Read all text from a file
Java 11 added the readString() method to read small files as a String
, preserving line terminators:
String content = Files.readString(path, StandardCharsets.US_ASCII);
For versions between Java 7 and 11, here's a compact, robust idiom, wrapped up in a utility method:
static String readFile(String path, Charset encoding)
throws IOException
{
byte[] encoded = Files.readAllBytes(Paths.get(path));
return new String(encoded, encoding);
}
Read lines of text from a file
Java 7 added a convenience method to read a file as lines of text, represented as a List<String>
. This approach is "lossy" because the line separators are stripped from the end of each line.
List<String> lines = Files.readAllLines(Paths.get(path), encoding);
Java 8 added the Files.lines()
method to produce a Stream<String>
. Again, this method is lossy because line separators are stripped. If an IOException
is encountered while reading the file, it is wrapped in an UncheckedIOException
, since Stream
doesn't accept lambdas that throw checked exceptions.
try (Stream<String> lines = Files.lines(path, encoding)) {
lines.forEach(System.out::println);
}
This Stream
does need a close()
call; this is poorly documented on the API, and I suspect many people don't even notice Stream
has a close()
method. Be sure to use an ARM-block as shown.
If you are working with a source other than a file, you can use the lines()
method in BufferedReader
instead.
Memory utilization
The first method, that preserves line breaks, can temporarily require memory several times the size of the file, because for a short time the raw file contents (a byte array), and the decoded characters (each of which is 16 bits even if encoded as 8 bits in the file) reside in memory at once. It is safest to apply to files that you know to be small relative to the available memory.
The second method, reading lines, is usually more memory efficient, because the input byte buffer for decoding doesn't need to contain the entire file. However, it's still not suitable for files that are very large relative to available memory.
For reading large files, you need a different design for your program, one that reads a chunk of text from a stream, processes it, and then moves on to the next, reusing the same fixed-sized memory block. Here, "large" depends on the computer specs. Nowadays, this threshold might be many gigabytes of RAM. The third method, using a Stream<String>
is one way to do this, if your input "records" happen to be individual lines. (Using the readLine()
method of BufferedReader
is the procedural equivalent to this approach.)
Character encoding
One thing that is missing from the sample in the original post is the character encoding. There are some special cases where the platform default is what you want, but they are rare, and you should be able justify your choice.
The StandardCharsets
class defines some constants for the encodings required of all Java runtimes:
String content = readFile("test.txt", StandardCharsets.UTF_8);
The platform default is available from the Charset
class itself:
String content = readFile("test.txt", Charset.defaultCharset());
Note: This answer largely replaces my Java 6 version. The utility of Java 7 safely simplifies the code, and the old answer, which used a mapped byte buffer, prevented the file that was read from being deleted until the mapped buffer was garbage collected. You can view the old version via the "edited" link on this answer.
Note that each of the code samples below may throw IOException
. Try/catch/finally blocks have been omitted for brevity. See this tutorial for information about exception handling.
Note that each of the code samples below will overwrite the file if it already exists
Creating a text file:
PrintWriter writer = new PrintWriter("the-file-name.txt", "UTF-8");
writer.println("The first line");
writer.println("The second line");
writer.close();
Creating a binary file:
byte data[] = ...
FileOutputStream out = new FileOutputStream("the-file-name");
out.write(data);
out.close();
Java 7+ users can use the Files
class to write to files:
Creating a text file:
List<String> lines = Arrays.asList("The first line", "The second line");
Path file = Paths.get("the-file-name.txt");
Files.write(file, lines, StandardCharsets.UTF_8);
//Files.write(file, lines, StandardCharsets.UTF_8, StandardOpenOption.APPEND);
Creating a binary file:
byte data[] = ...
Path file = Paths.get("the-file-name");
Files.write(file, data);
//Files.write(file, data, StandardOpenOption.APPEND);
Best Answer
Because Dalvik's FileInputStream will close itself when it is garbage collected (this is also true for OpenJDK/Oracle) it is less common than you'd think to actually leak file descriptors. Of course, the file descriptors will be "leaked" until the GC runs so depending on your program it could take a while before they are reclaimed.
To accomplish a more permanent leak you will have to prevent the stream from being garbage collected by keeping a reference to it somewhere in memory.
Here's a short example that loads a properties file every 1 second and keeps track of every time it has changed:
Because of the lazy loading we keep the InputStream in the PropertiesFile which will be kept whenever we create a new Revision and since we never close the stream we will be leaking file descriptors here.
Now, these open file descriptors will be closed by the OS when the program terminates, but as long as the program is running it will continue to leak file descriptors as can be seen by using lsof:
And if we force the GC to run we can see that most of these are returned:
The remaining two descriptors are the ones stored in the Revisions.
I ran this on my Ubuntu machine but the output would look similar if run on Android.