Given the following code:

List<String> strList = new ArrayList<>();

strList.add("Java");

strList.add("Python");

strList.add("Php");



Stream<String> strStream = strList.stream();



Stream<String> jFilter = strStream.filter(str -> str.startsWith("J"));



strList.add("JavaScript"); // element added after filter creation

strList.add("JQuery"); // element added after filter creation



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

which outputs:

[Java, JavaScript, JQuery]

Why do JavaScript and JQuery appear in the filtered result even though they were added after creating the filtered stream?

Short Answer

You're assuming after this point:

Stream<String> jFilter = strStream.filter(str -> str.startsWith("J"));

That a new stream of the elements starting with "J" are returned i.e. only Java. However this is not the case;

streams are lazy i.e. they don't perform any logic unless told otherwise by a terminal operation.

The actual execution of the stream pipeline starts on the toArray() call and since the list was modified before the terminal toArray() operation commenced the result will be [Java, JavaScript, JQuery].

Longer Answer

here's part of the documentation which mentions this:

For well-behaved stream sources, the source can be modified before
the terminal operation commences and those modifications will be
reflected in the covered elements. For example, consider the following
code:

 List<String> l = new ArrayList(Arrays.asList("one", "two"));

 Stream<String> sl = l.stream();

 l.add("three");

 String s = sl.collect(joining(" "));  

First a list is created consisting of two strings: "one"; and "two". Then a stream is created
from that list. Next the list is modified by adding a third string:
"three". Finally the elements of the stream are collected and joined
together. Since the list was modified before the terminal collect
operation commenced the result will be a string of "one two three".
All the streams returned from JDK collections, and most other JDK
classes, are well-behaved in this manner;

Until the statement

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

runs, the stream doesn't do anything. A terminal operation (toArray in the example) is required for the stream to be traversed and your intermediate operations (filter in this case) to be executed.

In this case, what you can do is, for example, capture the size of the list before adding other elements:

int maxSize = strList.size();

Stream<String> jFilter = strStream.limit(maxSize)

                                  .filter(str -> str.startsWith("J"));

where limit(maxSize) will not allow more than the initial elements to go through the pipeline.

Its because the stream never got evaluated. you never called a "Terminal operation" on that stream for it to get executed as they're lazy.

Look at a modification of your code and the output. The filtering actually takes place when you call the Terminal Operator.

 public static void main(String args){

         List<String> strList = new ArrayList<>();

    strList.add("Java");

    strList.add("Python");

    strList.add("Php");



    Stream<String> strStream = strList.stream();



    Stream<String> jFilter = strStream.filter(str -> {

        System.out.println("Filtering" + str);

        return str.startsWith("J");

        });



 System.out.println("After Stream creation");

    strList.add("JavaScript"); // element added after filter creation

    strList.add("JQuery"); // element added after filter creation



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



     }

Output:

After Stream creation

FilteringJava

FilteringPython

FilteringPhp

FilteringJavaScript

FilteringJQuery

[Java, JavaScript, JQuery]

As explained in the official documentation at ,https://docs.oracle.com/javase/8/docs/api/java/util/stream/package-summary.html, streams have no storage, and so are more like iterators than collections, and are evaluated lazily.

So, nothing really happens with respect to the stream until you invoke the terminal operation toArray()

@Hadi J's comment but it should be answer according to the rules.

Because streams are lazy and when you call terminal operation it executed.