passing a char 2d array into a void function in c - c

Question

I'm new to c but I have been trying for ages to try and get this to work even though it seems so simple.
So below is what I am aiming to do which is working but I want to make 2 functions: fillseats() and printseatingplan()[for now I just want them all blank];
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main()
{
char seatingplan[15][15];
memset(seatingplan,'.',sizeof seatingplan);
for (int k = 0; k < 15; ++k)
{
for(int j = 0; j < 15; ++j)
{
printf("%c",seatingplan[k][j]);
}
printf("\n");
}
}
So whenever I try to run these functions without pointers it gives me these �`:�ij:�
i�d:�iH#=�i
�B��ik:�
how can i use pointers to fix this
can I pass this array into the functions where printseatingplan()
void printseatingplan(char array[15][15])
{
for (int k = 0; k < 15; ++k)
{
for(int j = 0; j < 15; ++j)
{
printf("%c",array[k][j]);
}
printf("\n");
}
}
and then fillseats() does:
void fillseats(char array[15][15])
{
memset(array,'.',sizeof array);
}

Answer 1

sizeof will give you only the size of the pointer, not the whole array.
You need to pass the sizes to the function. I would use pointer to array:
void fillseats(size_t rows, size_t cols, char (*array)[cols])
{
memset(array,'.',rows * sizeof(*array));
}
void printseatingplan(size_t rows, size_t cols, char (*array)[cols])
{
for (size_t row = 0; row < rows; row++)
{
for(size_t col = 0; col < cols; col++)
{
printf("%c",array[row][col]);
}
printf("\n");
}
}

Answer 2

char array[15][15] when used as parameter to a function "decays" into a pointer to the first element, in this case equivalent to char (*array)[15]). If you do sizeof array instead the function, you get the size of a pointer (4 or 8 etc). If you do sizeof *array you just get the size of one dimension, 15.
A simple way to fix it:
#include <stdio.h>
#include <string.h>
void fillseats (size_t x, size_t y, char array[x][y])
{
memset(array, '.', sizeof(char[x][y]));
}
void printseatingplan (size_t x, size_t y, char array[x][y])
{
for (size_t i = 0; i < x; i++)
{
for(size_t j = 0; j < y; j++)
{
printf("%c",array[i][j]);
}
printf("\n");
}
}
int main (void)
{
char seatingplan[15][15];
fillseats(15, 15, seatingplan);
printseatingplan(15, 15, seatingplan);
}

Answer 3

The size of array is bound to its type. The problem is that the parameters of array type decay to pointers. To prevent it, you can pass a pointer to an array. The pointer don't decay thus the essential part of the array type prevails.
void printseatingplan(char (*array)[15][15])
{
for (int k = 0; k < 15; ++k)
{
for(int j = 0; j < 15; ++j)
{
printf("%c", (*array)[k][j]);
}
printf("\n");
}
}
void fillseats(char (*array)[15][15])
{
memset(*array,'.',sizeof *array);
}
int main (void)
{
char seatingplan[15][15];
fillseats(&seatingplan);
printseatingplan(&seatingplan);
}