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?

1
18
9/10/2009 2:53:20 AM

Accepted Answer

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.

48
9/10/2009 3:30:43 AM

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];
    }

}

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