How to convert 2D array to single dimensional array?

Answer 1

You don't need to physically convert the array, you simply need to point at the first element and read the elements sequentially. The following code shows how to achieve this, as well as how to physically convert a 2D array to a separate 1D array. As can be seen, there is no practical benefit to making the conversion.

Example

#include

int main()

{

const int rows = 2;

const int cols = 3;

const int size = rows * cols;

// Create a static 2D array, intialise and print the matrix

int arr2[rows][cols];

std::cout << "2D array:\n" << std::endl;

for( int r=0; r

{

for( int c=0; c

{

arr2[r][c]= (r+1) * (c+1);

std::cout << arr2[r][c] << " ";

}

std::cout << std::endl;

}

std::cout << std::endl;

// Efficient conversion to 1D array:

// Point to the first element then print sequential values.

int* ptr = (int*) arr2;

std::cout << "1D array (using pointer):\n" << std::endl;

for( int i=0; i

std::cout << *ptr++ << " ";

std::cout << "\n" << std::endl;

// Inefficient conversion to 1D array:

// Copy the 2D array to a separate 1D array

int arr1[size];

ptr = (int*) arr2;

for( int i=0; i

arr1[i]=*ptr++;

// Print the copy:

std::cout << "1D array (using copy):\n" << std::endl;

for( int i=0; i

std::cout << arr1[i] << " ";

std::cout << "\n" << std::endl;

return( 0 );

}

Output

2D array:

1 2 3

2 4 6

1D array (using pointer):

1 2 3 2 4 6

1D array (using copy):

1 2 3 2 4 6

Answer 2

All arrays are one-dimensional. Consider the following C-style array:

int array2d[5][4];

for (int row=0; row<5; ++row) for (int col=0; col<4; ++col) array2d[row][col] = row * col;

Given that the elements of array2d are allocated contiguously, we can easily convert it to a one-dimensional array of type int[20] simply by casting the start address to the appropriate type (pointer to int):

int* array1d = static_cast(&array2d);

for (int i=0; i<20; ++i) array1d[i] = i;