Question

我正在使用Spectra C++库，它建立在Eigen C++库之上。

我需要根据方法中给定的SelectionRule和GeigsMode参数构建一个正确的SymGEigsSolver对象。 SymGEigsSolver的定义如下：

template<typename Scalar, int SelectionRule, typename OpType, typename BOpType, int GEigsMode>
class Spectra::SymGEigsSolver< Scalar, SelectionRule, OpType, BOpType, GEigsMode >

但是当我尝试编译以下函数时：

typedef SparseMatrix<double, Eigen::RowMajor> SpMat;
typedef Spectra::SparseSymMatProd<double, Eigen::Upper, Eigen::RowMajor> SpSymMatProd;
typedef Spectra::SparseCholesky<double, Eigen::Upper> SpCholesky;
typedef Spectra::SparseRegularInverse<double, Eigen::Upper, Eigen::RowMajor> SpRegularInverse;


template < typename Scalar,
    int SelectionRule,
    typename OpType,
    typename BOpType,
    int GEigsMode >
    SymGEigsSolver<Scalar, SelectionRule, OpType, BOpType, GEigsMode>& CreateSolverEngine
    (const SpMat& A, const SpMat&  B, int selectRule, int gMode,  int nMode, int ncv)
{
    SpSymMatProd op(A);

    if (gMode == Spectra::GEIGS_CHOLESKY)
    {
        SpCholesky  Bop(B);
        SymGEigsSolver<Scalar, SelectionRule, SpSymMatProd, SpCholesky, GEigsMode>
            ges1(op, Bop, nMode, ncv);
        return ges1;
    }
    if (gMode == Spectra::GEIGS_REGULAR_INVERSE)
    {
        SpRegularInverse Bop2(B);
        SymGEigsSolver<Scalar, SelectionRule, SpSymMatProd, SpRegularInverse, GEigsMode>
            ges(op, Bop2, nMode, ncv);
        return ges;
    }

    throw std::out_of_range("out of range for "+gMode);


}

我遇到了编译错误，例如：

Error   C2664   'Spectra::SymGEigsSolver<Scalar,SelectionRule,OpType,BOpType,GEigsMode>::SymGEigsSolver(OpType *,BOpType *,int,int)'
: cannot convert parameter 1
from 'SpSymMatProd' to 'SpSymMatProd *'

和

C2440   'return' : cannot convert from
'Spectra::SymGEigsSolver<Scalar,SelectionRule,OpType,BOpType,GEigsMode>'
to
'Spectra::SymGEigsSolver<Scalar,SelectionRule,OpType,BOpType,GEigsMode>&'

并且

C2664   Spectra::SymGEigsSolver<Scalar,SelectionRule,OpType,BOpType,GEigsMode>::SymGEigsSolver(OpType *,BOpType *,int,int)'
: cannot convert parameter 1 from 'SpSymMatProd' to 'SpSymMatProd *'

Answer 1

我自己找出了答案 - 我在方法名称上输入了&。这是正确的解决方案：

typedef SparseMatrix<double, Eigen::RowMajor> SpMat;
typedef Spectra::SparseSymMatProd<double, Eigen::Upper, Eigen::RowMajor> SpSymMatProd;
typedef Spectra::SparseCholesky<double, Eigen::Upper> SpCholesky;
typedef Spectra::SparseRegularInverse<double, Eigen::Upper, Eigen::RowMajor> SpRegularInverse;


template < typename Scalar,
    int SelectionRule,
    typename OpType,
    typename BOpType,
    int GEigsMode >
    SymGEigsSolver<Scalar, SelectionRule, OpType, BOpType, GEigsMode> CreateSolverEngine
    (const SpMat& A, const SpMat&  B, int selectRule, int gMode,  int nMode, int ncv)
{
    SpSymMatProd op(A);

    if (gMode == Spectra::GEIGS_CHOLESKY)
    {
        SpCholesky  Bop(B);
        SymGEigsSolver<Scalar, SelectionRule, SpSymMatProd, SpCholesky, GEigsMode>
            ges1(op, Bop, nMode, ncv);
        return ges1;
    }
    if (gMode == Spectra::GEIGS_REGULAR_INVERSE)
    {
        SpRegularInverse Bop2(B);
        SymGEigsSolver<Scalar, SelectionRule, SpSymMatProd, SpRegularInverse, GEigsMode>
            ges(op, Bop2, nMode, ncv);
        return ges;
    }

    throw std::out_of_range("out of range for "+gMode);


}

在C ++中构造模板函数时编译错误C2664和C2440

1 个答案: