为什么我必须初始化变量？

Question

我的代码正在检查给定的字符串是否为某种格式。字符串的第一个字符必须是大写字母，字符串的其余部分必须是从 1 到给定维度的任意数字。

如果字符串的第一个字符包含来自字母数组的字符串，则代码会检查字符串的其余部分是否包含来自 numbers 数组的数字。要成为有效坐标，两个条件都必须为真，如果其中一个为假，则不是有效坐标。我想返回布尔值 isValidCoordinate 但 IntelliJ 告诉我我必须初始化 isValid 坐标。为什么我必须初始化它，布尔表达式取决于'if'条件。

谢谢。

public static boolean validCoordinate(String coordinate, int dimension) {
        boolean isValidCoordinate;
        String [] alphabet = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
        int [] numbers = new int [dimension];
        int one = 1;
        for(int i = 0; i < dimension; i++){
            numbers[i] = one + i;
        }
        for(int i = 0; i < dimension; i++){
            if((Character.toString(coordinate.charAt(0))).contains(alphabet[i])) {
                for(int j = 0; j < dimension; j++) {
                    if ((coordinate.substring(1)).contains(Integer.toString(numbers[j]))) {
                        isValidCoordinate = true;
                    }
                    else {
                        isValidCoordinate = false;
                    }
                }

            }
            else {
                isValidCoordinate = false;
            }

        }

        return isValidCoordinate;
    }

这是我的最终代码：

public static boolean validCoordinate(String coordinate, int dimension) {
        boolean isValidCoordinate;
        String [] alphabet = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
        int [] numbers = new int [dimension];
        int one = 1;
        for(int i = 0; i < dimension; i++){
            numbers[i] = one + i;
        }
        for(int i = 0; i < dimension; i++){
            if((Character.toString(coordinate.charAt(0))).contains(alphabet[i])) {
                for(int j = 0; j < dimension; j++) {
                    if ((coordinate.substring(1)).contains(Integer.toString(numbers[j]))) {
                        isValidCoordinate = true;
                    }
                    else {
                        isValidCoordinate = false;
                    }
                }

            }
            else {
                isValidCoordinate = false;
            }

        }

        return true;
    }

Answer 1

for 循环内的所有情况都被覆盖，但dimension 为 0 时的情况除外。

该方法将直接转到return isValidCoordinate;，其中变量尚未初始化。

Answer 2

可能永远不会进入您的 for 循环（例如，如果 dimension == 0）。在这种情况下，您永远不会为 isValidCoordinate 赋值，但您会尝试在方法的最后一条语句中返回该变量的值。

在这种情况下isValidCoordinate 的值是多少？

它没有任何价值。

因此，编译器强制您为isValidCoordinate 分配一个初始值，以确保它在被访问之前有一个值。

编辑：

根据您的 cmets，我建议您消除 boolean 变量，并改用 return 语句：

public static boolean validCoordinate(String coordinate, int dimension) {
    String [] alphabet = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
    int [] numbers = new int [dimension];
    int one = 1;
    for(int i = 0; i < dimension; i++){
        numbers[i] = one + i;
    }
    for(int i = 0; i < dimension; i++){
        if((Character.toString(coordinate.charAt(0))).contains(alphabet[i])) {
            for(int j = 0; j < dimension; j++) {
                if ((coordinate.substring(1)).contains(Integer.toString(numbers[j]))) {
                    return true;
                }
            }
        }
    }

    return false;
}

这样，当boolean 变量设置为true 时，您不必担心会跳出嵌套循环。

Answer 3

import java.util.regex.*;

public static boolean validCoordinate(String coordinate, int dimension)
{
    Pattern p = Pattern.compile("[A-Z]+");
    Matcher m = p.matcher(coordinate);
    return m.matches();
}