How to pass 2D array (matrix) in a function in C?

Question

I need to do this to persist operations on the matrix as well. Does that mean that it needs to be passed by reference?

Will this suffice?

void operate_on_matrix(char matrix[][20]);

Answer 1

C does not really have multi-dimensional arrays, but there are several ways to simulate them. The way to pass such arrays to a function depends on the way used to simulate the multiple dimensions:

1) Use an array of arrays. This can only be used if your array bounds are fully determined at compile time, or if your compiler supports VLA's:

#define ROWS 4
#define COLS 5

void func(int array[ROWS][COLS])
{
  int i, j;

  for (i=0; i<ROWS; i++)
  {
    for (j=0; j<COLS; j++)
    {
      array[i][j] = i*j;
    }
  }
}

void func_vla(int rows, int cols, int array[rows][cols])
{
  int i, j;

  for (i=0; i<rows; i++)
  {
    for (j=0; j<cols; j++)
    {
      array[i][j] = i*j;
    }
  }
}

int main()
{
  int x[ROWS][COLS];

  func(x);
  func_vla(ROWS, COLS, x);
}

2) Use a (dynamically allocated) array of pointers to (dynamically allocated) arrays. This is used mostly when the array bounds are not known until runtime.

void func(int** array, int rows, int cols)
{
  int i, j;

  for (i=0; i<rows; i++)
  {
    for (j=0; j<cols; j++)
    {
      array[i][j] = i*j;
    }
  }
}

int main()
{
  int rows, cols, i;
  int **x;

  /* obtain values for rows & cols */

  /* allocate the array */
  x = malloc(rows * sizeof *x);
  for (i=0; i<rows; i++)
  {
    x[i] = malloc(cols * sizeof *x[i]);
  }

  /* use the array */
  func(x, rows, cols);

  /* deallocate the array */
  for (i=0; i<rows; i++)
  {
    free(x[i]);
  }
  free(x);
}

3) Use a 1-dimensional array and fixup the indices. This can be used with both statically allocated (fixed-size) and dynamically allocated arrays:

void func(int* array, int rows, int cols)
{
  int i, j;

  for (i=0; i<rows; i++)
  {
    for (j=0; j<cols; j++)
    {
      array[i*cols+j]=i*j;
    }
  }
}

int main()
{
  int rows, cols;
  int *x;

  /* obtain values for rows & cols */

  /* allocate the array */
  x = malloc(rows * cols * sizeof *x);

  /* use the array */
  func(x, rows, cols);

  /* deallocate the array */
  free(x);
}

4) Use a dynamically allocated VLA. One advantage of this over option 2 is that there is a single memory allocation; another is that less memory is needed because the array of pointers is not required.

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

extern void func_vla(int rows, int cols, int array[rows][cols]);
extern void get_rows_cols(int *rows, int *cols);
extern void dump_array(const char *tag, int rows, int cols, int array[rows][cols]);

void func_vla(int rows, int cols, int array[rows][cols])
{
    for (int i = 0; i < rows; i++)
    {
        for (int j = 0; j < cols; j++)
        {
            array[i][j] = (i + 1) * (j + 1);
        }
    }
}

int main(void)
{
    int rows, cols;

    get_rows_cols(&rows, &cols);

    int (*array)[cols] = malloc(rows * cols * sizeof(array[0][0]));
    /* error check omitted */

    func_vla(rows, cols, array);
    dump_array("After initialization", rows, cols, array);

    free(array);
    return 0;
}

void dump_array(const char *tag, int rows, int cols, int array[rows][cols])
{
    printf("%s (%dx%d):\n", tag, rows, cols);
    for (int i = 0; i < rows; i++)
    {
        for (int j = 0; j < cols; j++)
            printf("%4d", array[i][j]);
        putchar('\n');
    }
}

void get_rows_cols(int *rows, int *cols)
{
    srand(time(0));           // Only acceptable because it is called once
    *rows = 5 + rand() % 10;
    *cols = 3 + rand() % 12;
}

(See srand() — why call it only once?.)

Answer 2

Easiest Way in Passing A Variable-Length 2D Array

Most clean technique for both C & C++ is: pass 2D array like a 1D array, then use as 2D inside the function.

#include <stdio.h>

void func(int row, int col, int* matrix){
    int i, j;
    for(i=0; i<row; i++){
        for(j=0; j<col; j++){
            printf("%d ", *(matrix + i*col + j)); // or better: printf("%d ", *matrix++);
        }
        printf("\n");
    }
}

int main(){
    int matrix[2][3] = { {0, 1, 2}, {3, 4, 5} };
    func(2, 3, matrix[0]);

    return 0;
}

_{Internally, no matter how many dimensions an array has, C/C++ always maintains a 1D array. And so, we can pass any multi-dimensional array like this.}

Answer 3

I don't know what you mean by "data dont get lost". Here's how you pass a normal 2D array to a function:

void myfunc(int arr[M][N]) { // M is optional, but N is required
  ..
}

int main() {
  int somearr[M][N];
  ...
  myfunc(somearr);
  ...
}

Answer 4

2D array:

int sum(int array[][COLS], int rows)
{

}

3D array:

int sum(int array[][B][C], int A)
{

}

4D array:

int sum(int array[][B][C][D], int A)
{

}

and nD array:

int sum(int ar[][B][C][D][E][F].....[N], int A)
{

}