how to fill two dimensional array(as a pointer) in a function?

Question

I declared this in the main:

char** board;
int N;
  .
  .
  .
  fill_Board(board, N);

then I scanf the number N and using malloc I did dynamic allocation for the char** board like that:

  void creat_board(char*** board,int N)
  {
   int i=0;
    (*board)=(char**)malloc(N*sizeof(char*)); 
     if (board==NULL)
     {
       printf("malloc failed!\n");
       exit(1);
     }
    for (i=0;i<N;i++)
       {
        (*board)[i]=(char*)malloc(N*sizeof(char)); /*add the 
        if((*board)[i]==NULL)
       {
         printf("malloc failed!\n");
         exit(1);
       }
     }
 }

I want now to fill the board and to print it but its not working..here is what I tried to fill:

Void fill_Board(char** board,int N)
  {
    int i=0,j=0;
    
        for(i=0;i<N;i++)
        {
           for(j=0;j<N;j++)
           {
            board[i][j]='S';  
            }                  
         }
     }

when I tried to print it the values in the array (board) didnt change at all.why it that happening??

Answer 1

use pointer to array:

void fill_Board(size_t N, char (*board)[N][N])
{
    for(size_t i = 0; i < N; i++)
    {
        for(size_t j = 0; j < N; j++)
        {
            (*board)[i][j]='S';  
        }                  
    }
}

or more simple:

void fill_Board(size_t N, char board[N][N])
{
    for(size_t i = 0; i < N; i++)
    {
        for(size_t j = 0; j < N; j++)
        {
            board[i][j]='S';  
        }                  
    }
}

to dynamically allocate memory:

void *create_Board(size_t N)
{
    return calloc(1, sizeof(char[N][N]));
}