Question

我尝试为Android编写MINPACK-Fortran端口，但我无法理解在JNI中正确使用双数组。在论坛上研究这个问题，我意识到在使用数组时JNI需要大量无用的代码。我想错了，所以把代码带到这里。

在此示例中，调用过程 hybrd1 _ （Fortran sub，由MINPACK提供）来计算非线性方程组。用户（Java端）提供类aSolver，其中包含方法 solveStep 。 solveStep 有两个参数（double的数组），其中x - 参数的矢量; f - NLES中右侧的矢量。问题是，我不知道 hybrd1 _ 程序将哪些向量传输到 solveStep ，而迭代会继续。但我知道，这些向量可以在x，fvec或wa（aSolver对象中的字段）中找到。 hybrd1 _ 调用 rhs 过程（cdecl） - 过程中我需要做很多可怕的事情才能将两个数组转移到Java，然后将它们带回c ++。正如您可以看到 rhs 过程非常频繁地调用，这些JNI转换正在破坏整个过程。

C ++方

//struct which allow me to talk with Java from rhs procedure
struct fake_n
{
    int n;
    jobject aSolver;
    jmethodID meth;
    JNIEnv* env;
};

//iterator. the place where i need help
void rhs(int * n, double * x, double * f, int * flag)
{
    fake_n* f_n = (fake_n *) n;
    // creating array parametres 
    jdoubleArray x_row = ((*f_n).env)->NewDoubleArray((*f_n).n);
    jdoubleArray f_row = ((*f_n).env)->NewDoubleArray((*f_n).n);
    // filling array parametres 
    ((*f_n).env)->SetDoubleArrayRegion(x_row, 0, (*f_n).n, x);
    // launch object method
    (*f_n).env->CallVoidMethod((*f_n).aSolver, (*f_n).meth, x_row, f_row);
    // copy result back to c++ - why do i need? 
    jdouble *res = ((*f_n).env)->GetDoubleArrayElements(f_row, NULL);
    for (int i =0; i< (*f_n).n; i++) { f[i] = res[i]; }        
}

JNIEXPORT void JNICALL Java_com_example_myapplication_MainActivity_hybrd_1NLES(JNIEnv* env,
                                                            jclass clazz,
                                                            jobject aSolver)
{
    jclass cls = env->GetObjectClass(aSolver);
    jmethodID solvestep = env->GetMethodID(cls, "solveStep", "([D[D)V");
    if (solvestep == 0) {
        return;
    }

    jfieldID afields[4];
    char* atype[4] = {(char *) "[I", (char *)"[D", (char *)"[D", (char *)"[D"};
    char* aname[4] = {(char *)"n", (char *)"x", (char *)"fvec", (char *)"wa"};
    jobject objs[4];

    for (int i = 0; i < 4; i++) {
        afields[i] = env->GetFieldID(cls, aname[i], atype[i]);
        objs[i] = env->GetObjectField(aSolver, afields[i]);
    }
    jintArray *na = reinterpret_cast<jintArray *>(&(objs[0]));
    jdoubleArray *xa = reinterpret_cast<jdoubleArray *>(&(objs[1]));
    jdoubleArray *fveca = reinterpret_cast<jdoubleArray *>(&(objs[2]));
    jdoubleArray *waa = reinterpret_cast<jdoubleArray *>(&(objs[3]));

    int *n = env->GetIntArrayElements(*na, NULL);
    double *x = env->GetDoubleArrayElements(*xa, NULL);
    double *fvec = env->GetDoubleArrayElements(*fveca, NULL);
    double *wa = env->GetDoubleArrayElements(*waa, NULL);

    fake_n f_n;
    f_n.n = *n;
    f_n.aSolver = aSolver;
    f_n.meth = solvestep;
    f_n.env = env;

    // tol, lwa, info - local variables
    // this procedure calls the iterator (rhs) a lot of times   
    hybrd1_(rhs, (int *)&f_n, x, fvec, &tol, &info, wa, &lwa);

    // should i ?
    env->ReleaseIntArrayElements(*na, n, 0);
    env->ReleaseDoubleArrayElements(*xa, x, 0);
    env->ReleaseDoubleArrayElements(*fveca, fvec, 0);
    env->ReleaseDoubleArrayElements(*waa, wa, 0);
}

