我正在尝试使用下载的tqli算法计算矩阵的特征值。作为一个简单的例子，我试图测试它的4x4单位矩阵：
double ze[4][4];
double *zepointer;
ze[0][0] = 1;
ze[1][1] = 1;
ze[2][2] = 1;
ze[3][3] = 1;
ze[0][1] = 0;
ze[1][0] = 0;
ze[0][2] = 0;
ze[2][0] = 0;
ze[0][3] = 0;
ze[3][0] = 0;
ze[1][2] = 0;
ze[2][1] = 0;
ze[1][3] = 0;
ze[3][1] = 0;
ze[2][3] = 0;
ze[3][2] = 0;
zepointer = &ze;
tqli(de,ee,4,zepointer);
除此之外，我希望有一种更聪明的方法来生成一个单位矩阵，我的编译器会给我一个警告说“双＆＃39;间接水平不同于双重（）[4] [4]＆＃39;对于该行＆＃39; zepointer =＆amp; ze;＆＃39;。我查看this question并在那里说警告来自于在宣布之前的引用。但是，我无法看到我在代码片段中完成了这项工作。
TQLI是一种计算特征值和特征向量的算法。我是从here下载的。
答案 0 :(得分：0)
我正在尝试使用下载的方法计算矩阵的特征值 tqli算法
尝试投射double**无法获得double[4][4]指针。
您必须使用不同的方法为double **函数提供所需的tqli指针。
您可以使用以下方法之一传递数据。
第一种方法：
1）为指针double **aa的指针分配内存，并使用指向a数组中行向量的指针初始化它们。
第二种方法：
2）为指向矩阵行的指针分配内存（如上所述）。为行的内容分配内存。复制ze。
测试程序：
#include <stdlib.h>
#include <stdio.h>
#define M_SIZE 4 // size of the nxn matrix
void tqli(double d[], double e[], int n, double **z)
{
int i,j;
for(i=0; i<n; i++)
{
for(j=0; j<n; j++)
{
double d = z[i][j];
printf(" z[%i][%i] = %f \n", i, j, d );
}
printf("\n");
}
}
int main()
{
int i,j;
double ze[M_SIZE][M_SIZE];
double **zepointer;
ze[0][0] = 1;
ze[1][1] = 1;
ze[2][2] = 1;
ze[3][3] = 1;
ze[0][1] = 0;
ze[1][0] = 0;
ze[0][2] = 0;
ze[2][0] = 0;
ze[0][3] = 0;
ze[3][0] = 0;
ze[1][2] = 0;
ze[2][1] = 0;
ze[1][3] = 0;
ze[3][1] = 0;
ze[2][3] = 0;
ze[3][2] = 0;
double de[M_SIZE] = {0};
double ee[M_SIZE] = {0};
// 1.---------------------
printf(" First array is { {6,7,8,9}, {1,6,1,5}, {6,2,2,4}, {1,2,3,4} } \n");
double a[M_SIZE][M_SIZE]={ {6,7,8,9}, {1,6,1,5}, {6,2,2,4}, {1,2,3,4} };
double **aa = malloc( M_SIZE * sizeof(double*)); // allocate memory to hold the pointers to all rows
for(i=0; i<M_SIZE; i++)
aa[i] = a[i]; // initialize pointers to the `a` rows
tqli(de,ee,M_SIZE,aa); // test
free(aa); // free memory for row pointers
// 2.--------------------
printf(" Second array is { {1,0,0,0}, {0,1,0,0}, {0,0,1,0}, {0,0,0,1} } \n");
zepointer = malloc( M_SIZE * sizeof(double*)); // allocate memory to hold the pointers to all rows
// allocate memory for every row to hold column elements
for(i=0; i<M_SIZE; i++)
zepointer[i] = malloc (M_SIZE * sizeof (double )) ;
// assign values from your ze[][] matrix
for(i=0; i<M_SIZE; i++)
for(j=0; j<M_SIZE; j++)
zepointer[i][j] = ze[i][j];
tqli(de,ee,M_SIZE,zepointer); // test
// cleanup
for(i=0; i<M_SIZE; i++) // free memory allocated for row elements
free(zepointer[i]) ;
free(zepointer); // free memory for the row pointers
return 0;
}
输出：
First method: array is { {6,7,8,9}, {1,6,1,5}, {6,2,2,4}, {1,2,3,4} }
z[0][0] = 6.000000
z[0][1] = 7.000000
z[0][2] = 8.000000
z[0][3] = 9.000000
z[1][0] = 1.000000
z[1][1] = 6.000000
z[1][2] = 1.000000
z[1][3] = 5.000000
z[2][0] = 6.000000
z[2][1] = 2.000000
z[2][2] = 2.000000
z[2][3] = 4.000000
z[3][0] = 1.000000
z[3][1] = 2.000000
z[3][2] = 3.000000
z[3][3] = 4.000000
Second method: array is { {1,0,0,0}, {0,1,0,0}, {0,0,1,0}, {0,0,0,1} }
z[0][0] = 1.000000
z[0][1] = 0.000000
z[0][2] = 0.000000
z[0][3] = 0.000000
z[1][0] = 0.000000
z[1][1] = 1.000000
z[1][2] = 0.000000
z[1][3] = 0.000000
z[2][0] = 0.000000
z[2][1] = 0.000000
z[2][2] = 1.000000
z[2][3] = 0.000000
z[3][0] = 0.000000
z[3][1] = 0.000000
z[3][2] = 0.000000
z[3][3] = 1.000000
答案 1 :(得分：0)
从函数的定义
void tqli(double d[], double e[], int n, double **z)
我得出结论，函数期望double的2维被传递为“锯齿状”，即“散乱”数组。
要创建这样的东西，请执行以下操作：
const size_t rows = 4;
const size_t columns = 4;
/* Allocate as many pointers to column's 1st element as we have rows. */
double ** z = malloc(rows * sizeof *z);
/* Loop over all rows (pointers to column's 1st element). */
for (size_t r = 0; r < rows; ++r)
{
/* Allocate to each row (pointer to column's 1st element) as many doubles as we have columns. */
z[r] = malloc(columns * sizeof *z[r]);
}
/* Assign value to z's elements. */
for (size_t r = 0; r < rows; ++r)
{
for (size_t c = 0; c < columns; ++c)
{
z[r][c] = r * c;
}
}
tqli(..., z);
/* Free the stuff,
by calling free() on every item we received from malloc(). */
for (size_t i = 0; i < rows; ++i)
{
free(z[i]);
}
free(z);
添加正确的错误检查留给读者。