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No, it won't. That statement doesn't turn an array back into a single
number.
I think what you want is to take a single price value from the
ValueWhen array and pass that to the function. In fact, it looks to me
like all you need is:
Price = Low[start];
GP
--- In amibroker@xxxxxxxxxxxxxxx, "wongloktim" <wlt@xxx> wrote:
>
> Even I define Hlow=0, Do Loop also can'nt work
>
>
> --- In amibroker@xxxxxxxxxxxxxxx, "gp_sydney" <gp.investment@>
> wrote:
> >
> > The problem is that the Price parameter passed to the function is
> an
> > array, since ValueWhen returns an array, and that makes HLow an
> array
> > and you can't test arrays in an "if" statement.
> >
> > GP
> >
> >
> > --- In amibroker@xxxxxxxxxxxxxxx, "wongloktim" <wlt@> wrote:
> > >
> > > Please Help
> > >
> > > I found this cause Do Loop problem:{for(i=bars+1;
> i<BarCount;i++)}
> > >
> >
> > ///////////////////////////////////////////////////////////////////
> /
> > > SetChartOptions(0,chartShowArrows|chartShowDates);
> > > _N(Title = FullName()+StrFormat(" {{NAME}} - {{INTERVAL}}
> {{DATE}}
> > > Open %g, Hi %g, Lo %g, Close %g (%.1f%%) {{VALUES}}", O, H, L,
> C,
> > > SelectedValue( ROC( C, 1 ) ) ));
> > > Plot( C,"Price",colorBlack,64 );
> > >
> > > function HigherLow(Bars,Price)
> > > {
> > > HLow=Price;
> > > HLowArr=Null;
> > > if(Bars>0)
> > > {
> > > for(i=bars+1; i<BarCount;i++)
> > > {
> > > if(Low[i] >= HLow)
> > > {
> > > HLow=Low[i];
> > > HLowArr[i]=HLow;
> > > }
> > > else
> > > HlowArr[i]=HLow;
> > > }
> > > }
> > > return HlowArr;
> > > }
> > > Start = SelectedValue(BarIndex());
> > > Price=ValueWhen(BarIndex()==start,L,1);
> > > Test=HigherLow(Start,Price);
> > > Plot(IIf(start,Test,Null),"StopLoss",colorRed,4);
> > >
> > > ///////////////////////////////////////////////////////////
> > >
> >
>
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