C - 指向 double 和 double 组的指针之间的细微差别。将一个转换为另一个？

Question

我一直在为物理学研究项目编写程序。该程序是用 C 语言编写的，但使用了一个 fortran 函数(称为“zgesv”，来自 LAPACK 和 BLAS 库)。

这个想法是求解一个方程组。 LHS.X = vector “X”的 RHS。 int INFO 被传递给 zgesv。如果方程无法求解(即 LHS 是单数)，info 应该返回一个值 != 0；

尝试通过将我的双 * 传递给求解函数(解决方案 1，在以下代码中)以“正常”方式运行我的程序，INFO 返回为 0 - 即使 LHS 是单数。不仅如此，如果我打印出解决方案，那将是数字的灾难——有些小，有些为零，有些很大。

如果我通过以下方式手动创建 LHS 和 RHS LHS[] = {值 1, 值 2, ...}; RHS[] = {值 1, 值 2, ...};然后 INFO 按预期返回 3，解等于 RHS(这也是我所期望的。)

如果我声明数组 LHS2[] = {值 1, 值 2, ...}; RHS2[] = {值 1, 值 2, ...};并将它们逐个元素复制到 LHS 和 RHS 中，然后 INFO 返回为 8(这对我来说很奇怪，它与之前的情况不同.)，解等于RHS。

我觉得这一定是两种声明数组的方式之间的根本区别。我真的无法访问函数 zgesv 以使其采用我想要的类型，因为 A) 它是科学界的标准，并且 B) 它是用 fortran 编写的——我从未学过。

谁能解释一下这是怎么回事？是否有一个简单的(最好是计算成本低的)修复方法？我应该只将我的双 * 复制到数组 [] 中吗？

这是我的程序(为测试而修改):

包括

#include <stdlib.h>
#include <math.h>
#define PI 3.1415926535897932384626433832795029L

#define ERROR_VALUE 911.911


int* getA(int N, char* argv[])
{
  int i;
  int* AMatrix;
  AMatrix = malloc(N * N * sizeof(int));

  if (AMatrix == NULL)
  {
    printf("Failed to allocate memory for AMatrix. Exiting.");
    exit (EXIT_FAILURE);
  }

  for (i = 0; i < N * N; i++)
  {
    AMatrix[i] = atoi(argv[i + 1]);
  }

  return AMatrix;

}

double* generateLHS(int N, int* AMatrix, int TAPs[], long double kal)
{

  double S, C;
  S = sinl(kal);
  C = cosl(kal);
  printf("According to C, Sin(Pi/2) = %.25lf and Cos(Pi/2) = %.25lf", S, C);
  //       S = 1;
  //       C = 0;
  double* LHS;
  LHS = malloc(N * N * 2 * sizeof(double));

  if (LHS == NULL)
  {
    printf("Failed to allocate memory for LHS. Exiting.");
    exit (EXIT_FAILURE);
  }

  int i;
  for (i = 0; i < N * N; i++)
  {
    LHS[2 * i] = -1 * AMatrix[i];  
    LHS[(2 * i) + 1] = 0;
  }


  for (i = 0; i <= 2 * N * N - 2; i = i + (2 * N) + 2)
  {
    LHS[i] = LHS[i] + (2 * C); 
  }

  int j;
  for (i = 0; i <= 3; i++)
  {
    j = 2 * N * TAPs[i] + 2 * TAPs[i];
    LHS[j] = LHS[j] - C;
    LHS[j + 1] = LHS[j + 1] - S;
  }

  return LHS;
}

double* generateRHS(int N, int inputTailAttachmentPoint, long double kal)
{

  double* RHS;
  RHS = malloc(2 * N * sizeof(double));

  int i;
  for (i = 0; i < 2 * N; i++)
  {
    RHS[i] = 0.0;
  }
  RHS[2 * inputTailAttachmentPoint + 1] = - 2 * sin(kal);
  return RHS; 
}


double* solveUsingLUD(int N, double* LHS, double* RHS)
{
  int INFO; /*Info is changed by ZGELSD to 0 if the computation was carried out successfully. Else it changes to some none-zero integer. */
  int ione = 1;
  int LDA = N;
  int LDB = N;
  int n = N;
  int* IPV = malloc(N * sizeof(int));

  if (IPV == NULL)
  {
    printf("Failed to allocate memory for IPV. Exiting.");
    exit (EXIT_FAILURE);
  }

