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请问这个异或0x8000和除以32768的目的分别是什么呢？

kun chen7
Prodigy 82 points


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;