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I've been looking at the following code and it brought up this question: Is Stop1 presumed to be zero right off the bat, or do I have to call out Stop1:=0?
Stop1:=If( PREV < L,
{then} If(( H - 3*ATR(10) ) >= PREV,
{then} ( H - 3*ATR(10) ),
{else} PREV),
{else (L <= PREV)}
( H - 3*ATR(10) ));
Thanks
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Rank: Advanced Member
Groups: Registered, Registered Users, Subscribers
Joined: 7/25/2005(UTC)
Posts: 1,042
Was thanked: 57 time(s) in 54 post(s)
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Yes, an initial PREV value of zero is assumed. You can’t predefine a variable with PREV in it but what works in some situations is that the variable can be “seeded” on the first bar from within – see the following exponential moving average formula as an example of this.
{Exponential Moving Average}
Pds:=Input("Periods",1,1000,10);
Rate:=2/(Min(Cum(1),Pds)+1);
If(Cum(1)=0,C,PREV*(1-Rate)+C*Rate);
Your formula can be reduced to just two PREVs as follows.
X:=H-3*ATR(10);
If(PREV<L,Max(X,PREV),X);
In some cases a PREV-based variable can be rewritten to remove PREV altogether, although I don’t think it’s possible with your example.
Roy
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