正如您可能会注意到这背后有一个双重存储。任何读/写值都被转换为平面索引,因此它存储在一个大的单维数组中。
我需要知道我是否可以在不同的线程中写入此类对象中的不同索引?我正在使用operator()来设置值。在对对象进行实例化时预定义边界。即:矩阵m(行,列); 我在boost线程中使用它并遇到问题。我需要放任何范围的锁吗?但那是我相信的性能成本。
class matrix
{
public:
matrix(size_t rows, size_t cols):_rows(rows),_cols(cols),_data(0)
{
reset(rows, cols);
}
matrix():_data(0),_rows(0),_cols(0){}
~matrix()
{
if (0 != _data)
{
delete [] _data;
}
}
matrix(const matrix& copythis)
:_data(0),_rows(0),_cols(0)
{
if (0 != _data)
{
delete [] _data;
}
_rows = copythis.rows();
_cols = copythis.cols();
_data = new double [_rows * _cols];
if (0 == _data)
{
NL_THROW("Insufficient memory to create a cloned matrix of double of " << _rows << " X " << _cols);
}
memcpy(_data, copythis._data, _rows * _cols * sizeof(double) );
}
public:
const matrix& operator = (const matrix& copythis)
{
if (0 != _data)
{
delete [] _data;
}
_rows = copythis.rows();
_cols = copythis.cols();
_data = new double [_rows * _cols];
if (0 == _data)
{
NL_THROW("Insufficient memory to create a cloned matrix of double of " << _rows << " X " << _cols);
}
memcpy(_data, copythis._data, _rows * _cols * sizeof(double) );
return (*this);
}
double* cArray()
{
if (0 == _data)
{
NL_THROW("Matrix is not initialised");
}
return _data;
}
void reset(size_t rows, size_t cols)
{
if (0 != _data)
{
delete [] _data;
}
_rows = rows;
_cols = cols;
_data = new double[rows * cols];
if (0 == _data)
{
NL_THROW("Insufficient memory to create a matrix of double of " << _rows << " X " << _cols);
}
memset(_data, 0, sizeof(double) * _rows * _cols);
}
double& operator () (size_t rowIndex, size_t colIndex) const
{
if (rowIndex >= _rows)
{
NL_THROW("Row index " << rowIndex << " out of range(" << _rows - 1 << ")");
}
if (colIndex >= _cols)
{
NL_THROW("Column index " << colIndex << " out of range(" << _cols - 1 << ")");
}
size_t flatIndex = colIndex + rowIndex * _cols;
return _data[flatIndex];
}
Array rowSlice(size_t rowIndex) const
{
if (rowIndex >= _rows)
{
NL_THROW("Cannot slice matrix, required row: " << rowIndex << " is out of range(" << (_rows - 1) << ")");
}
Array retval(_data + rowIndex * _cols, _cols);
/*
for(size_t i = 0; i < _cols; i++)
{
retval[i] = operator()(rowIndex, i);
}
*/
return retval;
}
Array colSlice(size_t colIndex) const
{
if (colIndex >= _cols)
{
NL_THROW("Cannot slice matrix, required row: " << colIndex << " is out of range(" << (_cols - 1) << ")");
}
Array retval(_rows);
for(size_t i = 0; i < _rows; i++)
{
retval[i] = operator()(i, colIndex);
}
return retval;
}
void fill(double value)
{
for(size_t rowIndex = 0; rowIndex < _rows; rowIndex++)
{
for(size_t colIndex = 0; colIndex < _cols; colIndex++)
{
size_t flatIndex = colIndex + rowIndex * _cols;
_data[flatIndex] = value;
}
}
}
bool isEmpty() const
{
if (0 == _rows) return true;
if (0 == _cols) return true;
return false;
}
size_t rows() const {return _rows;}
size_t cols() const {return _cols;}
private:
double* _data;
size_t _rows;
size_t _cols;
};
答案 0 :(得分:1)
如果您可以保证所有线程永远不会在矩阵的同一元素上发生冲突,那么您就不会需要锁定。在实践中,期望这样的保证是不现实的,因此您需要锁定。
答案 1 :(得分:1)
编写此类的方式,除了getter rows()
和cols()
之外,您需要锁定每个方法,几乎在每个方法的最广泛范围内
你甚至不能依赖size_t
类型的原子来保存任何锁,因为reset()
非原子地设置了_row
和_col
。任何需要了解_row
和_col
的操作,例如isEmpty()
,如果从另一个线程调用reset()
,则会遇到麻烦。