  zgesv_(&n, &ione, LHS, &LDA, IPV, RHS, &LDB, &INFO);
  free(IPV);

  if (INFO != 0)
  {

    printf("\n ERROR: info = %d\n", INFO); 
  }
  return RHS;
}


void printComplexVectors(int numberOfRows, double* matrix)
{
  int i;

  for (i = 0; i < 2 * numberOfRows - 1; i = i + 2)
  {
    printf("%f + %f*i    \n", matrix[i], matrix[i + 1]);
  }
  printf("\n");
}


int main(int argc, char* argv[])
{
  int N = 8;
  int* AMatrix;
  AMatrix = getA(N, argv);
  int TAPs[]={4,4,4,3};
  long double kal = PI/2;




  double *LHS, *RHS;
  LHS = generateLHS(N, AMatrix, TAPs,kal);
  int i;
  RHS = generateRHS(N, TAPs[0],kal);





  printf("\n LHS = \n{{");
  for (i = 0; i < 2 * N * N - 1;)
  {
    printf("%lf + ", LHS[i]);
    i = i + 1;
    printf("%lfI", LHS[i]);
    i = i + 1;

    if ((int)(.5 * i) % N == 0)
    {
      printf("},\n{"); 
    }
    else
    {
      printf(",");
    }
  }
  printf("}");


  double LHS2[] = {0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,-1.000000,0.000000,0.000000,0.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,-0.000000,-3.000000,0.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,-1.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.00000};
  double RHS2[] ={0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000,-2.000000,0.000000,0.000000,0.000000,0.000000,0.000000,0.000000};

  printf("comparing LHS and LHS2\n");
  for (i = 0; i < 2 * N * N;)
  {
    if (LHS[i] != LHS2[i]) {
      printf( "LHS difference at index %d\n", i);
      printf("LHS[%d] = %.16lf\n", i, LHS[i]);
  printf("LHS2[%d] = %.16lf\n", i, LHS2[i]);
  printf("The difference is %.16lf\n", LHS[i] - LHS2[i]);
    }
    i = i + 1;
  }

  printf("\n");
  printf("comparing RHS and RHS2\n");
  for (i = 0; i < 2 * N;)
  {
    if (RHS[i] != RHS2[i]) {
      printf( "RHS difference at index %d\n", i);
      printf("RHS[%d] = %.16lf\n", i, RHS[i]);
  printf("RHS2[%d] = %.16lf\n", i, RHS2[i]);
  printf("The difference is %.16lf", RHS[i] - RHS2[i]);

    }
    i = i + 1;
  }

  printf("\n");


  double *solution;

  solution = solveUsingLUD(N,LHS,RHS);
  printf("\n Solution = \n{");
  for (i = 0; i < 2 * N - 1;)
  {
    printf("{%.16lf + ", solution[i]);
    i = i + 1;
    printf("%.16lfI},", solution[i]);
    i = i + 1;
    printf("\n");
    }
    solution = solveUsingLUD(N,LHS2,RHS2);


    printf("Solution2 = \n{");
    for (i = 0; i < 2 * N - 1;)
    {
      printf("{%lf + ", solution[i]);
      i = i + 1;
      printf("%lfI},", solution[i]);
      i = i + 1;
      printf("\n");
    }


    for (i = 0; i < 2 * N * N;)
    {
      LHS[i] = LHS2[i];
      i = i + 1;
    }


    for (i = 0; i < 2 * N;)
    {
      RHS[i] = RHS2[i];
      i = i + 1;
    }
    solution = solveUsingLUD(N,LHS,RHS);

    printf("Solution3 = \n{");
    for (i = 0; i < 2 * N - 1;)
    {
      printf("{%lf + ", solution[i]);
      i = i + 1;
      printf("%lfI},", solution[i]);
      i = i + 1;
      printf("\n");
    }
    return 0;
  }

我用的是编译行

gcc -lm -llapack -lblas PiecesOfCprogarm.c -Wall -g

然后我执行:

