Closed federicoorta closed 2 years ago
Hi @federicoorta, an empty bit should be inserted between bits.
1 0 0 1 0 1 x 0 1 1 0 1 1
c1 a1 c2 a2 c4 a4 b1 d1 b2 d2 b4 d4
The result should remain as 6615
Oh, I checked again. You are right, the extra zero bit is already included in the message, and the code was fixed by a different PR. Thank you @federicoorta.
ME=E112B600000000 where TC=11100 SubType=001 Emergency=000 A4.A2.A1=1.1.0= 6 B4.B2.B1=1.0.1= 5 C4.C2.C1=0.0.1= 1 D4.D2.D1=0.1.1= 3
Thus, the expected Mode3/A Code is '6513'.