如何在Java中打印出List的所有元素？

Question

I am trying to print out all the elements of a List , however it is printing the pointer of the Object rather than the value.

This is my printing code...

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

Could anyone please help me why it isn't printing the value of the elements.

Answer 1

The following is compact and avoids the loop in your example code (and gives you nice commas):

System.out.println(Arrays.toString(list.toArray()));

However, as others have pointed out, if you don't have sensible toString() methods implemented for the objects inside the list, you will get the object pointers (hash codes, in fact) you're observing. This is true whether they're in a list or not.

Answer 2

Here is some example about getting print out the list component:

public class ListExample {

    public static void main(String[] args) {
        List<Model> models = new ArrayList<>();

        // TODO: First create your model and add to models ArrayList, to prevent NullPointerException for trying this example

        // Print the name from the list....
        for(Model model : models) {
            System.out.println(model.getName());
        }

        // Or like this...
        for(int i = 0; i < models.size(); i++) {
            System.out.println(models.get(i).getName());
        }
    }
}

class Model {

    private String name;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

Answer 3

从 Java 8 开始，List 继承了默认的“forEach”方法，您可以将其与方法引用“System.out::println”结合，如下所示：

list.forEach(System.out::println);

Answer 4

System.out.println(list);//toString() is easy and good enough for debugging.

toString() of AbstractCollection will be clean and easy enough to do that . AbstractList is a subclass of AbstractCollection , so no need to for loop and no toArray() needed.

Returns a string representation of this collection. The string representation consists of a list of the collection's elements in the order they are returned by its iterator, enclosed in square brackets ("[]"). Adjacent elements are separated by the characters ", " (comma and space). Elements are converted to strings as by String.valueOf(Object).

If you are using any custom object in your list, say Student , you need to override its toString() method(it is always good to override this method) to have a meaningful output

See the below example:

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;
    }
}

output:

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

Answer 5

使用String.join()例如：

System.out.print(String.join("\n", list));

Answer 6

Java 8 Streams 方法...

list.stream().forEach(System.out::println);

Answer 7

The objects in the list must have toString implemented for them to print something meaningful to screen.

Here's a quick test to see the differences:

public class Test {

    public class T1 {
        public Integer x;
    }

    public class T2 {
        public Integer x;

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

    public void run() {
        T1 t1 = new T1();
        t1.x = 5;
        System.out.println(t1);

        T2 t2 = new T2();
        t2.x = 5;
        System.out.println(t2);

    }

    public static void main(String[] args) {        
        new Test().run();
    }
}

And when this executes, the results printed to screen are:

t1 = Test$T1@19821f
t2 = 5

Since T1 does not override the toString method, its instance t1 prints out as something that isn't very useful. On the other hand, T2 overrides toString, so we control what it prints when it is used in I/O, and we see something a little better on screen.

Answer 8

Or you could simply use the Apache Commons utilities:

https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/ArrayUtils.html#toString-java.lang.Object-

List<MyObject> myObjects = ...
System.out.println(ArrayUtils.toString(myObjects));

Answer 9

Consider a List<String> stringList which can be printed in many ways using Java 8 constructs:

stringList.forEach(System.out::println);                            // 1) Iterable.forEach
stringList.stream().forEach(System.out::println);                   // 2) Stream.forEach (order maintained generally but doc does not guarantee)
stringList.stream().forEachOrdered(System.out::println);            // 3) Stream.forEachOrdered (order maintained always)
stringList.parallelStream().forEach(System.out::println);           // 4) Parallel version of Stream.forEach (order not maintained)
stringList.parallelStream().forEachOrdered(System.out::println);    // 5) Parallel version ofStream.forEachOrdered (order maintained always)

How are these approaches different from each other?

First Approach ( Iterable.forEach )- The iterator of the collection is generally used and that is designed to be fail-fast which means it will throw ConcurrentModificationException if the underlying collection is structurally modified during the iteration. As mentioned in the doc for ArrayList :

A structural modification is any operation that adds or deletes one or more elements, or explicitly resizes the backing array; merely setting the value of an element is not a structural modification.

So it means for ArrayList.forEach setting the value is allowed without any issue. And in case of concurrent collection eg ConcurrentLinkedQueue the iterator would be weakly-consistent which means the actions passed in forEach are allowed to make even structural changes without ConcurrentModificationException exception being thrown. But here the modifications might or might not be visible in that iteration.

Second Approach ( Stream.forEach )- The order is undefined. Though it may not occur for sequential streams but the specification does not guarantee it. Also the action is required to be non-interfering in nature. As mentioned in doc :

The behavior of this operation is explicitly nondeterministic. For parallel stream pipelines, this operation does not guarantee to respect the encounter order of the stream, as doing so would sacrifice the benefit of parallelism.

Third Approach ( Stream.forEachOrdered )- The action would be performed in the encounter order of the stream. So whenever order matters use forEachOrdered without a second thought. As mentioned in the doc :

Performs an action for each element of this stream, in the encounter order of the stream if the stream has a defined encounter order.

