Dynamic allocation of memory to 2 dimensional Array in a Single malloc Call

[neeru29]

To allocate a 2D array of integers, in such a way that the array elements can be accessed with a[i][j] syntax i.e., should not use pointers to access the values.

Note: The function should perform only a single malloc call.

The function declaration is:

int **alloc_array( int nrows, int ncols);

[weaknessforcats]

Read: http://bytes.com/topic/c/insights/77...rrays-revealed

[neeru29]

I had tried with the following statement to allocate memory to 2D array:

arr2D = (int **)malloc(row * sizeof(int *) + row * col * sizeof(int));
 
for(i = 0; i< row; i++)
{
 for(j=0; j< col; j++)
 {
  scanf("%d", &a[i][j]); // here the memory segmentation is displayed.
 }
}
 

Please help me out

[donbock]

Are you sure you can't use two malloc calls to set this up?

[neeru29]

i got it by using multiple malloc calls.. i want it by single malloc call

[weaknessforcats]

Remember, there is no such thing as a 2D array.

And an int** does not point to a 2D array.

If you have a 9 element array, the elements are in memory as

1 2 3 4 5 6 7 8 9

and you can't tell if this is a [9][1] or a [3][3].

So when you allocate you allocate the number of elements.

Let's say you need a [4][6] array of int. That's 24 ints. So allocate 24 ints.

malloc will return a pointer to the 24 ints.

Now a [4][6] array is an array of 4 elements where each element is an array of 6 int. Therefore typecast the return from malloc as a pointer to an array of 6 int:

int (*MyArray)[6] = (int(*)[6])malloc(24*sizeof(int));

You will need a separate variable to hold the 4 which is the number of elements.

Go back and read that article again. The final example is the one I am talking about here.