我有SquareMatrix
如此定义
方阵.h
#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
我正在尝试设置 Lptr
, Uptr
(可能还有 Pptr
)调用 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,它向我抛出一个段错误,最后一行是我尝试操作 Lptr
的第一行.
我尝试更改 Lptr
指向的对象 ,我知道我将无法引用该函数并将指针设置为等于该引用。换句话说,我的编译器(GNU GCC 编译器)不允许 this->Lptr = &SquareMatrix::identityMatrix(rowCount);
因为它会抛出 -fpermissive 类型错误。
备注: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);
}
您认为您可以做些什么?
我想我有两个选择:
- 将对象扔到堆上,然后尝试使用函数对其进行设置(这看起来毫无用处,因为您正在通过将其扔到堆上来重新定义刚刚定义的对象)
- 获得 c++11(或类似的东西)
- 使该函数成为返回
std::vector<SquareMatrix*>
的辅助函数大小为二(包含指向两个所需的SquareMatrix
值的指针),并创建一个调用辅助函数并设置Lptr
的函数,Uptr
到返回的各个元素vector
.
我的选择就这么有限吗?
最佳答案
*Uptr = SquareMatrix(UData);
LUDecompose()
是问题所在。
您不能将指针设置为函数返回时正在销毁的对象。然后指针是一个悬挂指针,无论何时您尝试使用它,它都会发生段错误。
您需要执行 Uptr = new SquareMatrix(UData);
。然后在您的析构函数中,调用 delete Uptr;
。
如果您可以访问 C++11,则可以使用 std::unique_ptr
或任何指针容器/包装器。
您的选择示例:
#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 vector 中。
关于c++ - 取消引用指向其类内对象的指针,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20590187/