Question

嘿我正在完成这个tic tac toe项目，我的checkWin方法在我的棋盘类中有一个错误，其中winner = board [0] [i];出现为Int和String的不兼容错误。我已经通过使用Integer.toString（）命令在我的其他电路板上修复了这个问题，但它不适用于此。有任何想法吗？这是checkWin方法的代码。

public boolean checkWin()
{
    {
        int i;  // i = column
        int j; // j = row
        int count; 
        int winner; 

          winner = empty; // nobody has won yet

// Check all rows to see if same player has occupied every square.

        for (j = 0; j < boardSize; j ++)
{
        count = 0;
    if (board[j][0] != Integer.toString(empty))

    for (i = 0; i < boardSize; i ++)
    if (board[j][0] == board[j][i])
    count ++;
    if (count == boardSize)
    winner = (board[j][0]);
}

// Check all columns to see if same player has occupied every square.

    for (i = 0; i < boardSize; i ++)
{
    count = 0;
    if (board[0][i] != Integer.toString(empty))
    for (j = 0; j < boardSize; j ++)
    if (board[0][i] == board[j][i])
    count ++;
    if (count == boardSize)
    winner = board[0][i];
}

// Check diagonal from top-left to bottom-right.

    count = 0;
    if (board[0][0] != Integer.toString(empty))
    for (j = 0; j < boardSize; j ++)
    if (board[0][0] == board[j][j])
    count ++;
if (count == boardSize)
winner = board[0][0];

// Check diagonal from top-right to bottom-left.

count = 0;
if (board[0][boardSize-1] != Integer.toString(empty))
for (j = 0; j < boardSize; j ++)
if (board[0][boardSize-1] == board[j][boardSize-j-1])
count ++;
if (count == boardSize)
winner = board[0][boardSize-1];

// Did we find a winner?

if (winner != empty)
{
if (winner == Xstr)

System.out.println("\nCongratulations! P1 You win!");
else if (winner == Ostr)

System.out.println("\nCongratulations! P2 You win!");
else


return true; 
}



}

Answer 1

winner = board[0][i];

winner是一个int基本类型，board是多维字符串数组。

您正在尝试为int分配字符串，因此Int和String 的不兼容错误。

String[][] board;

在其索引处有字符串，当您尝试访问board[0][i]时，您正在检索字符串。

如果你的电路板数组包含数字的字符串表示，如

boards= {{"1"},{"2"}};

然后使用Integer.parseInt（），它将String作为参数并返回一个Integer。

winner = Integer.parseInt(board[0][i]);

但请注意，如果传递给parseInt的String不是字符串的有效整数表示，则它会抛出 NumberFormatException 。

类型Int和字符串不兼容

1 个答案: