How to pass a 2D array by pointer in C?

In main(), the variable "array" is declared as

char array[50][50];

This is a 2500 byte piece of data. When main()'s "array" is passed about, it is a pointer to the beginning of that data. It is a pointer to a char expected to be organized in rows of 50.

Yet in function printarray(), you declare

 char **array

"array" here is a pointer to a char *pointer.

@Lucus suggestion of void printarray( char array[][50], int SIZE ) works, except that it is not generic in that your SIZE parameter must be 50.

Idea: defeat (yeech) the type of parameter array in printarray()

void printarray(void *array, int SIZE ){
    int i;
    int j;
    char *charArray = (char *) array;

    for( j = 0; j < SIZE; j++ ){
        for( i = 0; i < SIZE; i ++){
            printf( "%c ", charArray[j*SIZE + i] );
        }
        printf( "\n" );
    }
}

A more elegant solution is to make the "array" in main() an array of pointers.

// Your original printarray()
void printarray(char **array, int SIZE ){
    int i;
    int j;
    for( j = 0; j < SIZE; j++ ){
        for( i = 0; i < SIZE; i ++){
            printf( "%c ", array[j][i] );
        }
        printf( "\n" );
    }
}

// main()
char **array;
int SIZE;
// Initialization of SIZE is not shown, but let's assume SIZE = 50;
// Allocate table
array = (char **) malloc(SIZE * sizeof(char*));
  // Note: cleaner alternative syntax
  // array = malloc(sizeof *array * SIZE);
// Allocate rows
for (int row = 0; row<SIZE; row++) {
  // Note: sizeof(char) is 1. (@Carl Norum)
  // Shown here to help show difference between this malloc() and the above one.
  array[row] = (char *) malloc(SIZE * sizeof(char));
    // Note: cleaner alternative syntax
    // array[row] = malloc(sizeof(**array) * SIZE);
  }
// Initialize each element.
for (int row = 0; row<SIZE; row++) {
  for (int col = 0; col<SIZE; col++) {
    array[row][col] = 'a';  // or whatever value you want
  }
}
// Print it
printarray(array, SIZE);
...

Since C99 supports dynamic-sized arrays, the following style is simply more convenient to pass a 2-dim array:

void printarray( void *array0, int SIZE ){
    char (*array)[SIZE] = array0;
    int i;
    int j;
    for( j = 0; j < SIZE; j++ ){
        for( i = 0; i < SIZE; i ++){
            printf( "%c ", array[j][i] );
        }
        printf( "\n" );
    }
}

none of the answers here were what I was looking for, so I'm posting my simple solution to the problem

#include <iostream>

using namespace std;

void example(int* mat, int dim0, int dim1){
    
    for(int i = 0; i < dim0; ++i) {
         for(int j = 0; j < dim1; ++j) {
             auto cur_index = i * dim1 + j;
            cout<< *(mat + cur_index) << endl;
        }
    }
}

int main()
{
    const int dim0 = 3;
    const int dim1 = 2;

    int mat[dim0][dim1];
    
    for(int i = 0; i < dim0; ++i) {
         for(int j = 0; j < dim1; ++j) {
            mat[i][j] = i * dim1 + j;
        }
    }
    
    example(&(mat[0][0]), dim0, dim1);
    return 0;
}

You need to typecast the 2D array to a simple pointer by using (char*)array.

The corresponding dummy argument should just be char *array, rather than char **array. However, you need to know the memory layout to handle the 2D-array correctly. In this case, it should be array[i*SIZE+j] or *(array+i*SIZE+j). The resulting code is:

#include<stdio.h>
void printarray( char *array, int SIZE ){
    for( int j = 0; j < SIZE; j++ ){
        for( int i = 0; i < SIZE; i ++){
            printf( "%c ", array[i*SIZE+j] );
        }
        printf( "\n" );
    }
}


int main(){

  char array[50][50];
    printarray( (char*)array, 50 );
    return 0;
}

You can easily pass the 2d array using double pointer.

  void printarray( char **array, int n)
  {
     int i, j;
     for(i=0; i<n; i++ )
     {
         for(j=0; j<n; j++)
         {
            printf("%c ", array[i][j] );
         }
        printf( "\n" );
     }
  }

  int main()
  {
      int n = 2;
      int i, j;

      char **array = (char **) malloc(n * sizeof(char*));

      for (i=0; i<n; i++) 
      {
        array[i] = (char *) malloc(n* sizeof(char));
      }

     for (i=0; i<n; i++)
     {
       for (j=0; j<n; j++)
       {
           scanf("%c ", &array[i][j]);
       }
     }

     printarray(array, n);

     return 0;
  }

Full Code : Ideone