请问这个异或0x8000和除以32768的目的分别是什么呢?
iu1=iMaxPU*(float32)((long)((short)((Unit16)((long)AdTmp->iu1AdSum/(long)AdTmp->nAdSum)^0x8000))-(long)iu1Zerof)/32768;
请问这个异或0x8000和除以32768的目的分别是什么呢?
iu1=iMaxPU*(float32)((long)((short)((Unit16)((long)AdTmp->iu1AdSum/(long)AdTmp->nAdSum)^0x8000))-(long)iu1Zerof)/32768;