C-双精度指针与双精度数组之间的细微差别。将一个转换为另一个？

| 我一直在为物理研究项目开发程序。该程序是用C语言编写的，但是使用了fortran函数（它称为\“ zgesv \”，它来自LAPACK和BLAS库）。 这个想法是要解决一个方程组。 LHS.X =向量“ X”的RHS。 int INFO传递给zgesv。如果这些方程无法求解（即LHS是奇异的），则info应该返回值= 0；否则返回0。 通过将double *传递给Solve函数（以下代码中的解决方案1），尝试以“正常”的方式运行程序，即使LHS是单数，INFO也返回0。不仅如此，而且如果我打印出解决方案，那将是数字的灾难-有些小，有些零，有些大。 如果我通过手动创建LHS和RHS     LHS [] = {值1，值2，...};     RHS [] = {值1，值2，...}; 然后将INFO作为预期值返回3，并且解决方案等于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，...}; 并将它们逐个复制到LHS和RHS中，然后将INFO返回为8（对我来说，这与以前的情况不同是很奇怪的。），解决方案等于RHS。 我觉得这肯定是声明数组的两种方式之间的根本区别。我真的无法使用zgesv函数来使muck成为我想要的类型，因为A）它是科学界的标准，并且B）它是用fortran编写的-我还没有从来没有学过。 谁能解释这是怎么回事？是否有一个简单的（最好是计算便宜的）修复程序？我是否应该将我的double *复制到array []中？ 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;
  }
我用编译行 
gcc -lm -llapack -lblas PiecesOfCprogarm.c -Wall -g
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},
    
2020-02-14 5 条评论
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没有找到相关结果

    已邀请:

    师埠女

            
    PI
    不能精确地表示为
    double
    。因此ѭ6都不是。 因此，
    sin(kal)
    和
    cos(kal)
    不必分别等于1或0。 （如果要通过将它们分配给\“ int \”来解决此问题，请记住，进行这种转换时C通常会四舍五入。） [编辑] 其实我有一个更好的建议... 使用GCC编译Fortran代码时，您可能要使用
    -ffloat-store
    。 Fortran代码通常在编写时非常注意双精度数字的量化...但是默认情况下，x86上的中间值可以具有更高的精度（因为浮点单元内部使用80位）。通常，额外的精度只会有所帮助，但是对于某些数字敏感的代码（例如sin（PI / 2）），它可能会导致意外结果。 
    -ffloat-store
    避免了这种额外的精度。 [编辑2，回应评论] 为了获得更好的正弦和余弦精度，我建议以下几点。 首先，在
    main
    和其他函数的参数列表中，将
    kal
    声明为
    long double
    。 其次，调用
    sinl()
    和
    cosl()
    而不是
    sin()
    和
    cos()
    。 第三，像这样定义PI： 
    #define PI 3.1415926535897932384626433832795029L
    将其他所有内容（尤其是
    S
    和
    C
    ）保留为
    double
    。看看是否有帮助...     
    2020-02-14 0 0
    分享

    砷竣阿

            您的代码中有一些语法错误（括号和引号不匹配），但是我想当您在此处复制代码时，这些错误会漏掉。 我的猜测是以下可能会导致问题： 
    solveSystemByLUD(int N, double *LHS, double* RHS, ...)
{
...
}
    这声明了一个返回
    int
    的函数，但您正在返回
    double *
    。添加返回类型，看看是否有任何变化。 编辑： 在澄清之后，仍然存在一些缺陷-由于这一行，代码仍无法编译： 
       * 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.
   */
    缺少开头评论。 另外，您还要将sin（）和cos（）的结果分配给整数变量（默认情况下C和S为整数）： 
      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)
    为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数组，这将导致不确定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]
    ，该函数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，应检查用户是否提供了足够的参数（
    argc - 1 >= N * N
    ）。     
    2020-02-14 0 0
    分享

    稍惮

    首先，我来自Fortran方面，并且没有C经验，因此我所说的内容可能有意义，也可能无关紧要。 BLAS / LAPACK例程希望接收Fortran数组-本质上是连续内存块的第一个元素的地址。假定阵列布局以列为主，并且大小由LDA控制。基本上不会进行任何检查，因此，如果您发送的内容不符合这些期望，那么一切都会变得一团糟。 我从经验中发现（或更确切地说，看到某人这样做并复制了它），从C ++调用LAPACK例程的工作原理如下：如果您使用
    std::vector<double>
    存储矩阵，并使用
    &A[0]
    调用LAPACK调用（其中A是
    std::vector<double>
    ），它可以工作。我的理解（也许太天真了）是标准库向量是“有点像” C数组，因此，我想可以将其直接转换为C。 另外，您使用的是复杂的例程，这些例程可能以微妙的方式更改，也可能不会更改。我猜想Fortran的COMPLEX * 16等于
    struct{double real_part; double imag_part;}
     最后，可以从netlib处获得LAPACK例程的参考实现，请参见http://www.netlib.org/lapack/complex16/zgesv.f。     
    2020-02-14 0 0
    分享

    翱抹村

            这是另一个猜测： 我想知道是否在
    generateLHS()
    和/或
    generateRHS()
    内生成的
    double
    值具有很小的误差，当用
    printf()
    显示它们时显示为零（或小数部分至少为零），但是差异会影响
    zgesv_()
    的计算。 您可以在
    LHS2
    和
    RHS2
    的声明之后尝试将这些循环添加到测试运行中吗： 
    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);
    }
    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);
    }
    i = i + 1;
}
printf(\"\\n\");
        
    2020-02-14 0 0
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