在2D数组中找到元素的坐标

I think I must be losing it. I'm trying to locate the coordinates of an element in a 2D Array.

我想我一定是要失去它。我正在尝试在2D数组中找到元素的坐标。

I've dumbed the code down to as simple as I could and still can't get it right. I'm very new to Java

我把代码简化为尽可能简单，但仍然无法正确完成。我是Java的新手

Please tell me why the answer to this is always 42.0, no matter where I put the ' * '

请告诉我为什么答案总是42.0，无论我把'*'放在哪里

public static void main(String[] args) {
    locateStar(board);
}

static char[][] board = {
    { '.', '.', '.', '.' },
    { '.', '.', '.', '.' },
    { '.', '.', '.', '*' },
    { '.', '.', '.', '.' }
};

public static void locateStar(char[][] board) {
    double star = 0;
    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            if (board[i][j] == '*') {
                star = board[i][j];
            }
        }
    }
    System.out.println(star);
}

3 个解决方案

#1

Look closely at this line:

仔细看看这一行：

star = board[i][j];

You are assigning a char to a double. The value 42 is the ASCII value of an asterisk *. If you want to print out the coordinates (which are two values, not just one), then try this:

您正在为double分配char。值42是星号*的ASCII值。如果要打印出坐标（两个值，而不是一个），请尝试以下方法：

public static void locateStar(char[][] board) {
    int x, y;
    for (int i=0; i < board.length; i++) {
        for (int j=0; j < board[0].length; j++) {
            if (board[i][j] == '*') {
                x = i;
                y = j;
            }
        }
    }
    System.out.println("Found a star at (" + x + ", " + y + ")");
}

#2

Ahah it's kinda funny :) You find the star, and you assign the star to a double value:

啊，它有点好笑:)你找到了星星，你将星星指定为双倍值：

star = board[i][j];

That is, you assign '*' to a double, getting the ASCII value of the * character, which is - in fact - 42.

也就是说，你将'*'分配给double，得到*字符的ASCII值，实际上是-42。

Here's some code that shows that it's found:

这里有一些代码显示它被发现：

for (int i = 0; i < board.length; i++) {
    for (int j = 0; j < board[0].length; j++) {
        if (board[i][j] == '*') {
            System.out.println("Found at " + i + " " + j);
            break;
        }
    }
}

#3

The ASCII value 42 corresponds to the * symbol. Your code is in effect retrieving the value * and then casting it implicitly to a number, which will always be 42. You are not going to see a difference regardless of position, because you are only looking for the '*' value and not its position or anything else.

ASCII值42对应于*符号。你的代码实际上是检索值*然后隐式地将它强制转换为一个数字，它总是为42.你不会看到任何位置的差异，因为你只是寻找'*'值，而不是它的位置或其他任何东西。

#1