Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
664 views
in Technique[技术] by (71.8m points)

java - 反序列化ArrayList:[unchecked]需要未经检查的转换:ArrayList <Employee> 找到:ArrayList(Deserialize ArrayList: [unchecked] unchecked conversion required: ArrayList<Employee> found: ArrayList)

I am trying out serializing and deserializing a java application.

(我正在尝试序列化和反序列化Java应用程序。)

It's working but I would like to know if I am doing this process correctly.

(它正在工作,但是我想知道我是否正确地执行了此过程。)

I am serializing the ArrayList;

(我正在序列化ArrayList;)

is this the proper way or should I be serializing the Employee class, but how would I go about that considering we could have many employees?

(这是正确的方法还是应该序列化Employee类,但是考虑到我们可能有很多员工,我将如何处理呢?)

The way I am doing it causes this error which I would like to get rid of:

(我这样做的方式导致此错误,我想摆脱它:)

serialTest.java:40: warning: [unchecked] unchecked conversion employees = (ArrayList) ois.readObject();

(serialTest.java:40:警告:[unchecked]未经检查的转换雇员=(ArrayList)ois.readObject();)

required: ArrayList found: ArrayList 1 warning

(必需:找到ArrayList:ArrayList 1警告)

Here is the Employee class: package serialTest;

(这是Employee类:package serialTest;)

import java.io.Serializable;

(导入java.io.Serializable;)

public class Employee implements Serializable {

    int id;
    String firstName;
    String lastName;

    public Employee(int id, String firstName, String lastName) {
        super();
        this.id = id;
        this.firstName = firstName;
        this.lastName = lastName;
    }

    @Override
    public String toString() {
        return "Employee [id=" + id + ", firstName=" + firstName + ", lastName=" + lastName + "]";
    }
}

And here is the main class:

(这是主要的类:)

package serialTest;

import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.util.ArrayList;

public class serialTest {

    static ArrayList<Employee> employees = new ArrayList<>();

    public static void main(String[] args) {
        if (args.length > 0) {
            deSerialize();
        } else {
            employees.add(new Employee(1, "John", "Doe"));
            employees.add(new Employee(2, "Jane", "Doe"));
            serialize();
        }
    }

    private static void serialize() {
        System.out.println("Serializing...");
        try {
            try (FileOutputStream fos = new FileOutputStream("employeeData"); ObjectOutputStream oos = new ObjectOutputStream(fos)) {
                oos.writeObject(employees);
            }
        } catch (IOException ioe) {
        }
    }

    private static void deSerialize() {
        System.out.println("DeSerializing...");
        try {
            try (FileInputStream fis = new FileInputStream("employeeData");
                    ObjectInputStream ois = new ObjectInputStream(fis)) {

                employees = (ArrayList) ois.readObject();

            }
        } catch (IOException ioe) {
            System.out.println("File problems");
            return;
        } catch (ClassNotFoundException c) {
            System.out.println("Class problems");
            return;
        }

        for (Employee info : employees) {
            System.out.println(info);
        }
    }
}
  ask by James Mead translate from so

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

You will have to suppress that warning.

(您将不得不禁止该警告。)

Java doesn't know the actual generic type due to type erasure.

(由于类型擦除,Java不知道实际的泛型类型。)


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...