C language - Representing a matrix as an array of pointers

Question

I am trying to represent a matrix as an array of pointers, and then do certain actions on the matrix.

To simplify further, imagine a matrix with n rows and m columns,

I defined an int **matrix that points to the beginning of an array (of pointers) of size n. And every block in this array is a pointer that points to an array (of integers) of size m. The result is clear. you have n pointers, each of them pointing to an array of m integers, overall nxm values. So it should be possible to represent a matrix this way.

My problem is that I don't know how to access the integer values of the matrix. For example, suppose I wish to insert values to the matrix. How do I access matrix[i][j]?

Here is my code, you can see it is incomplete, I would appreciate help completing it.

int** create_matrix(int rows,int columns)
{
    int** matrix,i;
    matrix=(int**)malloc(rows*sizeof(int*));
    for(i=0;i<rows;i++)
        *(matrix)+i=(int*)malloc(columns*sizeof(int));
    for(i=0;i<rows;i++)
    {
        for(j=0;j<columns;j++)
            *(*(matrix)+i)

I want to insert random numbers into the matrix and then return it. I allocated the necesary memory, just need to insert the actual values.

You access it just the way you have written it: `matrix[i][j]`. — Some programmer dude, Aug 04 '14 at 12:14
Oh by the way, in C you [should not cast the return of `malloc`](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc). — Some programmer dude, Aug 04 '14 at 12:15
So pointer Arithmetic is not needed? and if i want to do it anyway? Just so I know how to do it — Oria Gruber, Aug 04 '14 at 12:15
Then you should remember (or learn) that `*(array_or_pointer + X)` is equivalent to `array_or_pointer[X]`. — Some programmer dude, Aug 04 '14 at 12:17
Oh, and to really blow your mind, since addition is [commutative](http://en.wikipedia.org/wiki/Commutativity) `*(array_or_pointer + X)` is equal to `*(X + array_or_pointer)` which leads to `array_or_pointer[X]` is the same as `X[array_or_pointer]`. :) — Some programmer dude, Aug 04 '14 at 12:19
I don't think your comment is correct mch. I mean, I know what matrix is. matrix points to an array of pointers. We have no idea what matrix+i points to. however, *(matrix)+i is matrix[i] — Oria Gruber, Aug 04 '14 at 12:20
`*(matrix)+i=(int*)malloc(columns*sizeof(int));` : `*(matrix)+i` is wrong. — BLUEPIXY, Aug 04 '14 at 12:21

score 1 · Answer 1 · answered Aug 04 '14 at 12:41

The general pattern for any type T is

T **arr = malloc( rows * sizeof *arr );
if ( arr )
{
  for ( size_t i = 0; i < rows; i++ )
  {
    arr[i] = malloc( columns * sizeof *arr[i] );
    if ( arr[i] )
    {
      for ( size_t j = 0; j < columns; j++ )
      {
        arr[i][j] = some_value;
      }
    }
    else
      // memory allocation failure
  }
}
else
  // memory allocation failure.

Array subscripts are defined in terms of pointer arithmetic: a[i] is defined as *(a + i). a[i][j] is defined as *(*(a + i) + j).

score 1 · Accepted Answer · answered Aug 04 '14 at 12:41

int** matrix,i,j;
matrix=malloc(rows*sizeof(int*));
for(i=0;i<rows;i++)
    *(matrix+i)=malloc(columns*sizeof(int)); //or matrix[i]=malloc(columns*sizeof(int));
for(i=0;i<rows;i++)
{
    for(j=0;j<columns;j++)
        *(*(matrix+i)+j)=random(); //or matrix[i][j]=random();
}
return matrix;

Note the change of the parentheses around matrix

Sathish · Answer 3 · 2014-08-04T14:31:02.700

Try the following code-

int** create_matrix(int rows,int columns)
{
   int** matrix,i;
   matrix=(int**)malloc(rows*sizeof(int*));
   for(i=0;i<rows;i++)
      matrix[i]=(int*)malloc(columns*sizeof(int)); // Fix 1
   for(i=0;i<rows;i++)
   {
      for(j=0;j<columns;j++)
         *(*(matrix+i)+j)=rand();
    }
    return matrix;
 }

Instead of doing arithmetic you can directly access it by matrix[i][j]. matrix[i][j] is equal to *(*(matrix+i)+j).

C language - Representing a matrix as an array of pointers

3 Answers3