incompatible with parameter of type "int *" - c

Question

This is my final version for gameoflife. I have this error bug I can't fix. Please help me to solve this one:
1 IntelliSense: argument of type "int (*)[50]" is incompatible with parameter of type "int *"
line 79 print(board, HEIGHT, WIDTH);
void print(int *board, int rows, int cols)
{
int x, y;
char c;
for (y = 0; y < rows; y++) {
for (x = 0; x < cols; x++) {
if (*(board + y*cols + x) == 1)
printf("X");
else
printf(" ");
}
printf("\n");
}
printf("Press any key to continue:\n");
getchar();
}

Answer 1

That error means you are trying to pass a function an argument of type "array of 50 pointers to int" while it should be getting an argument of type "pointer to int" (which could also be an array).
In your case, the signature of your print() function should probably change to get an int board[][WIDTH] instead of int * which it gets now.
The change of the signature also requires a change in the function code, so the line
if (*(board + y*cols + x) == 1)
Should be changed to:
if (board[y][x] == 1)

Answer 2

void print(int *board, int rows, int cols)
...
int board[HEIGHT][WIDTH];
...
print(board, HEIGHT, WIDTH);
Your print function first parameter is of type int *board but you are calling the function with an argument of type int (*)[50].
You should fix your print function prototype (and body) to work with a parameter of the right type (i.e., int (*)[50]).

Answer 3

you declare
void print(int *board, int rows, int cols)
but in the the other functions you have used
foo(int board[][WIDTH], int rows)
either keep same or declare as
(int **board, int rows, int cols )

Answer 4

Your print function is expecting a pointer to int, but you are passing a 2-dimensional array, which decays to a pointer to a 1-dimensional array. Either cast board to an int* or convert print() to handle a 2-d array.
Casting the call would look something like this:
print((int *)board, HEIGHT, WIDTH);
Converting print() to take a 2D array as argument would look something like this:
void print(int board[][WIDTH])
{
int x, y;
char c;
for (y = 0; y < HEIGHT; y++) {
for (x = 0; x < WIDTH; x++) {
if (board[y][x] == 1)
printf("X");
else
printf(" ");
}
printf("\n");
}
printf("Press any key to continue:\n");
getchar();
}

Answer 5

why don't you just do like what sverre told you to, since a multidimensional array is really an abstraction the compiler provides not to do the maths you did to find the row_th and col_th element
just call it like print((int *)board, HEIGHT, WIDTH);

Answer 6

Your board is a double array. It has both height and width. When you expect a *board argument in your print you actually expect an array with only one dimension. For the error to go away you have to pass int (*board)[WIDTH].
When you pass a two-dimensional array to a function you have to include the size of one of the dimensions. So your board[HEIGHT][WIDTH] has to be passed with, at least, the information about it's width. So : board[][WIDTH]. You can take those empty [] and write them down as a pointer in front of board.
So, in your main you call your function correctly, but you have to edit your function's head to be like :
void print(int (*board)[WIDTH], int rows, int cols){