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Tomasz,
When a*b is an array, we may have quite strange results.
Example
p=StochD();
Cond=MACD()>Signal();
p=IIf(Cond,RSI(),Ref(p,-1));
We have to suppose we use THE SAME array a*b, which, I think, it is
not clear in Keath´s question.
DT
--- In amibroker@xxxxxxxxxxxxxxx, "Tomasz Janeczko" <amibroker@xxxx>
wrote:
> Dimitris,
>
> You are right *IF* 'a' and 'b' are NOT placeholders of more complex
> statements that are evaluated each bar OR they are not arrays.
>
> If 'a' and 'b' are arrays your assumptions are not correct.
>
> I already sent response to this post but Yahoo somehow
> 'forgot' it, so here it is again:
>
> Assumption: a and b are arrays.
>
> Using AMA/AMA2:
> =============
>
> x = AMA( a * b, Condition == True );
>
> Using loops:
> ========
>
> for( i = 1; i < BarCount; i++ )
> {
> if( Condition[ i ] )
> {
> x[ i ] = a[ i ] * b[ i ];
> }
> else
> {
> x[ i ] = x[ i - 1 ];
> }
> }
>
>
>
>
> Best regards,
> Tomasz Janeczko
> amibroker.com
>
> ----- Original Message -----
> From: "DIMITRIS TSOKAKIS" <TSOKAKIS@xxxx>
> To: <amibroker@xxxxxxxxxxxxxxx>
> Sent: Thursday, May 08, 2003 1:43 PM
> Subject: [amibroker] Re: How to solve in AFL.....
>
>
> > Your x, as described, is always a*b. Suppose day1 with true
> > condition, then
> > day1
> > COND=TRUE
> > X=a*b
> > day2
> > COND=FALSE
> > X should be its previous value, ie a*b again
> > Agree?
> > DT
> > --- In amibroker@xxxxxxxxxxxxxxx, "DIMITRIS TSOKAKIS"
<TSOKAKIS@xxxx>
> > wrote:
> > > As it is written, Condition may be TRUE, FALSE or something
else ?
> > > DT
> > > --- In amibroker@xxxxxxxxxxxxxxx, "Keith Bennett"
<kbennett@xxxx>
> > > wrote:
> > > >
> > > > 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
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