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社区首页 >问答首页 >零和交叉版本2

问零和交叉版本2
EN

Code Review用户
提问于 2019-02-15 21:29:02
回答 1查看 67关注 0票数 3

这款游戏比以前的版本更方便用户--它有一个示例网格来演示输入，现在使用的是Xs和Os。

游戏现在可以检测到抽签。

改变了算法来检测胜利者，虽然游戏结束时仍然有一些组合错误地决定了胜利者或输家--但不知道如何解决这个问题，也不知道是什么原因造成的。

我很想知道如何改进游戏和代码的结构，以及如何避免提到的错误。

代码语言：javascript
复制
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

// Struct with all game state variables.
struct game_data {
    int win; // Either 0 or 1.
    int turns; // Ranges from 1 to 9(game end).
    int turn; // Either 0 or 1 where 0 is human player
    char grid[3][3];
};

//Initialising game state variables
struct game_data game = {
    0,
    1,
    0,
    { { ' ', ' ', ' ' },
      { ' ', ' ', ' ' },
      { ' ', ' ', ' ' } }
    };

void intro(void){

    printf("Welcome to NOUGHTS AND CROSSES\n\n");
    printf("The grid you will be playing on is 3x3 and your input will be determined by the co ordinates you put in, in the form 'row column'.\n\n");
    printf("For example an input of '1 1' will put a 'Z' on the first row on the first column. Like so:\n\n");

    printf(
        "+---+---+---+\n"
        "| Z |   |   |\n"
        "+---+---+---+\n"
        "|   |   |   |\n"
        " ---+---+---|\n"
        "|   |   |   |\n"
        "+---+---+---+\n"
        "\n");
}

void player_one_move(struct game_data* game)
{
    int y_val, x_val;
    printf("You are 'Crosses'. Please input co-ordinates in the form 'row column' for the 3x3 grid:\n");
    scanf(" %d %d", &y_val, &x_val);

    if (game->grid[y_val - 1][x_val - 1] == ' ') {
        game->grid[y_val - 1][x_val - 1] = 'X';
        printf("\nYour turn:\n\n");
    }
    else {
        player_one_move(game);
    }
}

void computer_move(struct game_data* game)
{
    int x_val = rand() % 3;
    int y_val = rand() % 3;

    if (game->grid[y_val][x_val] == ' ') {
        game->grid[y_val][x_val] = 'O';
        printf("\nComputer turn:\n\n");
    }
    else {
        computer_move(game);
    }
}

void update(struct game_data* game)
{

    printf(

        "+---+---+---+\n"
        "| %c | %c | %c |\n"
        "+---+---+---+\n"
        "| %c | %c | %c |\n"
        "----+---+---|\n"
        "| %c | %c | %c |\n"
        "+---+---+---+\n"
        "\n",
        game->grid[0][0], game->grid[0][1], game->grid[0][2],
        game->grid[1][0], game->grid[1][1], game->grid[1][2],
        game->grid[2][0], game->grid[2][1], game->grid[2][2]);
}

void game_event_won(struct game_data* game)
{

    int count;

    //BUGGY
    /*char current_mark;

    if (game->turn == 0) {  
        current_mark = 'X';}
    else{
        current_mark = 'O';
    }*/

    for (int y_val = 0; y_val < 3; y_val++) {
        for (int x_val = 0; x_val < 3; x_val++) {
            count = 0;
            while (game->grid[y_val][x_val] == 'X') {
                x_val++;
                count++;

            //BUGGY
            /*if (count == 3 && current_mark == 'X') {   
                game->win = 1;
                printf("You have WON\n");
            }
            if (count == 3 && current_mark == 'O') {
                game->win = 1;
                printf("You have LOST\n");
            }*/

            if (count == 3) {
                game->win = 1;
                printf("You have WON\n");
            }
          }
        }
    }
    for (int x_val = 0; x_val < 3; x_val++) {
         for (int y_val = 0; y_val < 3; y_val++) {
             count = 0;
             while (game->grid[y_val][x_val] == 'X') {
                 y_val++;
                 count++;

                 if (count == 3) {
                     game->win = 1;
                     printf("You have WON\n");
                 }
             }
         }
     }
     for (int y_val = 0; y_val < 3; y_val++) {
         count = 0;
         while (game->grid[y_val][y_val] == 'X') {
             count++;
             y_val++;

             if (count == 3) {
                  game->win = 1;
                  printf("You have won\n");
             }
         }
     }
     for (int y_val = 0; y_val < 3; y_val++) {
         count = 0;
         while (game->grid[y_val][2 - y_val] == 'X') {
             count++;
             y_val++;

             if (count == 3) {
                 game->win = 1;
                 printf("You have won\n");
             }
         }
     }
 }


