java - 对于不返回内容的方法是否强制使用 void ？

Question

我在某处看到了这段代码:

public class MyDatabaseHelper extends SQLiteOpenHelper {

    private static final String DATABASE_NAME = "DBName";

    private static final int DATABASE_VERSION = 2;

    // Database creation sql statement
    private static final String DATABASE_CREATE = "create table MyEmployees
                                 ( _id integer primary key,name text not null);";

    public MyDatabaseHelper(Context context) {
        super(context, DATABASE_NAME, null, DATABASE_VERSION);
    }

    // Method is called during creation of the database
    @Override
    public void onCreate(SQLiteDatabase database) {
        database.execSQL(DATABASE_CREATE);
    }

    // Method is called during an upgrade of the database,
    @Override
    public void onUpgrade(SQLiteDatabase database,int oldVersion,int newVersion){
        Log.w(MyDatabaseHelper.class.getName(),
                         "Upgrading database from version " + oldVersion + " to "
                         + newVersion + ", which will destroy all old data");
        database.execSQL("DROP TABLE IF EXISTS MyEmployees");
        onCreate(database);
    }
}

如您所见，MyDatabaseHelper 方法没有返回任何内容(不返回任何内容)。那么为什么作者不使用 void 呢？

Answer 1

它称为构造函数，用于在创建时初始化对象(在本例中，您希望使用 Context 初始化数据库助手)。请参阅https://docs.oracle.com/javase/tutorial/java/javaOO/constructors.html

帮助您理解构造函数的更好示例:

public class Person {

    public String name;
    public int age;

    public Person(String name, int age) {
        this.name = name;
        this.age = age;
    }

    @Override
    public String toString() {
        return String.format("My name is %s and I am %d years old", name, age);
    }
}


public class Main {
    public static void main(String[] args) {
        Person me = new Person("Niels", 20);
        System.out.println(me);
    }
}

当您执行上述代码时，您将在控制台中看到一条消息:我的名字是 Niels，我今年 20 岁。

如您所见，我们创建了一个 Person 对象，该对象采用 name 和 age 参数。