CALL FB 2 , DB122
IN0 :=M12.0
IN1 :=8
IN2 :=P#DB120.DBX 0.0 BYTE 10
IN3 :=
IN4 :=
IN5 :=W#16#1
IN6 :=P#DB120.DBX 0.0 BYTE 10
IN7 :=
IN8 :=
IN9 :=W#16#2
IN10 :=P#DB120.DBX 0.0 BYTE 10
IN11 :=
IN12 :=
IN13 :=W#16#F
IN14 :=P#DB120.DBX 0.0 BYTE 10
IN15 :=
IN16 :=
IN17 :=W#16#10
IN18 :=P#DB120.DBX 10.0 BYTE 10
IN19 :=
IN20 :=
IN21 :=
IN22 :=P#DB120.DBX 20.0 BYTE 10
IN23 :=
IN24 :=
IN25 :=
IN26 :=P#DB120.DBX 0.0 BYTE 10
IN27 :=
IN28 :=
IN29 :=W#16#3
IN30 :=P#DB120.DBX 0.0 BYTE 10
IN31 :=
IN32 :=
IN33 :=W#16#5
OUT34:=M26.0
OUT35:=M27.0
OUT36:=MW28
IO37 :=P#M 32.0 REAL 1
IO38 :=P#M 36.0 REAL 1
IO39 :=P#M 40.0 REAL 1
IO40 :=P#M 44.0 REAL 1
IO41 :=P#M 48.0 REAL 1
IO42 :=P#M 52.0 REAL 1
IO43 :=P#M 56.0 REAL 1
IO44 :=P#M 60.0 REAL 1
程序中的
IO37 :=P#M 32.0 REAL 1
IO38 :=P#M 36.0 REAL 1
IO39 :=P#M 40.0 REAL 1
IO40 :=P#M 44.0 REAL 1
IO41 :=P#M 48.0 REAL 1
IO42 :=P#M 52.0 REAL 1
IO43 :=P#M 56.0 REAL 1
IO44 :=P#M 60.0 REAL 1
为何解?跟R值有关系吗?
请大家指教!