这是JAVA方面

public class aSolver {
    public double[] x;
    public double[] fvec;
    public double[] wa;
    public int[] n;
    public int rc;

    public aSolver() {
        n = new int[1];
        n[0] = 2;
        x = new double[n[0]];
        fvec = new double[n[0]];
        wa = new double[n[0] * (3 * n[0] + 13) / 2];
        x[0] = 0;
        x[1] = 1;
        rc = 0;
        solveStep(x, fvec);
    }

    // method, which calling from iterator (rhs)
    public void solveStep(double[] xv, double[] fv ) {
        // it is important: fv != fvec
        // it is important: xv != x
        fv[0] = 2.0 * xv[0] + 3.0 * xv[1] + 6.0;
        fv[1] = 5.0 * xv[0] - 3.0 * xv[1] - 27.0;
        rc++;
    }
}

//----
aSolver s = new aSolver();
hybrd_NLES(s);
//----

public native void hybrd_NLES(aSolver solver);

书面代码似乎很麻烦且效率低下。此外，感觉 ReleaseDoubleArrayElements 中的 rhs 子需要，但我不明白为什么需要它。请帮我写一个高效的包装。

Answer 1

首先，根据定义，JNI代码确实很麻烦，因为您必须使用'env'来创建某些变量，并且您必须进行一些编码以获取Java代码所需的数据并与之通信它。我真的不能说如何使你的代码更有效率，因为我真的不明白你想做什么。你可能有一些不必要的复制，我不确定。如果您查看代码并仍然认为可以改进它，请进一步解释一下您的目标是什么。关于'ReleaseDoubleArrayElements'的问题 - NewDoubleArray仅创建本地引用，因此您不需要发布它。

Answer 2

我想我找到了一个解决方案，如果有人遇到类似问题，我会把它放在这里。我将所有工作数组放在一个并从Java传递给C ++。然后在 rhs 内部，我计算向量x和f的内存偏移量，并将它们带入 solveStep （现在我只发送给Java偏移量）。这是更改的代码：

C ++方

Java方

struct fake_n
{
    int n;

    jobject aSolver;
    jmethodID meth;
    JNIEnv* env;
    double* dbls;
};

void rhs(int * n, double * x, double * f, int * flag)
{
    fake_n* f_n = (fake_n *) n;
    // calculate the offsets to x and f vectors
    jint x_row = (jint)(x - (*f_n).dbls);
    jint f_row = (jint)(f - (*f_n).dbls);
    // and call the method of aSolver object
    (*f_n).env->CallVoidMethod((*f_n).aSolver, (*f_n).meth, x_row, f_row);
}
JNIEXPORT jint JNICALL Java_com_example_myapplication_MainActivity_hybrd_1NLES(JNIEnv* env,
                                                            jclass clazz,
                                                            jobject aSolver)
{
    jclass cls = env->GetObjectClass(aSolver);
    jmethodID solvestep = env->GetMethodID(cls, "solveStep", "(II)V");
    if (solvestep == 0) {
        return -1;
    }
    int info = 0;

    //JNIlist - helper class to connect to JNI and to hold the variables data
    //nothing unusual - jast a couple of simple wrappers
    //original code is shown above
    JNIlist * lJNI = new JNIlist(env, aSolver, cls);
    lJNI->add(JNI_DT_INT,          (char *)"n");
    lJNI->add(JNI_DT_DOUBLE_ARRAY, (char *)"dbls");
    lJNI->connectJNI();

    int n = *((int*)lJNI->data(0));
    int lwa = n * (3 * n + 13) / 2 + 1;
    double tol = rel_TOL;

    fake_n f_n;
    f_n.n = n;
    f_n.aSolver = aSolver;
    f_n.meth = solvestep;
    f_n.env = env;
    f_n.dbls = (double*)(lJNI->data(1)); //pointer to the REAL memory section
    // with the array of doubles "dbl" 
    //cause GetDoubleArrayElements with the iCopy parameter which equals JNI_FALSE
    // gives access to the real memory section

    hybrd1_(rhs, (int *)&f_n, f_n.dbls, f_n.dbls + n, &tol, &info, f_n.dbls + 2*n, &lwa);

    delete lJNI;

    return info;
}

使用双数组的最佳做法

2 个答案: