以下代码片段死于 Eigen 断言:
MatrixXd L;
VectorXd x, b;
...
ASSERT_MATRIX_EQ(L*x, b);
与,
template <typename DerivedL, typename DerivedR>
void ASSERT_MATRIX_EQ(const Eigen::DenseBase<DerivedL>& A, const Eigen::DenseBase<DerivedR>& B, double tol=1e-7) {
ASSERT_EQ(A.rows(), B.rows());
ASSERT_EQ(A.cols(), B.cols());
for(int i=0; i < A.rows(); i++) {
for(int j=0; j < A.cols(); j++) {
ASSERT_NEAR(A(i,j), B(i,j), tol);
}
}
}
它死于错误:
test_leq: /usr/include/eigen3/Eigen/src/Core/ProductBase.h:154: typename Base::CoeffReturnType Eigen::ProductBase<Eigen::GeneralProduct<Eigen::Matrix<double, -1, -1, 0, -1, -1>, Eigen::Matrix<double, -1, 1, 0, -1, 1>, 4>, Eigen::Matrix<double, -1, -1, 0, -1, -1>, Eigen::Matrix<double, -1, 1, 0, -1, 1> >::coeff(Index, Index) const: Assertion `this->rows() == 1 && this->cols() == 1' failed.
在调用 A(i,j)
时. (不过,我可以调用 cout << A << endl;
就好了。)
第 154 行,ProductBase.h
奇怪的是有断言
// restrict coeff accessors to 1x1 expressions. No need to care about mutators here since this isnt a Lvalue expression
typename Base::CoeffReturnType coeff(Index row, Index col) const
{
#ifdef EIGEN2_SUPPORT
return lhs().row(row).cwiseProduct(rhs().col(col).transpose()).sum();
#else
EIGEN_STATIC_ASSERT_SIZE_1x1(Derived)
eigen_assert(this->rows() == 1 && this->cols() == 1);
return derived().coeff(row,col);
#endif
}
我正在关注 Eigen's guide用于编写通用矩阵函数。如何正确编写此通用函数?
编辑:也很高兴知道为什么 ProductBase
需要一个 1x1 矩阵。
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