./a.out 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0

给出输出

According to C, Sin(Pi/2) = 1.0000000000000000000000000 and Cos(Pi/2) = -0.0000000000000000000271051
 LHS = 
{{-0.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I},
{0.000000 + 0.000000I,-0.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I},
{0.000000 + 0.000000I,0.000000 + 0.000000I,-0.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I},
{0.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,-0.000000 + -1.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I},
{-1.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + -3.000000I,0.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I},
{-1.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,-0.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I},
{0.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I,0.000000 + 0.000000I,-0.000000 + 0.000000I,0.000000 + 0.000000I},
{0.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I,-1.000000 + 0.000000I,0.000000 + 0.000000I,0.000000 + 0.000000I,-0.000000 + 0.000000I},
{}comparing LHS and LHS2
LHS difference at index 0
LHS[0] = -0.0000000000000000
LHS2[0] = 0.0000000000000000
The difference is -0.0000000000000000
LHS difference at index 18
LHS[18] = -0.0000000000000000
LHS2[18] = 0.0000000000000000
The difference is -0.0000000000000000
LHS difference at index 36
LHS[36] = -0.0000000000000000
LHS2[36] = 0.0000000000000000
The difference is -0.0000000000000000
LHS difference at index 54
LHS[54] = -0.0000000000000000
LHS2[54] = 0.0000000000000000
The difference is -0.0000000000000000
LHS difference at index 72
LHS[72] = 0.0000000000000000
LHS2[72] = -0.0000000000000000
The difference is 0.0000000000000000
LHS difference at index 90
LHS[90] = -0.0000000000000000
LHS2[90] = 0.0000000000000000
The difference is -0.0000000000000000
LHS difference at index 108
LHS[108] = -0.0000000000000000
LHS2[108] = 0.0000000000000000
The difference is -0.0000000000000000
LHS difference at index 126
LHS[126] = -0.0000000000000000
LHS2[126] = 0.0000000000000000
The difference is -0.0000000000000000

comparing RHS and RHS2


 Solution = 
{{1.0000000000000000 + -0.0000000000000000I},
{-1.0000000000000000 + -0.0000000000000000I},
{-0.0000000000000000 + 0.0000000000000000I},
{-0.0000000000000000 + 0.0000000000000000I},
{0.6000000000000000 + 0.2000000000000000I},
{-0.0000000000000000 + -0.0000000000000000I},
{-6854258945071195.0000000000000000 + 4042255275298396.0000000000000000I},
{6854258945071195.0000000000000000 + -4042255275298396.0000000000000000I},

 ERROR: info = 3
Solution2 = 
{{0.000000 + 0.000000I},
{0.000000 + 0.000000I},
{0.000000 + 0.000000I},
{0.000000 + 0.000000I},
{0.000000 + -2.000000I},
{0.000000 + 0.000000I},
{0.000000 + 0.000000I},
{0.000000 + 0.000000I},

 ERROR: info = 8
Solution3 = 
{{0.000000 + 0.000000I},
{0.000000 + 0.000000I},
{0.000000 + 0.000000I},
{0.000000 + 0.000000I},
{0.000000 + -2.000000I},
{0.000000 + 0.000000I},
{0.000000 + 0.000000I},
{0.000000 + 0.000000I},

Answer 1

您的代码中存在一些语法错误(不匹配的括号和引号)，但我想这些是您在此处复制代码时出现的。

我的猜测是以下可能导致问题:

solveSystemByLUD(int N, double *LHS, double* RHS, ...)
{
...
}

这声明了一个返回 int 的函数，但您返回的是 double *。添加返回类型并查看是否有任何变化。

编辑:
在您澄清之后仍然存在一些缺陷 - 由于这一行代码仍然无法编译:

   * SUBROUTINE ZGESV( N, Nrhs, A, LDA, IPIV, B, LDB, INFO ) Solves A * X = B and stores the answer in B.   If the solver worked, INFO is set to 0.
   */

缺少开场白。

您还将 sin() 和 cos() 的结果分配给整数变量(默认情况下 C 和 S 是整数):

  S = sin(kal);
  C = cos(kal);

从您的代码来看，这不是您想要的。

最后要注意的是，在 LHS2 中缺少最后一个元素(sizeof(LHS2)/sizeof(double) 是 127 而不是 2 * 8 * 8 = 128).这意味着您正在读取和写入超出数组末尾的位置，这会导致未定义的行为并可能导致您看到的问题。

编辑 2:
还有一件事:您正在读取函数 getA() 中的 argv[0]，它始终是可执行文件的路径。您应该从 argv[1] 开始阅读。为了安全起见，您应该检查用户是否提供了足够的参数 (argc - 1 >= N * N)。