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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
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
> >
> >
> >
> > Send BUG REPORTS to bugs@xxxx
> > Send SUGGESTIONS to suggest@xxxx
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> >
> >
> >
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