Why is Files.lines (and similar Streams) not automatically closed?

JavaStreamJava 8Java StreamResource Leak

Java Problem Overview


The javadoc for Stream states:

> Streams have a BaseStream.close() method and implement AutoCloseable, but nearly all stream instances do not actually need to be closed after use. Generally, only streams whose source is an IO channel (such as those returned by Files.lines(Path, Charset)) will require closing. Most streams are backed by collections, arrays, or generating functions, which require no special resource management. (If a stream does require closing, it can be declared as a resource in a try-with-resources statement.)

Therefore, the vast majority of the time one can use Streams in a one-liner, like collection.stream().forEach(System.out::println); but for Files.lines and other resource-backed streams, one must use a try-with-resources statement or else leak resources.

This strikes me as error-prone and unnecessary. As Streams can only be iterated once, it seems to me that there is no a situation where the output of Files.lines should not be closed as soon as it has been iterated, and therefore the implementation should simply call close implicitly at the end of any terminal operation. Am I mistaken?

Java Solutions


Solution 1 - Java

Yes, this was a deliberate decision. We considered both alternatives.

The operating design principle here is "whoever acquires the resource should release the resource". Files don't auto-close when you read to EOF; we expect files to be closed explicitly by whoever opened them. Streams that are backed by IO resources are the same.

Fortunately, the language provides a mechanism for automating this for you: try-with-resources. Because Stream implements AutoCloseable, you can do:

try (Stream<String> s = Files.lines(...)) {
    s.forEach(...);
}

The argument that "it would be really convenient to auto-close so I could write it as a one-liner" is nice, but would mostly be the tail wagging the dog. If you opened a file or other resource, you should also be prepared to close it. Effective and consistent resource management trumps "I want to write this in one line", and we chose not to distort the design just to preserve the one-line-ness.

Solution 2 - Java

I have more specific example in addition to @BrianGoetz answer. Don't forget that the Stream has escape-hatch methods like iterator(). Suppose you are doing this:

Iterator<String> iterator = Files.lines(path).iterator();

After that you may call hasNext() and next() several times, then just abandon this iterator: Iterator interface perfectly supports such use. There's no way to explicitly close the Iterator, the only object you can close here is the Stream. So this way it would work perfectly fine:

try(Stream<String> stream = Files.lines(path)) {
    Iterator<String> iterator = stream.iterator();
    // use iterator in any way you want and abandon it at any moment
} // file is correctly closed here.

Solution 3 - Java

In addition if you want "one line write". You can just do this:

Files.readAllLines(source).stream().forEach(...);

You can use it if you are sure that you need entire file and the file is small. Because it isn't a lazy read.

Solution 4 - Java

If you're lazy like me and don't mind the "if an exception is raised, it will leave the file handle open" you could wrap the stream in an autoclosing stream, something like this (there may be other ways):

  static Stream<String> allLinesCloseAtEnd(String filename) throws IOException {
    Stream<String> lines = Files.lines(Paths.get(filename));
    Iterator<String> linesIter = lines.iterator();

    Iterator it = new Iterator() {
      @Override
      public boolean hasNext() {
        if (!linesIter.hasNext()) {
          lines.close(); // auto-close when reach end
          return false;
        }
        return true;
      }

      @Override
      public Object next() {
        return linesIter.next();
      }
    };
    return StreamSupport.stream(Spliterators.spliteratorUnknownSize(it, Spliterator.DISTINCT), false);
  }

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionMikeFHayView Question on Stackoverflow
Solution 1 - JavaBrian GoetzView Answer on Stackoverflow
Solution 2 - JavaTagir ValeevView Answer on Stackoverflow
Solution 3 - JavaMichael StarodynovView Answer on Stackoverflow
Solution 4 - JavarogerdpackView Answer on Stackoverflow