While iterating over a synchronized collection the first approach would take the collection's lock once and would hold it across all the calls to action method, but in case of streams they use collection's spliterator, which does not lock and relies on the already established rules of non-interference. In case collection backing the stream is modified during iteration a ConcurrentModificationException would be thrown or inconsistent result may occur.

Fourth Approach (Parallel Stream.forEach )- As already mentioned no guarantee to respect the encounter order as expected in case of parallel streams. It is possible that action is performed in different thread for different elements which can never be the case with forEachOrdered .

Fifth Approach (Parallel Stream.forEachOrdered )- The forEachOrdered will process the elements in the order specified by the source irrespective of the fact whether stream is sequential or parallel. So it makes no sense to use this with parallel streams.

Answer 10

I have faced similar problems. My code:

List<Integer> leaveDatesList = new ArrayList<>();

.....inserted value in list.......

Way 1: printing a list in a for loop

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

Way 2: printing the list in a forEach, for loop

for(Integer leave : leaveDatesList){
    System.out.println(leave);
}

Way 3: printing the list in java 8

leaveDatesList.forEach(System.out::println);

Answer 11

By using java 8 features.

import java.util.Arrays;
import java.util.List;

/**
 * @author AJMAL SHA
 *
 */
public class ForEach {

    public static void main(String[] args) {

        List<String> features = Arrays.asList("Lambdas", "Default Method", "Stream API", "Date and Time API");
        (features).forEach(System.out::println);

    }

}

Answer 12

1. You haven't specified what kind of elements the list contains, if it is a primitive data type then you can print out the elements.

2. But if the elements are objects then as Kshitij Mehta mentioned you need to implement (override) the method "toString" within that object - if it is not already implemented - and let it return something meaning full from within the object, example:

    class Person { private String firstName; private String lastName; @Override public String toString() { return this.firstName + " " + this.lastName; } }

Answer 13

对于字符串数组的列表

list.forEach(s -> System.out.println(Arrays.toString((String[]) s )));

Answer 14

For loop to print the content of a list :

List<String> myList = new ArrayList<String>();
myList.add("AA");
myList.add("BB");

for ( String elem : myList ) {
  System.out.println("Element : "+elem);
}

Result :

Element : AA
Element : BB

If you want to print in a single line (just for information) :

String strList = String.join(", ", myList);
System.out.println("Elements : "+strList);

Result :

Elements : AA, BB

Answer 15

System.out.println(list); works for me.

Here is a full example:

import java.util.List;    
import java.util.ArrayList;

public class HelloWorld {
     public static void main(String[] args) {
        final List<String> list = new ArrayList<>();
        list.add("Hello");
        list.add("World");
        System.out.println(list);
     }
}

It will print [Hello, World] .

Answer 16

I wrote a dump function, which basicly prints out the public members of an object if it has not overriden toString(). One could easily expand it to call getters. Javadoc:

Dumps an given Object to System.out, using the following rules:

• If the Object is Iterable, all of its components are dumped.
• If the Object or one of its superclasses overrides toString(), the "toString" is dumped
• Else the method is called recursively for all public members of the Object

/**
 * 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();
}

Answer 17

    list.stream().map(x -> x.getName()).forEach(System.out::println);

Answer 18

It depends on what type of objects stored in the List , and whether it has implementation for toString() method. System.out.println(list) should print all the standard java object types ( String , Long , Integer etc). In case, if we are using custom object types, then we need to override toString() method of our custom object.

Example:

class Employee {
 private String name;
 private long id;

 @Override
 public String toString() {
   return "name: " + this.name() + 
           ", id: " + this.id();
 }  
}

Test:

class TestPrintList {
   public static void main(String[] args) {
     Employee employee1 =new Employee("test1", 123);
     Employee employee2 =new Employee("test2", 453);
     List<Employee> employees = new ArrayList(2);
     employee.add(employee1);
     employee.add(employee2);
     System.out.println(employees);
   }
}

Answer 19

I happen to be working on this now...

List<Integer> a = Arrays.asList(1, 2, 3);
List<Integer> b = Arrays.asList(3, 4);
List<int[]> pairs = a.stream()
  .flatMap(x -> b.stream().map(y -> new int[]{x, y}))
  .collect(Collectors.toList());

Consumer<int[]> pretty = xs -> System.out.printf("\n(%d,%d)", xs[0], xs[1]);
pairs.forEach(pretty);

Answer 20

public static void main(String[] args) {
        answer(10,60);

    }
    public static void answer(int m,int k){
        AtomicInteger n = new AtomicInteger(m);
        Stream<Integer> stream = Stream.generate(() -> n.incrementAndGet()).limit(k);
        System.out.println(Arrays.toString(stream.toArray()));
    }

Answer 21

try to override toString() method as you want that the element will be printend. so the method to print can be this:

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

Answer 22

Solusion of your problem for java 11 is:

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

System.out.println(toPrint);

Answer 23

   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;
     }
   }