Question

我正在编写一个程序，它通过一个函数（upperhess）发送一个数组，该函数将它从一个方形的nxn矩阵变成一个上层的hessenberg矩阵。这运行正常，但后来我需要通过另一个函数发送数组，我遇到了问题。我需要使double等于数组的第一个元素。但是，它将double（mu）设置为等于零。我在下面加了我的代码

#include <stdio.h>
#include <math.h>


int idx(int r, int c, int n)
{
    return r * n + c;
}

void transpose(double *a, int n, double *at)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            at[(idx (i, j, n))] = a[(idx (j, i, n))];
        }
    }
}

void matrix_multiplication(double *a, double *b, int n, double *combo)
{

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            combo[(idx(i, j, n))] = 0;
            for (int k = 0; k < n; k++) {
                combo[(idx(i, j, n))] +=  a[(idx(i, k, n))] * b[(idx(k, j, n))];
            }
        }
    }

}

void identity(double *a, int n)
{
    for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i == j) {
                    a[(idx(i, j, n))] = 1;
                }
                else {
                    a[(idx(i, j, n))] = 0;
                }
            }
        }
}

void upperhes(double *a, int n, double *u, double *b)
{

    //Sets b equal to a
    for (int i = 0; i < (n*n); i++) {
        b[i] = a[i];
    }

    //Sets u equal to the identity matrix
    identity(u, n);

    int times = 0;
    for (int i = 0; i < (n-2); i++) {
        for (int j = n-1; j > (1 + times); j--) {
            double c, s, r;

            r = sqrt((b[(idx(j-1,i,n))] * b[(idx(j-1,i,n))]) + (b[(idx(j,i,n))] * b[(idx(j,i,n))]));
            if (r < 0) {
                r = (-1 * r);
            }
            if (r < pow(10,-50)) {
                c = 1;
                s = 0;
            }
            else {
                c = b[(idx(j-1, i, n))] / r;
                s = (-1 * b[(idx(j, i, n))]) / r;
            }

            //store takes the original values of specific points in matrix b and stores them, so the values of b can be manipulated
            double store[6];

            store[0] = b[(idx(j-1, i,n))];
            store[1] = b[(idx(j,i,n))];
            store[4] = b[(idx(i,j-1,n))];
            store[5] = b[(idx(i,j,n))];


            b[(idx (j,i,n))] = (s * store[0]) + (c * store[1]);
            b[(idx(j-1,i,n))] = (c * store[0]) + (-s * store[1]);

            for (int k= i+1; k<n; k++) {

                store[2] = b[(idx(j-1,k,n))];
                store[3] = b[(idx(j,k,n))];

                b[(idx(j-1,k,n))] = (c * store[2]) - (s * store[3]);
                b[(idx(j,k,n))] = (s * store[2]) + (c * store[3]);
            }


            b[(idx (i, j, n))] = (s * store[4]) + (c * store[5]);
            b[(idx (i, j-1, n))] = (c * store[4]) - (s * store[5]);

            for (int k = i + 1; k < n; k++) {

                store[2] = b[(idx(k, j-1, n))];
                store[3] = b[(idx(k, j, n))];

                b[idx(k, j-1, n)] = (c * store[2]) - (s * store[3]);
                b[idx(k, j, n)] = (s * store[2]) + (c * store[3]);
            }


            store[0] = u[(idx(j-1, 0, n))];
            store[1] = u[(idx(j, 0, n))];

            u[idx (j, 0, n)] = (s * store[0]) + (c * store[1]);
            u[idx(j-1, 0, n)] = (c * store[0]) + (-s * store[1]);

            for (int k= 1; k<n; k++) {

                store[2] = u[idx(j-1, k, n)];
                store[3] = u[idx(j, k, n)];

                u[idx(j-1, k, n)] = (c * store[2]) - (s * store[3]);
                u[idx(j, k, n)] = (s * store[2]) + (c * store[3]);
            }
        }

        times++;

    }

    //Sets ut equal to the transpose of u
    double ut[n*n];
    transpose(u, n, ut);

    //Multiplies u and a together.
    double ua[(n*n)];
    matrix_multiplication(u, a, n, ua);

    double b_check[(n*n)];
    matrix_multiplication(ua, ut, n, b_check);

    //Prints out the matrix!
    for(int i=0; i<n; i++){
        for(int j =0; j<n; j++){
            printf("(%+.3f)\t", b[(idx (i,j,n))]);
        }
        printf("\n\n");
    }
    printf("\n");

}

void two_through_five(double *b, int n, double *eigenvalues, int total)
{

    for (int times = 0; times < 100; times++){

        double mu = b[0];

        ..........
    }
}

void qr_symmetric(double *a, int n, double *b)
{

    //Creates a tridiagonal matrix by using a method that creates upper Hessenberg matrices on a symmetrical matrix a.
    double u[n * n];
    upperhes(a, n, u, b);

    //Creates an array to store the eigenvalues
    double eigenvalues[n];

    two_through_five(b, n, eigenvalues, n);
    for (int i = 0; i < n; i++) {
        printf("%d\t", eigenvalues[n]);
    }
    printf("\n");

}

int main(void)
{
    int n;
    n = 4;
    double a[(n*n)];
    a[0] = 1;
    a[1] = 2;
    a[2] = 3;
    a[3] = 4;
    a[4] = 2;
    a[5] = 6;
    a[6] = 7;
    a[7] = 8;
    a[8] = 3;
    a[9] = 7;
    a[10] = 11;
    a[11] = 15;
    a[12] = 4;
    a[13] = 8;
    a[14] = 15;
    a[15] = 16;
    double b[n * n];
    qr_symmetric(a, n, b);
    return 0; 
}

Answer 1

你的代码应该工作。例如，这对我有用：

void two_through_five(double *b, int n, double *eigenvalues, int total)
{
    int times;
    for (times = 0; times < 100; times++)
    {
        double mu = b[0];
        printf("%f\n", mu);
    }
}

int main()
{

   int n = 18;
   int i =0;

   double * b = malloc(n*sizeof(double));

   for( i=0; i<n;i++) b[i] = 15.0;

   double eigenvalues[n];

   two_through_five(b, n, eigenvalues, n);

   return 0;
}

Outupts：

所以我的判断是，你确定b [0]不是0吗？

获取数组索引的问题[C]

1 个答案: