我试图打印出一个列表的所有元素,但它是打印对象的指针,而不是值。

这是我的打印代码…

for(int i=0;i<list.size();i++){
    System.out.println(list.get(i));
} 

有人能告诉我为什么它不打印元素的值吗?


当前回答

我写了一个转储函数,它基本上打印出对象的公共成员,如果它没有覆盖toString()。可以很容易地将其扩展为调用getter。 Javadoc:

将给定对象转储到系统。Out,使用以下规则: 如果对象是Iterable,则它的所有组件都将被转储。 如果Object或它的一个超类覆盖了toString(),则“toString”将被转储 否则,该方法将为对象的所有公共成员递归调用

/**
 * Dumps an given Object to System.out, using the following rules:<br>
 * <ul>
 * <li> If the Object is {@link Iterable}, all of its components are dumped.</li>
 * <li> If the Object or one of its superclasses overrides {@link #toString()}, the "toString" is dumped</li>
 * <li> Else the method is called recursively for all public members of the Object </li>
 * </ul>
 * @param input
 * @throws Exception
 */
public static void dump(Object input) throws Exception{
    dump(input, 0);
}

private static void dump(Object input, int depth) throws Exception{
    if(input==null){
        System.out.print("null\n"+indent(depth));
        return;
    }

    Class<? extends Object> clazz = input.getClass();
    System.out.print(clazz.getSimpleName()+" ");
    if(input instanceof Iterable<?>){
        for(Object o: ((Iterable<?>)input)){
            System.out.print("\n"+indent(depth+1));
            dump(o, depth+1);
        }
    }else if(clazz.getMethod("toString").getDeclaringClass().equals(Object.class)){
        Field[] fields = clazz.getFields();
        if(fields.length == 0){
            System.out.print(input+"\n"+indent(depth));
        }
        System.out.print("\n"+indent(depth+1));
        for(Field field: fields){
            Object o = field.get(input);
            String s = "|- "+field.getName()+": ";
            System.out.print(s);
            dump(o, depth+1);
        }
    }else{

        System.out.print(input+"\n"+indent(depth));
    }
}

private static String indent(int depth) {
    StringBuilder sb = new StringBuilder();
    for(int i=0; i<depth; i++)
        sb.append("  ");
    return sb.toString();
}

其他回答

java 11问题的解决方案是:

String separator = ", ";
String toPrint = list.stream().map(o -> String.valueOf(o)).collect(Collectors.joining(separator));

System.out.println(toPrint);
   List<String> textList=  messageList.stream()
                            .map(Message::getText)
                            .collect(Collectors.toList());

        textList.stream().forEach(System.out::println);
        public class Message  {

        String name;
        String text;

        public Message(String name, String text) {
            this.name = name;
            this.text = text;
        }

        public String getName() {
            return name;
        }

      public String getText() {
        return text;
     }
   }
System.out.println(list);//toString() is easy and good enough for debugging.

AbstractCollection的toString()将非常干净和容易做到这一点。AbstractList是AbstractCollection的子类,因此不需要for循环,也不需要toArray()。

返回此集合的字符串表示形式。字符串表示形式由集合元素的列表组成 它们由迭代器返回的顺序,用方括号括起来 (“[]”)。相邻元素用字符“,”(逗号)分隔 和空间)。通过将元素转换为字符串 String.valueOf(对象)。

如果您正在使用列表中的任何自定义对象,例如Student,则需要重写它的toString()方法(重写这个方法总是好的)以获得有意义的输出

请看下面的例子:

public class TestPrintElements {

    public static void main(String[] args) {

        //Element is String, Integer,or other primitive type
        List<String> sList = new ArrayList<String>();
        sList.add("string1");
        sList.add("string2");
        System.out.println(sList);

        //Element is custom type
        Student st1=new Student(15,"Tom");
        Student st2=new Student(16,"Kate");
        List<Student> stList=new ArrayList<Student>();
        stList.add(st1);
        stList.add(st2);
        System.out.println(stList);
   }
}


public  class Student{
    private int age;
    private String name;

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

    @Override
    public String toString(){
        return "student "+name+", age:" +age;
    }
}

输出:

[string1, string2]
[student Tom age:15, student Kate age:16]

使用String.join () 例如:

System.out.print(String.join("\n", list));
    list.stream().map(x -> x.getName()).forEach(System.out::println);