A06B-0238-B605 TAK //Result -> Accu 2 把8.0放入ACCU2
ENT //Accu 2 -> Accu 3 再把8.0放入ACCU3中
L #TEMP //例如#TEMP數(shù)值是20.0
L #LIMIT_Start
-R //Accu 3 -> Accu 2 計(jì)算結(jié)果是18.0，并且把ACCU3內(nèi)容8.0放入
ACCU2中去（對(duì)于有4個(gè)累加器的CPU才會(huì)執(zhí)行此動(dòng)作）
A06B-0238-B605 TAK //把18.0放入ACCU2,把8.0放入ACCU1
/R //計(jì)算結(jié)果=8.0/18.0
L #IN //Eingang laden
L 2#1
AD //Eingang maskieren
A06B-0238-B605 SRD 0 //Bit ganz nach rechts auf start f黵 schieben
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