主题：macd顶底背离

是的可以这样，历史上满足条件的都可以显示

mm1:=sumbars(cross(macd,0),1);

nn1:=sumbars(cross(0,macd),1);

 

当前为绿住时的最低diff:

l1:=llv(diff,nn1);

if macd<0 then ll_diff1:=l1;

 

当前为绿住时之前一个绿住最低diff:

l2:=ref(llv(diff,nn1),mm1);

if macd<0 then ll_diff2:=l2;

 

当前为红柱时的最高diff:

h1:=hhv(diff,mm1);

if macd>0 then hh_diff1:=h1;

 

当前为红柱时之前一个红柱的最高diff:

h2:=ref(hhv(diff,mm1),nn1);

if macd>0 then hh_diff2:=h2;

if语句吃了好多资源，举步维艰

在哪里呀