C：如何将数组传递给 function？

Question

I'm a beginner in C trying to make a Tic Tac Toe game. So first I want to start off by displaying the board showing the available positions (0-8). Therefore I'm trying to pass all elements in numArray to board so I can put each number in the board, but that's where I'm getting stuck as no numbers are being displayed.

Would mean a lot if someone could help me out here:))

Output:

  |  | 
---|---|---
  |  | 
---|---|---
  |  | 
[Player 1] Pick a number: 

int main() {
    int numArray[] = {0, 1, 2, 3, 4, 5, 6, 7, 8};
    int playerOneCount = 0, playerTwoCount = 0;

    board(numArray);
    playerOne(numArray);
    return 0;
}

void board(int array[]) {
    printf(
    " %c | %c | %c\n"
    "---|---|---\n"
    " %c | %c | %c\n"
    "---|---|---\n"
    " %c | %c | %c\n", array[0], array[1], array[2], array[3], array[4],
                       array[5], array[6], array[7], array[8]);
    return;
}

Expected output:

 0 | 1 | 2 
---|---|---
 3 | 4 | 5
---|---|---
 6 | 7 | 8 
[Player 1] Pick a number: 

Answer 1

Arrays aren't first-class values, which means they cannot be passed around like that. You can, however, pass a pointer to the first element of the array, which will let you access the array in pretty much the same way:

int main (void) {
  int numArray[] = {0, 1, 2, 3, 4, 5, 6, 7, 8};
  // do whatever with the array
  board(numArray);
  // etc.
}

void board (int * arr) {
  // here you can access the array elements directly, as in arr[3]
}

Note that you can write the function's parameter as int arr[] ; this is completely equivalent to int * arr , and I'd recommend getting familiarized with pointers if you haven't yet.