How do we allocate a 2-D array using One malloc statement

I have been asked in an interview how do i allocate a 2-D array and below was my solution to it.

#include <stdlib.h>

int **array;
array = malloc(nrows * sizeof(int *));

for(i = 0; i < nrows; i++)
{
    array[i] = malloc(ncolumns * sizeof(int));
    if(array[i] == NULL)
    {
        fprintf(stderr, "out of memory\n");
        exit or return
    }
}

I thought I had done a good job but then he asked me to do it using one malloc() statement not two. I don't have any idea how to achieve it.

Can anyone suggest me some idea to do it in single malloc()?

Just compute the total amount of memory needed for both nrows row-pointers, and the actual data, add it all up, and do a single call:

int **array = malloc(nrows * sizeof *array + (nrows * (ncolumns * sizeof **array));

If you think this looks too complex, you can split it up and make it a bit self-documenting by naming the different terms of the size expression:

int **array; /* Declare this first so we can use it with sizeof. */
const size_t row_pointers_bytes = nrows * sizeof *array;
const size_t row_elements_bytes = ncolumns * sizeof **array;
array = malloc(row_pointers_bytes + nrows * row_elements_bytes);

You then need to go through and initialize the row pointers so that each row's pointer points at the first element for that particular row:

size_t i;
int * const data = array + nrows;
for(i = 0; i < nrows; i++)
  array[i] = data + i * ncolumns;

Note that the resulting structure is subtly different from what you get if you do e.g. int array[nrows][ncolumns], because we have explicit row pointers, meaning that for an array allocated like this, there's no real requirement that all rows have the same number of columns.

It also means that an access like array[2][3] does something distinct from a similar-looking access into an actual 2d array. In this case, the innermost access happens first, and array[2] reads out a pointer from the 3rd element in array. That pointer is then treatet as the base of a (column) array, into which we index to get the fourth element.

In contrast, for something like

int array2[4][3];

which is a "packed" proper 2d array taking up just 12 integers' worth of space, an access like array[3][2] simply breaks down to adding an offset to the base address to get at the element.

int **array = malloc (nrows * sizeof(int *) + (nrows * (ncolumns * sizeof(int)));

This works because in C, arrays are just all the elements one after another as a bunch of bytes. There is no metadata or anything. malloc() does not know whether it is allocating for use as chars, ints or lines in an array.

Then, you have to initialize:

int *offs = &array[nrows]; /*  same as int *offs = array + nrows; */
for (i = 0; i < nrows; i++, offs += ncolumns) {
    array[i] = offs;
}

Here's another approach.

If you know the number of columns at compile time, you can do something like this:

#define COLS ... // integer value > 0
...
size_t rows;
int (*arr)[COLS];
...              // get number of rows
arr = malloc(sizeof *arr * rows);
if (arr)
{
  size_t i, j;
  for (i = 0; i < rows; i++)
    for (j = 0; j < COLS; j++)
      arr[i][j] = ...;
}

If you're working in C99, you can use a pointer to a VLA:

size_t rows, cols;
...               // get rows and cols
int (*arr)[cols] = malloc(sizeof *arr * rows);
if (arr)
{
  size_t i, j;
  for (i = 0; i < rows; i++)
    for (j = 0; j < cols; j++)
      arr[i][j] = ...;
}