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58
2025.09.28_cpp_openblas_test/openblas_test.cpp
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2025.09.28_cpp_openblas_test/openblas_test.cpp
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#include <iostream>
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#include <vector>
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#include <cmath>
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extern "C" {
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void dgetrf_(int* m, int* n, double* a, int* lda, int* ipiv, int* info);
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void dgetri_(int* n, double* a, int* lda, int* ipiv, double* work, int* lwork, int* info);
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}
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int main() {
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const int N = 3;
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double A_data[] = {
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1, 0, 5, // 第1列
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2, 1, 6, // 第2列
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3, 4, 0 // 第3列
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};
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std::vector<double> A(A_data, A_data + 9);
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std::cout << "Original matrix (column-major):\n";
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for (int i = 0; i < N; ++i) {
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for (int j = 0; j < N; ++j) {
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std::cout << A[j*N + i] << " ";
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}
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std::cout << "\n";
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}
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// LU 分解
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std::vector<int> ipiv(N);
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int info;
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int n = N;
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int lda = N;
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dgetrf_(&n, &n, &A[0], &lda, &ipiv[0], &info);
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if (info != 0) {
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std::cerr << "LU decomposition failed! (info=" << info << ")\n";
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return 1;
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}
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// 计算逆矩阵
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std::vector<double> work(N);
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int lwork = N; // 关键:必须是可变变量
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dgetri_(&n, &A[0], &lda, &ipiv[0], &work[0], &lwork, &info);
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if (info != 0) {
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std::cerr << "Matrix inversion failed! (info=" << info << ")\n";
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return 1;
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}
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std::cout << "\nInverse matrix:\n";
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for (int i = 0; i < N; ++i) {
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for (int j = 0; j < N; ++j) {
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std::cout << A[j*N + i] << " ";
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}
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std::cout << "\n";
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}
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return 0;
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}
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19
2025.09.28_cpp_openblas_test/openblas_test_wtih_eigen.cpp
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2025.09.28_cpp_openblas_test/openblas_test_wtih_eigen.cpp
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#define EIGEN_USE_BLAS
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#include <iostream>
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#include "Eigen/Dense" // #include <Eigen/Dense>
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int main() {
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Eigen::MatrixXd A(4, 4);
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A << 2, 1, 1, 0,
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1, 3, 1, 1,
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1, 1, 4, 1,
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0, 1, 1, 5;
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Eigen::MatrixXd A_inv = A.inverse();
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std::cout << "A:\n" << A << std::endl;
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std::cout << "\nA_inv:\n" << A_inv << std::endl;
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std::cout << "\nA*A_inv:\n" << A * A_inv << std::endl;
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return 0;
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}
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25
2025.09.28_fortran_openblas_test/openblas_test.f90
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25
2025.09.28_fortran_openblas_test/openblas_test.f90
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program invert_simple
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implicit none
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integer, parameter :: n = 2
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double precision :: a(n,n)
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integer :: ipiv(n)
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double precision :: work(n) ! 最小工作空间(LWORK = N)
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integer :: info
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! 初始化一个简单的可逆矩阵: [[2, 1], [1, 2]]
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a(1,1) = 2.0d0; a(1,2) = 1.0d0
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a(2,1) = 1.0d0; a(2,2) = 2.0d0
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! LU 分解
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call DGETRF(n, n, a, n, ipiv, info)
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if (info /= 0) stop 'DGETRF failed'
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! 求逆(使用最小工作空间 LWORK = N)
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call DGETRI(n, a, n, ipiv, work, n, info)
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if (info /= 0) stop 'DGETRI failed'
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! 输出结果(应为 [[0.6667, -0.3333], [-0.3333, 0.6667]])
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print *, 'Inverse:'
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print *, a(1,1), a(1,2)
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print *, a(2,1), a(2,2)
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end program
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