在 Java 中传递一个 HashMap 作为参数
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Pass a HashMap as parameter in Java
提问by Fosh
I have the main class Librarywhere I create:
我有Library我创建的主类：
HashMap<String, HashSet<String>> students_books = new HashMap<String, HashSet<String>>();
Then I'll have class Studentwhere I'll make a constructor who will take the HashMap as a parameter like this:
然后我将有一个类Student，我将在其中创建一个将 HashMap 作为参数的构造函数，如下所示：
public class Student {
private Student(HashMap students_books){
then, back in my main class (Library) I create a studentobject where I want to put the HashMap as parameter:
然后，回到我的主类（库）中，我创建了一个student对象，我想在其中将 HashMap 作为参数：
Student student = new Student(*HashMap as parameter*);
What I'm not finding is how I can do this and how the Studentclass knows what type of HashMap I'm passing, for example, <String, HashSet<String>>
我没有找到的是我如何做到这一点以及Student班级如何知道我正在传递的 HashMap 类型，例如，<String, HashSet<String>>
采纳答案by Phi Luu
knows what type of HashMap I'm passing
知道我传递的是什么类型的 HashMap
Firstly, your method is not a constructor since it has return type, remove its return type and public your constructor. Then just force them to pass what type of HashMap you want by doing this
首先，您的方法不是构造函数，因为它具有返回类型，删除其返回类型并公开您的构造函数。然后只需通过执行此操作强制他们传递您想要的 HashMap 类型
public class Student {
public Student(HashMap<String, HashSet<String>> students_books){
and then passing them like this
然后像这样传递它们
HashMap<String, HashSet<String>> students_books = new HashMap<String, HashSet<String>>();
Student student = new Student(students_books);
回答by rpfun12
In your Student class constructor(which currently a method, since it has a return type), the argument is not utilizing generic type.
在您的 Student 类构造函数（当前是一个方法，因为它具有返回类型）中，该参数未使用泛型类型。
Please change it to the following.
请将其更改为以下内容。
public StudentHashMap(HashMap<String, HashSet<String>> students_books){
}
This will ensure type safetey during compile time
这将确保编译时的类型安全
回答by Vinod Krishnan
To answer your question - "How to pass HashMap as a parameter" and how Student class know the type I am providing a more generic and standard way of doing this
回答你的问题——“如何将 HashMap 作为参数传递”以及 Student 类如何知道类型我提供了一种更通用和标准的方法来做到这一点
Map<K,V> books = new HashMap<K,V>(); // K and V are Key and Value types
Student student = new Student(books); // pass the map to the constructor
..
//Student Constructor
public Student(Map<K,V> books){
..
}
回答by Dilshod Shaymanov
What you did is a private method, not a constructor. change your method to this:
你所做的是一个私有方法，而不是一个构造函数。将您的方法更改为：
public Student(HashMap<String, HashSet<String>> student_books) {
//your code here
}
I this way it will a true constructor you wanted, hope this will help
我这样它将成为您想要的真正构造函数，希望这会有所帮助
