how to allocate dynamic memory to int a[4][3] array

Question

how to allocate run time memory to an array of size[4][3]?

i.e int a[4][3]

If need is to allocate memory to an array at run time than how to allocate memory to 2D array or 3D array.

Answer 1

Editing the answer based on comments. Allocate separately for each dimension. For a 2D array a 2 level allocation is required.

*a = (int**)malloc(numberOfRows*sizeof(int*));
for(int i=0; i<numberOfRows; i++)
    {
    (*arr)[i] = (int*)malloc(numberOfColumns*sizeof(int));
    }

Answer 2

The simplest way to allocate dynamically an array of type int[4][3] is the following

int ( *a )[3] = new int[4][3];

// some stuff using the array

delete []a;

Another way is to allocate several arrays. For example

int **a = new int * [4];

for ( size_t i = 0; i < 4; i++ ) a[i] = new int[3];

// some stuff using the array

for ( size_t i = 0; i < 4; i++ ) delete []a[i];

delete []a;

Answer 3

What have you tried. new int[4][3] is a perfectly valid expression, and the results can be assigned to a variable with the appropriate type:

int (*array2D)[3] = new int[4][3];

Having said that: I can't really think of a case where this would be appropriate. Practically speaking, anytime you need a 2 dimensional array, you should define a class which implements it (using std::vector<int> for the actual memory).

Answer 4

A pure C approach is the following:

int (*size)[4][3];

size = malloc(sizeof *size);

/* Verify size is not NULL */

/* Example of access */
(*size)[1][2] = 89;

/* Do something useful */

/* Deallocate */
free(size);

The benefit is that you consume less memory by not allocating intermediate pointers, you deal with a single block of memory and deallocation is simpler. This is especially important if you start to have more than 2 dimensions.

The drawback is that the access syntax is more complicated, as you need to dereference a pointer before being able to index.

Answer 5

Use calloc, i guess this will do.

int **p;
p=(int**)calloc(4,sizeof(int));

Answer 6

In C you can use pointer to pointer

AS @Lundin mentioned this is not 2D array. It is a lookup table using pointers to fragmented memory areas allocated all over the heap.

You need to allocate how many pointers you need and then allocate each pointer. you can allocate fixed size or varaible size depending on your requirement

  //step-1: pointer to row
  int **a = malloc(sizeof(int *) * MAX_NUMBER_OF_POINTERS);
  //step-2: for each rows
  for(i = 0; i < MAX_NUMBER_OF_POINTERS; i++){
    //if you want to allocate variable sizes read them here
    a[i] = malloc(sizeof(int) * MAX_SIZE_FOR_EACH_POINTER);  // where as if you use character pointer always allocate one byte extra for null character
  } 

Where as if you want to allocate char pointers avoid using sizeof(char) inside for loop. because sizeof(char) == 1 and do not cast malloc result.

see How to declare a 2d array in C++ using new

Answer 7

You could use std::vector<> since it is a templated container (meaning array elements can be whatever type you need). std::vector<> allows for dynamic memory usage (you can change the size of the vector<> whenever you need to..the memory is allocated and free'd automatically).

For example:

#include <iostream>
#include <vector>

using namespace std; // saves you from having to write std:: in front of everthing

int main()
{
    vector<int> vA;
    vA.resize(4*3); // allocate memory for 12 elements

    // Or, if you prefer working with arrays of arrays (vectors of vectors)
    vector<vector<int> > vB;
    vB.resize(4);
    for (int i = 0; i < vB.size(); ++i)
        vB[i].resize(3);

    // Now you can access the elements the same as you would for an array
    cout << "The last element is " << vB[3][2] << endl;
}

Answer 8

You can use malloc() in c or new in c++ for dynamic memory allocation.