請(qǐng)教問題出在哪。怎么改寫代碼。
作者：金字塔 來源：cxh99.com 發(fā)布時(shí)間：2019年10月29日
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咨詢內(nèi)容：
序列模式下，主圖疊加。代碼如下：A1:EMA(CLOSE,5);A2:EMA(CLOSE,34);A44:EMA(CLOSE,44);C1:=CLOSE;K1:(C1<(A2-0.3*MINDIFF)&&C1<(A1-0.3*MINDIFF)&&C1<(A44-0.3*MINDIFF)),NODRAW;K2:(C1>(A2+0.3*MINDIFF)&&C1>(A1+0.3*MINDIFF)&&C1>(A44+0.3*MINDIFF)),NODRAW;G:IF(K2=0&&K1=1&&REF(K1,1)=0,-3,IF(K1=0&&K2=1&&REF(K2,1)=0,+3,0)),NODRAW;編譯能通過，應(yīng)用于圖的時(shí)候就顯示最后第七行解釋錯(cuò)誤，無論是序列模式還是逐k模式，無論選主圖 副圖。都是此行解釋錯(cuò)誤。最后一行也就是G:IF(K2=0&&K1=1&&REF(K1,1)=0,-3,IF(K1=0&&K2=1&&REF(K2,1)=0,+3,0)),NODRAW;問題出在哪？將最后一行注釋掉，//G:IF(K2=0&&K1=1&&REF(K1,1)=0,-3,IF(K1=0&&K2=1&&REF(K2,1)=0,+3,0)),NODRAW;則正常了。可是我就是需要最后一行得到最終的判定條件啊問題一：序列模式下，ref函數(shù)能夠應(yīng)用于if語句里面嗎問題二：if語句里再嵌套if語句嘛。
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金字塔客服：
//用下面代碼再試試
A1:EMA(CLOSE,5);
A2:EMA(CLOSE,34);
A44:EMA(CLOSE,44);
C1:=CLOSE;
K1:(C1<(A2-0.3*MINDIFF)&&C1<(A1-0.3*MINDIFF)&&C1<(A44-0.3*MINDIFF)),NODRAW;
K2:(C1>(A2+0.3*MINDIFF)&&C1>(A1+0.3*MINDIFF)&&C1>(A44+0.3*MINDIFF)),NODRAW;
G:IF(K2=0&&K1=1&&REF(K1,1)=0,-3,IF(K1=0&&K2=1&&REF(K2,1)=0,3,0)),NODRAW;
1、可以使用；
2、可以再嵌套
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