 // Repetition of previous function but for 'O's. Less concise but less buggy than previous implementation.
 void game_event_lost(struct game_data* game)
 {

      int count;

      for (int y_val = 0; y_val < 3; y_val++) {
           for (int x_val = 0; x_val < 3; x_val++) {
               count = 0;
               while (game->grid[y_val][x_val] == 'O') {
                    x_val++;
                    count++;


                    if (count == 3) {
                         game->win = 1;
                         printf("You have LOST\n");
                    }
              }
         }
     }
     for (int x_val = 0; x_val < 3; x_val++) {
     for (int y_val = 0; y_val < 3; y_val++) {
        count = 0;
        while (game->grid[y_val][x_val] == 'O') {
            y_val++;
            count++;

            if (count == 3) {
                game->win = 1;
                printf("You have LOST\n");
            }
        }
    }
  }
for (int y_val = 0; y_val < 3; y_val++) {
    count = 0;
    while (game->grid[y_val][y_val] == 'O') {
        count++;
        y_val++;

        if (count == 3) {
            game->win = 1;
            printf("You have LOST\n");
        }
    }
}
for (int y_val = 0; y_val < 3; y_val++) {
    count = 0;
    while (game->grid[y_val][2 - y_val] == 'O') {
        count++;
        y_val++;

        if (count == 3) {
            game->win = 1;
            printf("You have LOST\n");
            }
        }
    }
}


int main(void)
{

    srand((unsigned)time(0));

    intro();

    while (game.win == 0 ) {
        if (game.turn == 0) {
            player_one_move(&game);
            game.turns++;
            game.turn = 1;
        }

        else {
            game.turn = 0;
            computer_move(&game);
            game.turns++;
        }
        if (game.turns == 9 && game.win == 0){
            game.win = 1;
            printf("You have drawn\n");
            break;
        }
        update(&game);


        game_event_won(&game);
        game_event_lost(&game);

    }

    return 0;
}

编辑：

我已经更改了game_event_won函数。据我所知，它简单得多，没有bug，即使它更丑。可能会让它变短。

代码语言：javascript
复制
 void game_event_won(struct game_data* game)
    {

      if( ((game->grid[0][0] == game->grid[0][1]) && (game->grid[0][1] == game->grid[0][2]) && (game->grid[0][2] == 'X')) ||
          ((game->grid[1][0] == game->grid[1][1]) && (game->grid[1][1] == game->grid[1][2]) && (game->grid[1][2] == 'X')) ||
          ((game->grid[2][0] == game->grid[2][1]) && (game->grid[2][1] == game->grid[2][2]) && (game->grid[2][2] == 'X')) ||
          ((game->grid[0][0] == game->grid[1][0]) && (game->grid[1][0] == game->grid[2][0]) && (game->grid[2][0] == 'X')) ||
          ((game->grid[0][1] == game->grid[1][1]) && (game->grid[1][1] == game->grid[2][1]) && (game->grid[2][1] == 'X')) ||
          ((game->grid[0][2] == game->grid[1][2]) && (game->grid[1][2] == game->grid[2][2]) && (game->grid[2][2] == 'X')) ||
          ((game->grid[0][0] == game->grid[1][1]) && (game->grid[1][1] == game->grid[2][2]) && (game->grid[2][2] == 'X')) ||
          ((game->grid[0][2] == game->grid[1][1]) && (game->grid[1][1] == game->grid[2][0]) && (game->grid[2][0] == 'X')) ){
             game->win = 1;
             printf("You have WON\n");
       }

       if( ((game->grid[0][0] == game->grid[0][1]) && (game->grid[0][1] == game->grid[0][2]) && (game->grid[0][2] == 'O')) ||
           ((game->grid[1][0] == game->grid[1][1]) && (game->grid[1][1] == game->grid[1][2]) && (game->grid[1][2] == 'O')) ||
           ((game->grid[2][0] == game->grid[2][1]) && (game->grid[2][1] == game->grid[2][2]) && (game->grid[2][2] == 'O')) ||
           ((game->grid[0][0] == game->grid[1][0]) && (game->grid[1][0] == game->grid[2][0]) && (game->grid[2][0] == 'O')) ||
           ((game->grid[0][1] == game->grid[1][1]) && (game->grid[1][1] == game->grid[2][1]) && (game->grid[2][1] == 'O')) ||
           ((game->grid[0][2] == game->grid[1][2]) && (game->grid[1][2] == game->grid[2][2]) && (game->grid[2][2] == 'O')) ||
           ((game->grid[0][0] == game->grid[1][1]) && (game->grid[1][1] == game->grid[2][2]) && (game->grid[2][2] == 'O')) ||
           ((game->grid[0][2] == game->grid[1][1]) && (game->grid[1][1] == game->grid[2][0]) && (game->grid[2][0] == 'O')) ){
             game->win = 1;
             printf("You have LOST\n");
       }

