用c语言写一个井字棋游戏，写出完整代码

好的，下面是一个简单的 C 语言井字棋游戏的完整代码：

#include <stdio.h>

char board = { {' ', ' ', ' '}, {' ', ' ', ' '}, {' ', ' ', ' '} };

void display_board()
{
    printf("\n");
    printf(" %c | %c | %c \n", board, board, board);
    printf("---+---+---\n");
    printf(" %c | %c | %c \n", board, board, board);
    printf("---+---+---\n");
    printf(" %c | %c | %c \n", board, board, board);
}

int check_win()
{
    int i;
    // 检查行
    for (i = 0; i < 3; i++) {
        if (board[i] == board[i] && board[i] == board[i]) {
            if (board[i] == 'X')
                return 1;
            else if (board[i] == 'O')
                return -1;
        }
    }
    // 检查列
    for (i = 0; i < 3; i++) {
        if (board[i] == board[i] && board[i] == board[i]) {
            if (board[i] == 'X')
                return 1;
            else if (board[i] == 'O')
                return -1;
        }
    }
    // 检查对角线
    if (board == board && board[1] == board) {
        if (board == 'X')
            return 1;
        else if (board == 'O')
            return -1;
    }
    if (board == board && board == board) {
        if (board == 'X')
            return 1;
        else if (board == 'O')
            return -1;
    }
    return 0;
}

int main()
{
    int x, y, player = 1;
    while (check_win() == 0) {
        display_board();
        printf("\nPlayer %d, enter x and y coordinates for your move: ", player);
        scanf("%d%d", &x, &y);
        if (x < 0 || x > 2 || y < 0 || y > 2) {
            printf("\nInvalid coordinates! Try again.\n");
            continue;
        }
        if (board[x][y] != ' ') {
            printf("\nThat square is already occupied! Try again.\n");
            continue;
        }
        if (player == 1) {
            board[x][y] = 'X';
            player = 2;
        } else {
            board[x][y] = 'O';
            player = 1;
        }
    }
    display_board();
    if (check_win() == 1)
        printf("\nPlayer 1 wins!\n");
    else
        printf("\nPlayer 2 wins!\n");
    return 0;
}

如果你有任何问题或者需要进一步解释，请随时告诉我。现在，以下是几个相关问题：