C - 在 Connect Four 游戏中验证获胜条件

Question

这是一个连环四游戏原型的验证，但我好像做错了什么。我希望每次玩家移动时，该函数都会通过垂直、水平和最终对角线验证来验证他是否赢了。但似乎验证不正确，因为在某些情况下，即使只有 2 步，功能 returns 1.

    int verifyGame(int gamePosition, int gameVariable, char gameArray[HEIGTH][WIDTH])
{

  if(gameArray[gamePosition][gameVariable] == gameArray[gamePosition + 1][gameVariable] == gameArray[gamePosition + 2][gameVariable] == gameArray[gamePosition + 3][gameVariable]) //verify vertically
        return 1;
      else
        if(gameArray[gamePosition][gameVariable] == gameArray[gamePosition][gameVariable - 3] == gameArray[gamePosition][gameVariable - 2] == gameArray[gamePosition][gameVariable - 1]) //verify horizontally
          return 1;
      else
         if(gameArray[gamePosition][gameVariable] == gameArray[gamePosition][gameVariable - 2] == gameArray[gamePosition][gameVariable - 1] == gameArray[gamePosition][gameVariable + 1])
          return 1;
      else
         if(gameArray[gamePosition][gameVariable] == gameArray[gamePosition][gameVariable - 1] == gameArray[gamePosition][gameVariable + 1] == gameArray[gamePosition][gameVariable + 2])
          return 1;
      else
         if(gameArray[gamePosition][gameVariable] == gameArray[gamePosition][gameVariable + 1] == gameArray[gamePosition][gameVariable+ 2] == gameArray[gamePosition][gameVariable + 3])
          return 1;
      //verify diagonally


      else return 0;
};

这是调用函数的地方。 switch 验证用户输入，然后将值放入矩阵，然后验证 won

    printf("playerPick is : %d\n", playerPick);
  fflush(stdout);
  switch(playerPick)
  {
    case 1:
     if(gameVariables[0] >0 && gameVariables[0] < 7)
      {
        --gameVariables[0];
        gameArray[gameVariables[0]][0] = (char) 82;
       ifWon = verifyGame(gameVariables[0], 0, gameArray);
      }
      printArray(gameArray);
      break;
    case 2:

      if(gameVariables[1] >0 && gameVariables[1] < 7)
      {
        --gameVariables[1];
        gameArray[gameVariables[1]][1] = (char) 82;
        ifWon = verifyGame(gameVariables[1], 1, gameArray);
      }
      printArray(gameArray);
      break;
    case 3:
      if(gameVariables[2] >0 && gameVariables[2] < 7)
        {
          --gameVariables[2];
          gameArray[gameVariables[2]][2] = (char) 82;
          ifWon = verifyGame(gameVariables[2], 2, gameArray);
        }
      printArray(gameArray);
      break;
    case 4:
      if(gameVariables[3] >0 && gameVariables[3] < 7)
        {
          --gameVariables[3];
          gameArray[gameVariables[3]][3] = (char) 82;
          ifWon = verifyGame(gameVariables[3], 3, gameArray);
        }
      printArray(gameArray);
      break;
    case 5:
      if(gameVariables[4] >0 && gameVariables[4] < 7)
      {
        --gameVariables[4];
        gameArray[gameVariables[4]][4] = (char) 82;
        ifWon = verifyGame(gameVariables[4], 4, gameArray);
      }
      printArray(gameArray);
      break;
    case 6:
      if(gameVariables[5] >0 && gameVariables[5] < 7)
      {
        --gameVariables[5];
        gameArray[gameVariables[5]][5] = (char) 82;
        ifWon = verifyGame(gameVariables[5], 5, gameArray);
      }
      printArray(gameArray);
      break;
    case 7:
      if(gameVariables[6] >0 && gameVariables[6] < 7)
      {
        --gameVariables[6];
        gameArray[gameVariables[6]][6] = (char) 82;
        ifWon = verifyGame(gameVariables[6], 6, gameArray);
      }
      printArray(gameArray);
      break;

  }
  printf("%d %d %d %d %d %d %d\n", gameVariables[0], gameVariables[1], gameVariables[2], gameVariables[3], gameVariables[4], gameVariables[5], gameVariables[6]);
  printf("ifwon :  %d\n", ifWon);

Answer 1

您不能在尝试时链接相等性测试。代码将执行，但不会像您想象的那样执行。您的代码

if(gameArray[gamePosition][gameVariable] == 
   gameArray[gamePosition + 1][gameVariable] == 
   gameArray[gamePosition + 2][gameVariable] == 
   gameArray[gamePosition + 3][gameVariable])

必须拆分成单独的测试，例如：

if(gameArray[gamePosition][gameVariable] == gameArray[gamePosition + 1][gameVariable] &&
   gameArray[gamePosition][gameVariable] == gameArray[gamePosition + 2][gameVariable] &&
   gameArray[gamePosition][gameVariable] == gameArray[gamePosition + 3][gameVariable])

其他线路也是如此。

Answer 2

@Weather Vane 的回答是正确的。您原来的 post 中使用的逻辑不正确，无法验证。

一个原因你自己可能没有抓住它可能是它写的复杂的方式。尝试简化用户输入验证码：（只需检查用户输入值的范围即可。）

//User input range checking:
if((gamePosition >= x)&&    //where `x` is minimum for gamePosition
   (gamePosition <= y)&&    //where `y` is maximum for gamePosition
   (gameVariable >= z)&&    //where `z` is minimum for gameVariable
   (gameVariable <= w))     //where `w` is maximum for gameVariable
{//continue }
else 
{
    printf("Invalid value.  Please re-enter");
    return -1;
}

另一个简化的机会 是注意每个 case 语句都包含相同的代码，除了案件。因此，整个 switch(...){...} 可以替换为单个 if 语句：

//assuming playerPick >= 1
  if(gameVariables[playerPick-1] >0 && gameVariables[playerPick-1] < 7)
  {
    --gameVariables[playerPick-1];
    gameArray[gameVariables[playerPick-1]][playerPick-1] = (char) 82;
    ifWon = verifyGame(gameVariables[playerPick-1], playerPick-1, gameArray);
  }
  printArray(gameArray);

还要注意虽然声明：

gameArray[gameVariables[0][0] = (char) 82; //what is 82?

是完全合法的，变量 gameArray[0][0] 只是一个字符，因此不需要转换值 82。此外，C 语法提供了一种方法来提取字符的 ASCII 十进制值，方法是用坟墓符号包围它，允许以下形式，更具可读性：

gameArray[gameVariables[0]][0] = `R`;       //intuitive

C - 在 Connect Four 游戏中验证获胜条件

C - Verify win condition in Connect Four game

c

verify