Understanding the Eclipse classpath declarations

Question

I'm trying to understand the Eclipse classpath file, in particular, I want to know this:

1. How is the JRE using it (i.e. is the JVM reading this xml file directly, or is eclipse somehow feeding it into its internal compiler)?

2. How are the complex entries (like the ivy path below) parsed and incorporated into the JVM ClassLoader when I run my classes from my IDE?

Context: I have a strange bug which is that eclipse is using the "wrong" version of a class, whereas my ivy / ant build is using the correct version, and I thus want to tool Eclipse to better mimick the classloader used in my pure build. In order to do this, I'm thinking I will have to look at the Eclipse project/classpath files.

<?xml version="1.0" encoding="UTF-8"?>
<classpath>
    <classpathentry kind="src" path="src"/>
    <classpathentry kind="src" path="test"/>
    <classpathentry kind="con" path="org.eclipse.jdt.launching.JRE_CONTAINER"/>
    <classpathentry kind="lib" path="conf"/>
    <classpathentry kind="con" path="org.apache.ivyde.eclipse.cpcontainer.IVYDE_CONTAINER/?ivyXmlPath=ivy.xml&amp;confs=*"/>
    <classpathentry kind="output" path="build"/>
</classpath>

Answer 1

Two different things:

1) Project classpath is used to compile your code using Eclipse Java Compiler (ejc), so the file information is passed to the EJC.

2) When you create a launch configuration, you are actually declaring the classpath to run your application, which, by default, is based on your project classpath. This classpath is passed as an argument to the JVM like you would do it manually (java -cp ${classpathentries} yourmainclass). If you want to find out what is precisely the classpath of your launch configuration, launch your app/classes in debug mode, and in the Debug view, select your process and click on Properties where you will see the full classpath (all the jars/directories that are passed as argument to the JVM)

NB: I cannot see your ivy path stuff.