我已经这样定义了SquareMatrix
方阵
#ifndef SQUAREMATRIX_H
#define SQUAREMATRIX_H
#include "Matrix.h"
#include <vector>
class SquareMatrix : public Matrix
{
public:
SquareMatrix();
SquareMatrix(std::vector<std::vector<long double> >);
//~SquareMatrix(); //just in case there is dynamic memory explicitly used
//convenient member functions exclusive to SquareMatrix
bool isUpperDiagonalMatrix() const;
static SquareMatrix identityMatrix(unsigned);
void LUDecompose();
SquareMatrix *Lptr, *Uptr, *Pptr; //should be initialized by LUDecompose before using
protected:
void validateData();
private:
};
#endif // SQUAREMATRIX_H
我试图通过调用
Lptr来设置Uptr，Pptr（甚至是SquareMatrix::LUDecompose()）。定义如下：void SquareMatrix::LUDecompose()
{
unsigned rowCount = this->getRowCount();
//initialize L to identityMatrix
*this->Lptr = SquareMatrix::identityMatrix(rowCount);
//initialize U to sparse matrix with the first row containing the sole non-zero elements
std::vector<std::vector<long double> > UData(1, this->matrixData[0]); //making first rowVector the first rowVector of this
UData.insert(UData.end(), rowCount - 1, std::vector<long double>(rowCount,0)); //making all other rowVectors zero vectors
*Uptr = SquareMatrix(UData);
// attempting to perform LU decomposition
for (unsigned j = 0; j < rowCount; j++)
{
long double pivot = Uptr->matrixData[j][j];
//the pivot should be non-zero; throwing exception that should effectively cause function to return
if (pivot == 0)
throw MatrixArithmeticException(LU_DECOMPOSITION_FAILURE);
for (unsigned k = j+1; k < rowCount; k++)
{
if (j == 0)
{
//using *this to compute entries for L,U
this->Lptr->matrixData[k][j] = (this->matrixData[k][j])/pivot; //setting columns of L
long double multiplier = this->Lptr->matrixData[k][j];
//setting row of U
for (unsigned l = k; l < rowCount; l++)
{
Uptr->matrixData[k][l] = (this->matrixData[k][l])-multiplier*(this->matrixData[0][l]);
}
}
else
{
//using U to compute entries for L,U
//same procedure as before
this->Lptr->matrixData[k][j] = (Uptr->matrixData[k][j])/pivot;
long double multiplier = this->Lptr->matrixData[k][j];
for (unsigned l = k; l < rowCount; l++)
{
Uptr->matrixData[k][l] -= multiplier*(Uptr->matrixData[0][l]);
}
}
}
}
}
在尝试测试此功能时，它向我抛出了分段错误，最后一行是我尝试操纵
Lptr的第一行。我尝试更改
Lptr指向的对象，并且我知道我将无法引用该函数并将指针设置为与该引用相等。换句话说，我的编译器（GNU GCC编译器）不允许this->Lptr = &SquareMatrix::identityMatrix(rowCount);，因为它会引发-fpermissive类型错误。注意：
SquareMatrix::identityMatrix(unsigned)定义为：SquareMatrix SquareMatrix::identityMatrix(unsigned size)
{
std::vector<long double> rowVector(size, 0L);
std::vector<std::vector<long double> > identityMatrixData;
for (int i = 0; i < size; i++)
{
//setting the rowVector to zero-one vector
rowVector[i] = 1L;
if (i > 0) rowVector[i-1] = 0L;
//pushing rowVector into identityMatrixData
identityMatrixData.push_back(rowVector);
}
return SquareMatrix(identityMatrixData);
}
您认为您可以做什么？
我认为我有两种选择：
将对象扔到堆上，然后尝试使用函数进行设置（当您通过将对象扔到堆上重新定义刚刚定义的对象时，这似乎毫无用处）
得到c ++ 11（或类似的东西）
使该函数成为一个辅助函数，该函数返回大小为2的
std::vector<SquareMatrix*>（包含指向两个所需的SquareMatrix值的指针），并创建一个函数，该函数调用该辅助函数并将Lptr，Uptr设置为各个元素返回的vector。我的选择受限吗？
最佳答案
问题是*Uptr = SquareMatrix(UData);中的LUDecompose()。
您不能将指针设置为函数返回时将被破坏的对象。然后，该指针是一个悬空指针，每当您尝试使用它时，它都会出现段错误。
您需要执行Uptr = new SquareMatrix(UData);。然后在析构函数中调用delete Uptr;。
如果可以访问C ++ 11，则可以使用std::unique_ptr或任何指针容器/包装器。
选项示例：
#include <memory>
class Matrix
{
public:
Matrix() {}
virtual ~Matrix() {}
};
class SqMatrix : public Matrix //using raw pointers. You must remember to delete your pointers.
{
private:
SqMatrix* UPtr = nullptr;
public:
SqMatrix() : Matrix() {}
void InitPtrs() {delete UPtr; UPtr = new SqMatrix();}
~SqMatrix() {delete UPtr;}
};
class OMatrix : public Matrix //No need to worry about clean up.
{
private:
std::unique_ptr<OMatrix> OPtr;
public:
OMatrix() : Matrix() {}
void InitPtrs() {OPtr.reset(new OMatrix());}
~OMatrix() {}
};
另一种选择是将其存储在向量中。