1 Then
> >      perform some process here
> >     End If
'Perform some process here' has another If-then statement which only 
activates with a string like "+3(" in the first loop and ")3+" in the 
second loop. In each loop the Debugger jumps from the second If-then 
statement to the 'Loop Until Pos=0' step and nothing is processed 
in-between.
After posting the original problem, I rechecked 
ProbStr=(2+1/2)+(3+1/4), and it has the same problem. You can watch the 
value of Pos reset to a smaller value in the debugger. If you try  just 
(2+1/2), it works fine.
I have no idea of where to go from here.

Aubrey

> Message: 8
> Subject: InStr problems
> From: Aubrey Todd <atodd at copper dot net>
> Date: Thu, 27 May 2004 11:41:04 -0700
>
>
> I am having a problem with the InStr command in the following code:
>
>    Dim Pos as Integer
>
>    //Check for Justaposition on the left
>   //Checks for a number before the '(' and inserts a '*' if appropriate
>    Pos=-1
>    Do //This loop works fine
>      Pos=Pos+2
>      Pos=InStr(Pos,ProbStr,"(")
>      If Pos>1 Then
>       perform some process here
>      End If
>    Loop Until Pos=0
>
>    //Check for Justaposition on the right
>   //Checks for a number after the ')' and inserts a '*' if appropriate
>    Pos=-1
>    Do //This loop does not work all the time
>      Pos=Pos+2
>      Pos=InStr(Pos,ProbStr,")") //<===Problem occurs here
>      If Pos>1 Then
>       perform some process here
>      End If
>    Loop Until Pos=0
>
> ProbStr is a string property.
> If ProbStr=(2+1/2)+(3+1/4) both loops work correctly.
> If ProbStr=Dec(2+1/2)+Dec(3+1/4), the second loop does not exit because
> Pos never goes to zero.
> In the second loop,  Pos=InStr(Pos,ProbStr,")")=21 the second time
> through, then loop and  Pos=Pos+2=23, but when it hits
> Pos=InStr(Pos,ProbStr,")") again, Pos is set to 21 instead of zero. If
> I manually change the command to Pos=InStr(23,ProbStr,")"), it works
> fine and returns a zero.
>
> Anyone have a hint as to what I am doing wrong or why it does not work?
>
> Thanks,
> Aubrey
> PB G4 Mac OS 10.3.3
> RB 5.2.4
>
>
>
>
> ==================================================
>
> Message: 9
> Subject: Re: InStr problems
> From: "Joseph J. Strout" <joe at realsoftware dot com>
> Date: Thu, 27 May 2004 13:47:48 -0500
>
> At 11:41 AM -0700 5/27/04, Aubrey Todd wrote:
>
> >   Do //This loop does not work all the time
> >     Pos=Pos+2
> >     Pos=InStr(Pos,ProbStr,")") //<===Problem occurs here
> >     If Pos>1 Then
> >      perform some process here
> >     End If
> >   Loop Until Pos=0
> > ...
> > In the second loop,  Pos=InStr(Pos,ProbStr,")")=21 the second time
> > through, then loop and  Pos=Pos+2=23, but when it hits
> > Pos=InStr(Pos,ProbStr,")") again, Pos is set to 21 instead of zero.
> You've verified this by stepping through it with the debugger?
>
> >  If I manually change the command to Pos=InStr(23,ProbStr,")"), it
> > works fine and returns a zero.
> In that case, there has to be a some sort of confusion here; there is
> no difference to InStr between a literal 23 and a variable which
> contains 23.
>
> My guess is that the second time through the loop, Pos is not
> actually equal to 21.  Perhaps "perform some process here" is
> accidentally clobbering the value of Pos.
>
> Best,
> - Joe
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