  }

我还更改了主循环以适应它，防止错误绘制：

代码语言：javascript
复制
    int main(void)
    {

        srand((unsigned)time(0));

        intro();

        while (game.win == 0 ) {
        if (game.turn == 0) {
            player_one_move(&game);
            game.turns++;
            game.turn = 1;
        }

        else {
            game.turn = 0;
            computer_move(&game);
            game.turns++;
        }

        update(&game);

        game_event_won(&game);

        if (game.turns == 10 && game.win == 0){
            game.win = 1;
            printf("You have DRAWN\n");
            break;
        }


    }



    return 0;
}
beginner
c
game
tic-tac-toe
EN

回答 1

Code Review用户

回答已采纳

发布于 2019-02-16 19:52:13

输入验证

用户输入是邪恶的--不管它是邪恶的还是错误的。花费一些精力来验证它。

代码语言：javascript
复制
printf("You are 'Crosses'. Please input co-ordinates ...\n");
// scanf(" %d %d", &y_val, &x_val);
if (scanf(" %d %d", &y_val, &x_val) != 2) {
  puts("Invalid input");
  exit(EXIT_FAILURE);
}
if (y_val < 1 || y_val > 3 || x_val < 1 || x_val > 3) {
  puts("Out of range input");
  exit(EXIT_FAILURE);
}

更好的代码将有选择地允许重新输入。

删除//BUGGY注释

在这里张贴工作代码。对于您在解决问题时遇到的错误，请考虑堆栈溢出。

使用const

不改变其指向数据的函数使用const编写得更好。这允许更清晰的功能界面，更广泛的应用，更多的编译器时间检查和潜在的更好的优化。

代码语言：javascript
复制
// void update(struct game_data* game)
void update(const struct game_data* game)

重用代码

而不是使用if( ((game->grid[0][0] == game->grid[0][1]) ...使用'X'的8行代码，然后再用'O'生成8行类似的代码，形成一个辅助函数。

代码语言：javascript
复制
 char three_in_row(const char g[][3]) {
   if(g[0][0] != ' ' && g[0][0] == g[0][1] && g[0][1] == g[0][2]) return g[0][0];
   if(g[1][0] != ' ' && g[1][0] == g[1][1] && g[1][1] == g[1][2]) return g[1][0];
   if(g[2][0] != ' ' && g[2][0] == g[2][1] && g[2][1] == g[2][2]) return g[2][0];
   if(g[0][0] != ' ' && g[0][0] == g[1][0] && g[1][0] == g[2][0]) return g[0][0];
   if(g[0][1] != ' ' && g[0][1] == g[1][1] && g[1][1] == g[2][1]) return g[0][1];
   if(g[0][2] != ' ' && g[0][2] == g[1][2] && g[1][2] == g[2][2]) return g[0][2];
   if(g[0][0] != ' ' && g[0][0] == g[1][1] && g[1][1] == g[2][2]) return g[0][0];
   if(g[0][2] != ' ' && g[0][2] == g[1][1] && g[1][1] == g[2][0]) return g[0][2];
   return ' ';
 }

void game_event_won(struct game_data* game) {
  switch (three_in_row(game->grid)) {
    case 'X': game->win = 1; printf("You have WON\n"); break;
    case 'O': game->win = 1; printf("You have LOST\n"); break;
  }
}

状态评估

而不是game_event_won()，考虑一个从-100到100的TTT_rate(game)函数。

代码语言：javascript
复制
0: tie
100: X won
-100: O won
1...99: X winning
-1...-99: O winning

不需要递归

在computer_move(struct game_data* game)中使用循环

注释和代码doe非jibe

看起来范围是1.10。

代码语言：javascript
复制
int turns; // Ranges from 1 to 9(game end).
if (game.turns == 10 && game.win == 0){

格式

我希望while块会缩进。

提示:使用自动格式化程序。手动格式化是无效的。

代码语言：javascript
复制
    while (game.win == 0 ) {
    // if (game.turn == 0) {
      if (game.turn == 0) {
      ...

不需要全局数据

将struct game_data game从全局移动到main()内部。

为什么用户总是首先使用？

考虑允许用户或计算机先行。

高级

为X和O移动调用一个函数。函数对可以来自user_move()、computer_move()、smarter_computer_move1()、smarter_computer_move2()等，这将允许2播放，计算机算法1与算法2，用户与不同级别的游戏，等等。

票数 1
EN
页面原文内容由Code Review提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：

https://codereview.stackexchange.com/questions/213547

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