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},
    
已邀请:
        
PI
不能精确地表示为
double
。因此ѭ6都不是。 因此,
sin(kal)
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()
cosl()
而不是
sin()
cos()
。 第三,像这样定义PI:
#define PI 3.1415926535897932384626433832795029L
将其他所有内容(尤其是
S
C
)保留为
double
。看看是否有帮助...     
        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
)。     
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。     
        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
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\");
    

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