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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@xxxxxxxxx>
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
>
>
>
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-----------------------------------------
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