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Dimitris,
From: "DIMITRIS TSOKAKIS" <TSOKAKIS@xxxxxxxxx>
>
> Tomasz,
> Let us look at the question first :
> Your code is [I suppose] for the question
> x = IIF(Condition is TRUE, a * b, Ref(x,-1));
> Keith question was
> x = IIF(Condition is TRUE, a * b,
> IIF(Condition is FALSE, Ref(x,-1),NULL));
> I think it does not make sense because Condition is TRUE OR FALSE
> without 3rd possibility
yes, and especially if it is the same condition for the same bar :-)
> DT
>
>
> --- In amibroker@xxxxxxxxxxxxxxx, "Tomasz Janeczko" <amibroker@xxxx>
> wrote:
> > Hello,
> >
> > Using AMA/AMA2:
> >
> > x = AMA( a * b, Condition == True );
> >
> > Using loops:
> >
> > for( i = 1; i < BarCount; i++ )
> > {
> > if( Condition[ i ] )
> > {
> > x[ i ] = a * b;
> > }
> > else
> > {
> > x[ i ] = x[ i - 1 ];
> > }
> > }
> >
> >
> > Best regards,
> > Tomasz Janeczko
> > amibroker.com
> > ----- Original Message -----
> > From: "Keith Bennett" <kbennett@xxxx>
> > To: <amibroker@xxxxxxxxxxxxxxx>
> > Sent: Thursday, May 08, 2003 12:33 PM
> > Subject: [amibroker] How to solve in AFL.....
> >
> >
> > >
> > > Hi,
> > >
> > > The following situation frequently occurs, but it doesn't seem
> > > possible to solve directly due to the self reference.
> > >
> > > How do you tackle this in AFL:
> > >
> > > x = IIF(Condition is TRUE, a * b,
> > > IIF(Condition is FALSE, Ref(x,-1),NULL));
> > >
> > > Thanks
> > > Keith
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