Java 2D数组对象不能解析为变量吗？
Java 2D数组对象不能解析为变量吗？
提问于 2018-09-24 02:30:16
我有一个名为game的类，在这个类中，我使用默认构造函数来创建和初始化来自另一个类的对象的2D数组。但是，在我用来操作2D数组中对象的变量的另一种方法中，我遇到了一个错误，指出"gameBoard (对象)无法解析为变量“。该错误仅在第二个方法的底部遇到。
public game() {
// Start initialize game board
grid[][] gameBoard = new grid[9][9];
gameBoard[0][0] = new grid(" ", "N", true);
gameBoard[0][1] = new grid("A", "N", true);
gameBoard[0][2] = new grid("B", "N", true);
gameBoard[0][3] = new grid("C", "N", true);
gameBoard[0][4] = new grid("D", "N", true);
gameBoard[0][5] = new grid("E", "N", true);
gameBoard[0][6] = new grid("F", "N", true);
gameBoard[0][7] = new grid("G", "N", true);
gameBoard[0][8] = new grid("H", "N", true);
gameBoard[1][0] = new grid("1", "N", true);
gameBoard[2][0] = new grid("2", "N", true);
gameBoard[3][0] = new grid("3", "N", true);
gameBoard[4][0] = new grid("4", "N", true);
gameBoard[5][0] = new grid("5", "N", true);
gameBoard[6][0] = new grid("6", "N", true);
gameBoard[7][0] = new grid("7", "N", true);
gameBoard[8][0] = new grid("8", "N", true);
for (int x = 1; x < 9; x++) {
for (int y = 1; y < 9; y++) {
gameBoard[x][y] = new grid();
}
}
// End
}
public void placeUserShips() {
// Start place user ships
for (int x = 1; x <= 6; x++) {
System.out.print("Enter the coordinates of your ship #" + x + ": ");
String cell = input_scanner.nextLine();
String letter = cell.substring(0, 1);
switch (letter) {
case "A":
letterCol = 1;
break;
case "B":
letterCol = 2;
break;
case "C":
letterCol = 3;
break;
case "D":
letterCol = 4;
break;
case "E":
letterCol = 5;
break;
case "F":
letterCol = 6;
break;
case "G":
letterCol = 7;
break;
case "H":
letterCol = 8;
break;
default:
letterCol = 0;
break;
}
number = Integer.parseInt(cell.substring(1, 2));
if (gameBoard[letterCol][number].type == "" && number >= 1 && number <= 8) {
gameBoard[letterCol][number].type = "s";
} else {
System.out.println("Sorry, those coordinates are already in use. Try again.");
x -= 1;
}
}
// End
}回答 1
Stack Overflow用户
回答已采纳
发布于 2018-09-24 03:39:43
正如问题Java, “Variable name” cannot be resolved to a variable中所述，当您引用一个未定义的变量(或超出范围)时，就会发生这种错误。
您可以在game()中构造gameBoard，然后将其返回给placeUserShips()
public grid[][] game() {
// Start initialize game board
grid[][] gameBoard = new grid[9][9];
// ...
return gameBoard;
}
public void placeUserShips() {
// Start place user ships
grid[][] gameBoard = game();
// ...
}或者使用全局变量(在main中，您应该首先调用game )：
private grid[][] gameBoard;
public void game() {
// Start initialize game board
gameBoard = new grid[9][9];
// ...
}
public void placeUserShips() {
// Start place user ships
// ...
}页面原文内容由Stack Overflow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://stackoverflow.com/questions/52469090
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