Passing two-dimensional array via pointer

Question

How do I pass the m matrix to foo()? if I am not allowed to change the code or the prototype of foo()?

void foo(float **pm)
{
    int i,j;
    for (i = 0; i          


        

Answer 1

You can't. m is not compatible with the argument to foo. You'd have to use a temporary array of pointers.

int main()
{
    float m[4][4];
    int i,j;

    float *p[4];

    p[0] = m[0];
    p[1] = m[1];
    p[2] = m[2];
    p[3] = m[3];

    for (i = 0; i < 4; i++)
        for (j = 0; j < 4; j++)
            m[i][j] = i+j;


    foo(p);

Answer 2

If you can't change foo(), you will need to change m. Declare it as float **m, and allocate the memory appropriately. Then call foo(). Something like:

float **m = malloc(4 * sizeof(float *));
int i, j;
for (i = 0; i < 4; i++)
{
    m[i] = malloc(4 * sizeof(float));
    for (j = 0; j < 4; j++)
    {
        m[i][j] = i + j;
    }
}

Don't forget to free() afterwards!

Answer 3

typedef float Float4[4];

void foo(Float4 *pm)
{
  int i,j;
  for (i = 0; i < 4; i++)
    for (j = 0; j < 4; j++)
      printf("%f\n", pm[i][j]);
}

main()
{
  Float4 m[4];

  int i,j;
  for (i = 0; i < 4; i++)
    for (j = 0; j < 4; j++)
      m[i][j] = i+j;

  foo(m);
  return 0;
}

Answer 4

Using C99 which supports run-time sized arrays, the following is a possible way to pass a 2-dim array:

void foo(void *pm, int row, int col)
{
    float (*m)[col] = pm;

    for (int i = 0; i < row; i++)
        for (int j = 0; j < col; j++)
            printf("%4.1f%s", m[i][j], (j == col-1)?"\n":" ");

}

int main()
{
    float m[4][4];

    for (int i = 0; i < 4; i++)
        for (int j = 0; j < 4; j++)
            m[i][j] = i+j;

    foo(m, 4, 4);

    return 0;
}

Answer 5

• you dont need to do any changes in the main,but you function will work properly if you change the formal prototype of your function to (*pm)[4] or pm[][4] because **pm means pointer to pointer of integer while (*pm)[4] or pm[][4] means pointer to poiner of 4 integers .
  

  m here is also a pointer to pointer of 4 integers and not pointer to pointer of integers and hence not compatible.

  #include<stdio.h>
  void foo(float (*pm)[4])
  {
      int i,j;
      for (i = 0; i < 4; i++)
          for (j = 0; j < 4; j++)
              printf("%f\n", pm[i][j]);
  
  }
  
  int main ()
  {
      float m[4][4];
      int i,j;
      for (i = 0; i < 4; i++)
          for (j = 0; j < 4; j++)
                  m[i][j] = i+j;
  
      foo(m);
   }
  

Answer 6

If you insist on the above declaration of foo, i.e.

void foo(float **pm)

and on using a built-in 2D array, i.e.

float m[4][4];

then the only way to make your foo work with m is to create an extra "row index" array and pass it instead of m

...
float *m_rows[4] = { m[0], m[1], m[2], m[3] };
foo(m_rows);

There no way to pass m to foo directly. It is impossible. The parameter type float ** is hopelessly incompatible with the argument type float [4][4].

Also, since C99 the above can be expressed in a more compact fashion as

foo((float *[]) { m[0], m[1], m[2], m[3] });

P.S. If you look carefully, you'll that this is basically the same thing as what Carl Norum suggested in his answer. Except that Carl is malloc-ing the array memory, which is not absolutely necessary.