Java – How to change the default output from a Maven 2 / Cobertura instrument goal

NOTE:

The mentioned LATEST and RELEASE metaversions have been dropped for plugin dependencies in Maven 3 "for the sake of reproducible builds", over 6 years ago. (They still work perfectly fine for regular dependencies.) For plugin dependencies please refer to this Maven 3 compliant solution.

If you always want to use the newest version, Maven has two keywords you can use as an alternative to version ranges. You should use these options with care as you are no longer in control of the plugins/dependencies you are using.

When you depend on a plugin or a dependency, you can use the a version value of LATEST or RELEASE. LATEST refers to the latest released or snapshot version of a particular artifact, the most recently deployed artifact in a particular repository. RELEASE refers to the last non-snapshot release in the repository. In general, it is not a best practice to design software which depends on a non-specific version of an artifact. If you are developing software, you might want to use RELEASE or LATEST as a convenience so that you don't have to update version numbers when a new release of a third-party library is released. When you release software, you should always make sure that your project depends on specific versions to reduce the chances of your build or your project being affected by a software release not under your control. Use LATEST and RELEASE with caution, if at all.

See the POM Syntax section of the Maven book for more details. Or see this doc on Dependency Version Ranges, where:

• A square bracket ( [ & ] ) means "closed" (inclusive).
• A parenthesis ( ( & ) ) means "open" (exclusive).

Here's an example illustrating the various options. In the Maven repository, com.foo:my-foo has the following metadata:

<?xml version="1.0" encoding="UTF-8"?><metadata>
  <groupId>com.foo</groupId>
  <artifactId>my-foo</artifactId>
  <version>2.0.0</version>
  <versioning>
    <release>1.1.1</release>
    <versions>
      <version>1.0</version>
      <version>1.0.1</version>
      <version>1.1</version>
      <version>1.1.1</version>
      <version>2.0.0</version>
    </versions>
    <lastUpdated>20090722140000</lastUpdated>
  </versioning>
</metadata>

If a dependency on that artifact is required, you have the following options (other version ranges can be specified of course, just showing the relevant ones here):

Declare an exact version (will always resolve to 1.0.1):

<version>[1.0.1]</version>

Declare an explicit version (will always resolve to 1.0.1 unless a collision occurs, when Maven will select a matching version):

<version>1.0.1</version>

Declare a version range for all 1.x (will currently resolve to 1.1.1):

<version>[1.0.0,2.0.0)</version>

Declare an open-ended version range (will resolve to 2.0.0):

<version>[1.0.0,)</version>

Declare the version as LATEST (will resolve to 2.0.0) (removed from maven 3.x)

<version>LATEST</version>

Declare the version as RELEASE (will resolve to 1.1.1) (removed from maven 3.x):

<version>RELEASE</version>

Note that by default your own deployments will update the "latest" entry in the Maven metadata, but to update the "release" entry, you need to activate the "release-profile" from the Maven super POM. You can do this with either "-Prelease-profile" or "-DperformRelease=true"

It's worth emphasising that any approach that allows Maven to pick the dependency versions (LATEST, RELEASE, and version ranges) can leave you open to build time issues, as later versions can have different behaviour (for example the dependency plugin has previously switched a default value from true to false, with confusing results).

It is therefore generally a good idea to define exact versions in releases. As Tim's answer points out, the maven-versions-plugin is a handy tool for updating dependency versions, particularly the versions:use-latest-versions and versions:use-latest-releases goals.

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.