Question

您好我不知道如何实际查找但我很好奇，您可以根据您在构造函数中提供的参数更改对象。就像你有一个对象库，它有一个构造函数public Library（Map books）之类的东西。如果我创建Map ... new Hashmap，在其中放入5个键值，然后将该映射提供给Library的构造函数，则Library对象将具有一个包含5个键值的hashmap。如果我在hashmap中再插入2个键值会发生什么？构建的Library对象突然在其地图中有7个键值，还是只有最初的5个？

Answer 1

这取决于构造函数的作用。如果构造函数只将Map存储在实例变量中：

public class Library {
    private final Map<String, String> books;

    public Library(Map<String, String> books) {
        this.books = books;
    }

    public int size() {
        return books.size();
    }

    public static void main(String[] args) {
        Map<String, String> myBooks  = new HashMap<String,String>();
        myBooks.put("Jonathan Swift", "Gullivers Travels");
        myBooks.put("Robert Louis Stevenson", "Treasure Island");
        myBooks.put("William Shakespeare", "Hamlet");
        myBooks.put("J R R Tolkein", "The Hobbit");
        myBooks.put("Mary Shelley", "Frankenstein");
        Library library = new Library(myBooks);
        myBooks.put("J K Rowling", "Harry Potter and the Philosophers stone");
        System.out.println("Library contains" + library.size() + " books");
    }
}

然后，Library包含对传递给它的地图的引用，它将看到对该地图内容的任何更改。上面的代码应该打印出库中有6本书。

另一方面，如果构造函数制作了地图的副本：

public class Library {
    private final Map<String, String> books;

    public Library(Map<String, String> books) {
        this.books = new HashMap<String, String>(books);
    }

    public int size() {
        return books.size();
    }

    public static void main(String[] args) {
        Map<String, String> myBooks  = new HashMap<String,String>();
        myBooks.put("Jonathan Swift", "Gullivers Travels");
        myBooks.put("Robert Louis Stevenson", "Treasure Island");
        myBooks.put("William Shakespeare", "Hamlet");
        myBooks.put("J R R Tolkein", "The Hobbit");
        myBooks.put("Mary Shelley", "Frankenstein");
        Library library = new Library(myBooks);
        myBooks.put("J K Rowling", "Harry Potter and the Philosophers stone");
        System.out.println("Library contains" + library.size() + " books");
    }
}

然后Library有自己的地图副本，并且不会看到对原始地图的任何更改，因此上面的代码应该打印出库中有5本书。

Answer 2

是的，它会的。例如：

import java.util.Map;
import java.util.HashMap;

public class HelloWorld{

 public static void main(String []args){
    Map<String, String> books = new HashMap();
    books.put("Author1", "Book1");
    books.put("Author2", "Book1");
    Library l1 = new Library(books);
    books.put("Author3", "Book1");
    System.out.println(l1.toString());
 }
}

class Library {
private Map<String, String> books;

public Library(Map<String, String> books) {
    this.books = books;
}

@Override
public String toString() {
    return books.toString();
}

}

输出将是： {Author3 = Book1，Author2 = Book1，Author1 = Book1}

这是因为java不是将对象本身传递给构造函数，而是传递给该对象。这意味着从主要方法书籍地图和图书馆书籍地图实际上是一个对象。如果您不希望Library类从Library类外部更改，则可以根据构造函数输入创建新对象。说谎：

 public Library(Map<String, String> books) {
    this.books = new HashMap(books);
}

更改构造函数参数运行时

2 个答案: