阻止 Tic Tac Toe 覆盖 C 中的动作

Question

抱歉，我几个小时前刚问了一个问题，但这是我完成项目所需要做的最后一件事。我的程序允许覆盖移动。就像说如果玩家 1 选择第一个方块，它将允许玩家 2 也选择这个方块。其他所有功能都完美无缺我只是在执行此操作时失败了。

有什么想法吗？我只放了一部分代码，因为很遗憾，我的很多代码都是复制粘贴的。我真的很新，所以我还没有想出优化，我已经没有时间做这个项目了，所以任何帮助将不胜感激。

所以在这段代码中，我只包含了原始棋盘的打印和第一位玩家的第一步（如果有 2 个人）。这应该足以获得帮助，但如果有人想查看其余代码，请告诉我，因为我已经省略了大部分代码。谢谢！！！

#include <stdio.h>
#include <stdlib.h>

/* known bugs: Allows players to place a 1 or 2 on board even if there is already a 1 or 2 in that spot. */

int main()
{
    char board[3][3];
    int i,j,k,l, player, move;

    for(i=0;i<=3;i++) // prints initial board of 0's
    {
        for(j=0;j<=3;j++)
        {
            board[i][j] = '0';
        }
    }


    printf("Hello! Do you want to play alone or with a computer? \n\n"
           "Enter 1 for alone or 2 to play with a friend!\n");
    scanf("%d", &player);

    if (player == 2)
    {
        for(k=0;k<9;k++) // 9 moves max.
        {
            printf("\n\n"); // print board again.
            printf(" %c | %c | %c\n", board[0][0], board[0][1], board[0][2]);
            printf("---+---+---\n");
            printf(" %c | %c | %c\n", board[1][0], board[1][1], board[1][2]);
            printf("---+---+---\n");
            printf(" %c | %c | %c\n", board[2][0], board[2][1], board[2][2]);

            do
            {
                printf("Player 1, where would you like to move? Enter 1-9\n");
                scanf("%d", &move);

                if (move == 1)
                    board[0][0] = '1';
                if (move == 2)
                    board[0][1] = '1';
                if (move == 3)
                    board[0][2] = '1';
                if (move == 4)
                    board[1][0] = '1';
                if (move == 5)
                    board[1][1] = '1';
                if (move == 6)
                    board[1][2] = '1';
                if (move == 7)
                    board[2][0] = '1';
                if (move == 8)
                    board[2][1] = '1';
                if (move == 9)
                    board[2][2] = '1';

            }while(move>9 && move <1);

Answer 1

当您声明一个大小为 N 的数组时，索引将从零到 X 减一。

所以在你的例子中，board 数组的索引是 0 到 2。 main 函数中的循环从 0 到 3，即越界。

越界写入数组会导致 undefined behavior, and undefined behavior makes your program ill-formed. Undefined behavior (or UB) can cause any kind of problems, including seemingly work, just to the next moment lead to nasal demons。

Answer 2

我建议在最后一个 do while 循环中使用 switch case（参见示例）
我认为您只需要检查您引用的字段是否已被占用！

** check()**

void check (char *c, int *move){
    if(*c == '0'){
        *c = '1';
    }else {
        printf("\nThis field is already taken! Please choose another one.\n");
        /* since you repeat this as long as move is bigger than 9 or smaller
           than 1, this will have the user make another choice. */
        *move = 0; 
    }
}

** 你最后一次做**

do {
    printf("Player 1, where would you like to move? Enter 1-9\n");
    scanf("%d", &move);

    switch (move){
        case 1:
            check(&board[0][0], &move);
            break;
        case 2:
            check(&board[0][1], &move);
            break;
        case 3:
            check(&board[0][2], &move);
            break;
        case 4:
            check(&board[1][0], &move);
            break;
        case 5:
            check(&board[1][1], &move);
            break;
        case 6:
            check(&board[1][2], &move);
            break;
        case 7:
            check(&board[2][0], &move);
            break;
        case 8:
            check(&board[2][1], &move);
            break;
        case 9:
            check(&board[2][2], &move);
            break;
        default:
            printf("\nError! Choose a field between 1 and 9\n");
    }

}while(move>9 || move <1);

注意：正如其他人所说，您的 for 循环需要迭代直到 i/j < 3，因为您的数组大小为 3（索引 0、1、2）；

注意2：while语句必须是move > 9 OR move < 1，因为int不能同时大于9小于1，会死循环

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