The javac command - getting started

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Simple example

Assuming that the “HelloWorld.java” contains the following Java source:

public class HelloWorld {
    public static void main(String[] args) {
        System.out.println("Hello world!");
    }
}

(For an explanation of the above code, please refer to http://stackoverflow.com/documentation/java/84/compile-and-run-your-first-java-program#t=201608210035544002048 .)

We can compile the above file using this command:

$ javac HelloWorld.java

This produces a file called “HelloWorld.class”, which we can then run as follows:

$ java HelloWorld
Hello world!

The key points to note from this example are:

  1. The source filename “HelloWorld.java” must match the class name in the source file … which is HelloWorld. If they don’t match, you will get a compilation error.
  2. The bytecode filename “HelloWorld.class” corresponds to the classname. If you were to rename the “HelloWorld.class”, you would get an error when your tried to run it.
  3. When running a Java application using java, you supply the classname NOT the bytecode filename.

Example with packages

Most practical Java code uses packages to organize the namespace for classes and reduce the risk of accidental class name collision.

If we wanted to declare the HelloWorld class in a package call com.example, the “HelloWorld.java” would contain the following Java source:

package com.example;

public class HelloWorld {
    public static void main(String[] args) {
        System.out.println("Hello world!");
    }
}

This source code file needs to stored in a directory tree whose structure corresponds to the package naming.

.    # the current directory (for this example)
|
 ----com
     |
      ----example
          |
           ----HelloWorld.java

We can compile the above file using this command:

$ javac com/example/HelloWorld.java

This produces a file called “com/example/HelloWorld.class”; i.e. after compilation, the file structure should look like this:

.    # the current directory (for this example)
|
 ----com
     |
      ----example
          |
           ----HelloWorld.java
           ----HelloWorld.class

We can then run the application as follows:

$ java com.example.HelloWorld
Hello world!

Additional points to note from this example are:

  1. The directory structure must match the package name structure.
  2. When you run the class, the full class name must be supplied; i.e. “com.example.HelloWorld” not “HelloWorld”.
  3. You don’t have to compile and run Java code out of the current directory. We are just doing it here for illustration.

Compiling multiple files at once with ‘javac’.

If your application consists of multiple source code files (and most do!) you can compile them one at a time. Alternatively, you can compile multiple files at the same time by listing the pathnames:

$ javac Foo.java Bar.java

or using your command shell’s filename wildcard functionality ….

$ javac *.java
$ javac com/example/*.java
$ javac */**/*.java #Only works on Zsh or with globstar enabled on your shell

This will compile all Java source files in the current directory, in the “com/example” directory, and recursively in child directories respectively. A third alternative is to supply a list of source filenames (and compiler options) as a file. For example:

$ javac @sourcefiles

where the sourcefiles file contains:

Foo.java
Bar.java
com/example/HelloWorld.java

Note: compiling code like this is appropriate for small one-person projects, and for once-off programs. Beyond that, it is advisable to select and use a Java build tool. Alternatively, most programmers use a Java IDE (e.g. NetBeans, eclipse, IntelliJ IDEA) which offers an embedded compiler and incremental building of “projects”.

Commonly used ‘javac’ options

Here are a few options for the javac command that are likely to be useful to you

A more complete list of compiler options will be described in a separate example.

References

The definitive reference for the javac command is the Oracle manual page for javac.

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