The example you finally gave, -cp is a parameter to the command, which is java. Parameters are generally program-specific, in this case cp stands for Class Path, which is another location java will search to find the class files as they are needed by the program.

Using Java 6 or later, the classpath option supports wildcards. Note the following:

• Use straight quotes (")
• Use *, not *.jar

Windows

java -cp "Test.jar;lib/*" my.package.MainClass

Unix

java -cp "Test.jar:lib/*" my.package.MainClass

This is similar to Windows, but uses : instead of ;. If you cannot use wildcards, bash allows the following syntax (where lib is the directory containing all the Java archive files):

java -cp "$(printf %s: lib/*.jar)"

(Note that using a classpath is incompatible with the -jar option. See also: Execute jar file with multiple classpath libraries from command prompt)

Understanding Wildcards

From the Classpath document:

Class path entries can contain the basename wildcard character *, which is considered equivalent to specifying a list of all the files in the directory with the extension .jar or .JAR. For example, the class path entry foo/* specifies all JAR files in the directory named foo. A classpath entry consisting simply of * expands to a list of all the jar files in the current directory.

A class path entry that contains * will not match class files. To match both classes and JAR files in a single directory foo, use either foo;foo/* or foo/*;foo. The order chosen determines whether the classes and resources in foo are loaded before JAR files in foo, or vice versa.

Subdirectories are not searched recursively. For example, foo/* looks for JAR files only in foo, not in foo/bar, foo/baz, etc.

The order in which the JAR files in a directory are enumerated in the expanded class path is not specified and may vary from platform to platform and even from moment to moment on the same machine. A well-constructed application should not depend upon any particular order. If a specific order is required then the JAR files can be enumerated explicitly in the class path.

Expansion of wildcards is done early, prior to the invocation of a program's main method, rather than late, during the class-loading process itself. Each element of the input class path containing a wildcard is replaced by the (possibly empty) sequence of elements generated by enumerating the JAR files in the named directory. For example, if the directory foo contains a.jar, b.jar, and c.jar, then the class path foo/* is expanded into foo/a.jar;foo/b.jar;foo/c.jar, and that string would be the value of the system property java.class.path.

The CLASSPATH environment variable is not treated any differently from the -classpath (or -cp) command-line option. That is, wildcards are honored in all these cases. However, class path wildcards are not honored in the Class-Path jar-manifest header.

Note: due to a known bug in java 8, the windows examples must use a backslash preceding entries with a trailing asterisk: https://bugs.openjdk.java.net/browse/JDK-8131329

I prefer the first version to start a java application just because it has less pitfalls ("welcome to classpath hell"). The second one requires an executable jar file and the classpath for that application has to be defined inside the jar's manifest (all other classpath declaration will be silently ignored...). So with the second version you'd have to look into the jar, read the manifest and try to find out if the classpath entries are valid from where the jar is stored... That's avoidable.

I don't expect any performance advantages or disadvantages for either version. It's just telling the jvm which class to use for the main thread and where it can find the libraries.