c++ - 将JNI代码移植到java并理解jdouble * 用法

Question

我无法理解 jdouble * 的用法。我看到它用作数组和 double 。你能向我解释一下 InitiateBuffer 的作用以及 State->X = State->Buffer + OFFSET_X; 的结果是什么吗？

#undef OFFSET_X
#define OFFSET_X 0L
#undef OFFSET_Y
#define OFFSET_Y 500L
#undef OFFSET_Z
#define OFFSET_Z 1000L

typedef struct {    
  jdoubleArray BufferArray;
  jdouble *Buffer;    
  jdouble *X;
  jdouble *Y;
  jdouble *Z;
  jint Position;
} StateStructure;

void InitiateBuffer(StateStructure *State) {
  JNIEnv *JNI = State->JNI;
  State->BufferArray = (*JNI)->NewGlobalRef(JNI,                                                      (*JNI)->NewDoubleArray(JNI, SIZE_BUFFER));
  State->Buffer = (*JNI)->GetDoubleArrayElements(JNI, State->BufferArray,  NULL);
  Fill(State->Buffer, 0, SIZE_BUFFER, FP_NAN);
  (*JNI)->ReleaseDoubleArrayElements(JNI, State->BufferArray,  State->Buffer, JNI_COMMIT);
  State->TimeoutImpact = -1;
  State->TimeoutFalling = -1;
  State->Position = 0;
}

void InitiateSamples(StateStructure *State) {
  State->X = State->Buffer + OFFSET_X;
  State->Y = State->Buffer + OFFSET_Y;
  State->Z = State->Buffer + OFFSET_Z;
}

jdouble LPF(jdouble Value, jdouble *XV, jdouble *YV) {
  XV[0] = XV[1];
  XV[1] = XV[2];
  XV[2] = Value / FILTER_LPF_GAIN;
  YV[0] = YV[1];
  YV[1] = YV[2];
  YV[2] = (XV[0] + XV[2]) + 2 * XV[1] + (FILTER_FACTOR_0 * YV[0]) +    (FILTER_FACTOR_1 * YV[1]);
  return YV[2];
}

void Process() {
  jint At = State->Position;
  State->TimeoutFalling = EXPIRE(State->TimeoutFalling);
  State->TimeoutImpact = EXPIRE(State->TimeoutImpact);
  State->X_LPF[At] = LPF(State->X[At], State->XLPFXV, State->XLPFYV);
}

Answer 1

您发布的整体架构尝试将一个数组分为三个(或更多？)部分并单独访问它们。

您没有向我们展示所有代码(例如 XLPFXV 或 X_LPF )，但从您粘贴的内容我可以推断出以下关于 InitiateBuffer 的信息:

• State->BufferArray是一个新的Java double大小数组 SIZE_BUFFER
• State->Buffer是此数组的 C++ 友好别名
• 之后InitiateBuffer ，Java 数组填充了 NaN 值。

InitiateSamples在State->Buffer内设置三个指针，X指针从Buffer[0]开始，Y 指针指向 Buffer[500] ，Z 指针位于 Buffer[1000] 。然后可以使用它们进一步索引数组(例如 State->Y[42] 引用 State->Buffer[500+42] )。

希望有帮助。