How do you dynamically allocate a matrix?

Question

How do you dynamically allocate a 2D matrix in C++? I have tried based on what I already know:

#include <iostream>

int main(){
    int rows;
    int cols;
    int * arr;
    arr = new int[rows][cols];
 }

It works for one parameter, but now for two. What should I do?

Answer 1

A matrix is actually an array of arrays.

int rows = ..., cols = ...;
int** matrix = new int*[rows];
for (int i = 0; i < rows; ++i)
    matrix[i] = new int[cols];

Of course, to delete the matrix, you should do the following:

for (int i = 0; i < rows; ++i)
    delete [] matrix[i];
delete [] matrix;

---

I have just figured out another possibility:

int rows = ..., cols = ...;
int** matrix = new int*[rows];
if (rows)
{
    matrix[0] = new int[rows * cols];
    for (int i = 1; i < rows; ++i)
        matrix[i] = matrix[0] + i * cols;
}

Freeing this array is easier:

if (rows) delete [] matrix[0];
delete [] matrix;

This solution has the advantage of allocating a single big block of memory for all the elements, instead of several little chunks. The first solution I posted is a better example of the arrays of arrays concept, though.

Answer 2

 #include <iostream>

    int main(){
        int rows=4;
        int cols=4;
        int **arr;

        arr = new int*[rows];
        for(int i=0;i<rows;i++){
           arr[i]=new int[cols];
        }
        // statements

        for(int i=0;i<rows;i++){
           delete []arr[i];
        }
        delete []arr;
        return 0;
     }

Answer 3

or you can just allocate a 1D array but reference elements in a 2D fashion:

to address row 2, column 3 (top left corner is row 0, column 0):

arr[2 * MATRIX_WIDTH + 3]

where MATRIX_WIDTH is the number of elements in a row.

Answer 4

arr = new int[cols*rows];

If you either don't mind syntax

arr[row * cols + col] = Aij;

or use operator[] overaloading somewhere. This may be more cache-friendly than array of arrays, or may be not, more probably you shouldn't care about it. I just want to point out that a) array of arrays is not only solution, b) some operations are more easier to implement if matrix located in one block of memory. E.g.

for(int i=0;i < rows*cols;++i)
   matrix[i]=someOtherMatrix[i];

one line shorter than

for(int r=0;i < rows;++r)
  for(int c=0;i < cols;++s)
     matrix[r][c]=someOtherMatrix[r][c];

though adding rows to such matrix is more painful

Answer 5

You can also use std::vectors for achieving this:

using std::vector< std::vector<int> >

Example:

std::vector< std::vector<int> > a;

  //m * n is the size of the matrix

    int m = 2, n = 4;
    //Grow rows by m
    a.resize(m);
    for(int i = 0 ; i < m ; ++i)
    {
        //Grow Columns by n
        a[i].resize(n);
    }
    //Now you have matrix m*n with default values

    //you can use the Matrix, now
    a[1][0]=1;
    a[1][1]=2;
    a[1][2]=3;
    a[1][3]=4;

//OR
for(i = 0 ; i < m ; ++i)
{
    for(int j = 0 ; j < n ; ++j)
    {      //modify matrix
        int x = a[i][j];
    }

}

Answer 6

Try boost::multi_array

#include <boost/multi_array.hpp>

int main(){
    int rows;
    int cols;
    boost::multi_array<int, 2> arr(boost::extents[rows][cols] ;
}

Answer 7

The other answer describing arrays of arrays are correct.
BUT if you are planning of doing a anything mathematical with the arrays - or need something special like sparse matrices you should look at one of the many maths libs like TNT before re-inventing too many wheels

Answer 8

I have this grid class that can be used as a simple matrix if you don't need any mathematical operators.

/**
 * Represents a grid of values.
 * Indices are zero-based.
 */
template<class T>
class GenericGrid
{
 public:
  GenericGrid(size_t numRows, size_t numColumns);

  GenericGrid(size_t numRows, size_t numColumns, const T & inInitialValue);

  const T & get(size_t row, size_t col) const;

  T & get(size_t row, size_t col);

  void set(size_t row, size_t col, const T & inT);

  size_t numRows() const;

  size_t numColumns() const;

 private:
  size_t mNumRows;
  size_t mNumColumns;
  std::vector<T> mData;
};


template<class T>
GenericGrid<T>::GenericGrid(size_t numRows, size_t numColumns):
 mNumRows(numRows),
 mNumColumns(numColumns)
{
 mData.resize(numRows*numColumns);
}


template<class T>
GenericGrid<T>::GenericGrid(size_t numRows, size_t numColumns, const T & inInitialValue):
 mNumRows(numRows),
 mNumColumns(numColumns)
{
 mData.resize(numRows*numColumns, inInitialValue);
}


template<class T>
const T & GenericGrid<T>::get(size_t rowIdx, size_t colIdx) const
{
 return mData[rowIdx*mNumColumns + colIdx];
}


template<class T>
T & GenericGrid<T>::get(size_t rowIdx, size_t colIdx)
{
 return mData[rowIdx*mNumColumns + colIdx];
}


template<class T>
void GenericGrid<T>::set(size_t rowIdx, size_t colIdx, const T & inT)
{
 mData[rowIdx*mNumColumns + colIdx] = inT;
}


template<class T>
size_t GenericGrid<T>::numRows() const
{
 return mNumRows;
}


template<class T>
size_t GenericGrid<T>::numColumns() const
{
 return mNumColumns;
}