我一直在研究物理研究项目。该程序是用C语言编写的，但使用了fortran函数（它叫做“zgesv”，它来自LAPACK和BLAS库）。
这个想法是解决一个方程组。 LHS.X =向量“X”的RHS。 int INFO 传递给zgesv。如果无法求解方程（即LHS是单数），则应该返回一个值！= 0;
尝试通过将我的double *传递给solve函数（解决方案1，在以下代码中）将我的程序作为“正常”运行， 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，这是一个数字灾难 - 一些小，一些零，一些巨大。
如果我手动创建 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，...}; 并按元素将它们复制到 LHS 和 RHS 元素中，然后将 INFO 返回为8（我觉得它与前一个案例不同，这很奇怪。），解决方案等于 RHS 。
我觉得这必须是声明数组的两种方式之间的一些根本区别。我真的无法使用功能zgesv获取muck以使其采用我想要的类型，因为A）它是科学界的标准，而B）它是用fortran编写的 - 我从未学过它。
有谁可以解释这里发生了什么？是否有一个简单的（最好是计算上便宜的）修复？我应该将我的double *复制到数组[]中吗？
这是我的程序（为测试而修改）：
#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;
}
我使用编译行
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},
答案 0 :(得分：3)
你的代码中有一些语法错误（不匹配的括号和引号），但我想在你复制代码的时候会出现这些错误。
我的猜测是以下情况可能导致问题：
solveSystemByLUD(int N, double *LHS, double* RHS, ...)
{
...
}
这声明了一个返回int但返回double *的函数。添加返回类型，看看是否有任何变化。
修改：
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.
*/
缺少一个开场评论。
此外，您将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);
从您的代码判断，这不是您想要的。
最后要注意的是，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数组的末尾，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：您正在函数getA（）中读取argv[0]，它始终是可执行文件的路径。你应该从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）。
答案 1 :(得分：1)
首先：我来自Fortran方面而且我没有C经验，因此我说的可能相关或不相关。
BLAS / LAPACK例程期望接收Fortran数组---本质上是连续内存块的第一个元素的地址。假设数组布局是列主要的，并且大小由LDA控制。基本上没有进行任何检查，所以如果您发送的内容不符合这些期望，那么一切都会破裂。
我从经验中发现的（或者更确切地说 - 看到有人这样做并复制它）从C ++调用LAPACK例程是以下工作：如果使用std::vector<double>来存储矩阵，使用&A[0]（其中A是std::vector<double>）调用LAPACK调用，它可以工作。我的（可能太天真）理解是标准库向量是“有点像”C数组，所以我想这可以直接翻译成C.
此外，您正在使用复杂的例程，这些例程可能会或可能不会以微妙的方式更改某些内容。我猜想Fortran的COMPLEX * 16相当于struct{double real_part; double imag_part;}
最后，可以从netlib获得LAPACK例程的参考实现，这里http://www.netlib.org/lapack/complex16/zgesv.f
答案 2 :(得分：0)
这是另一个猜测：
我想知道在generateLHS()和/或generateRHS()内是否生成double值，其中非常小的错误显示为零（或小数部分中至少为零）您使用printf()显示它们，但这些差异会影响zgesv_()中的计算。
您可以尝试在LHS2和RHS2声明之后将这些循环添加到您的测试运行中：
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");
答案 3 :(得分：0)
PI不能完全代表double。因此PI/2也不是。{/ p>
因此sin(kal)和cos(kal)不一定分别等于1或0.
（如果您尝试通过将它们分配给“int”来解决此问题，请记住，在进行此类转换时，C通常会将向下舍入。）
[编辑]
其实我有更好的建议......
使用GCC编译Fortran代码时，您可能希望使用-ffloat-store。 Fortran代码通常是在仔细注意双精度数的量化的情况下编写的......但是默认情况下，x86上的中间值可以具有额外的精度（因为浮点单元在内部使用80位）。通常，额外的精度只会有所帮助，但对于某些数字敏感的代码（例如sin（PI / 2）），它可能会导致意外的结果。
-ffloat-store避免了这种额外的精确度。
[编辑2，回应评论]
为了从sin和cos中获得更好的精确度，我建议如下。
首先，将kal声明为long double，同时在main和参数列表中声明为其他函数。
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。看看它是否有帮助......