What is a classpath and how do I set it?

Question

I was just reading this line:

The first thing the format() method does is load a Velocity template from the classpath named output.vm

Please explain what was meant by classpath in this context, and how I should set the classpath.

Answer 1

When programming in Java, you make other classes available to the class you are writing by putting something like this at the top of your source file:

import org.javaguy.coolframework.MyClass;

Or sometimes you 'bulk import' stuff by saying:

import org.javaguy.coolframework.*;

So later in your program when you say:

MyClass mine = new MyClass();

The Java Virtual Machine will know where to find your compiled class.

It would be impractical to have the VM look through every folder on your machine, so you have to provide the VM a list of places to look. This is done by putting folder and jar files on your classpath.

Before we talk about how the classpath is set, let's talk about .class files, packages, and .jar files.

First, let's suppose that MyClass is something you built as part of your project, and it is in a directory in your project called output . The .class file would be at output/org/javaguy/coolframework/MyClass.class (along with every other file in that package). In order to get to that file, your path would simply need to contain the folder 'output', not the whole package structure, since your import statement provides all that information to the VM.

Now let's suppose that you bundle CoolFramework up into a .jar file, and put that CoolFramework.jar into a lib directory in your project. You would now need to put lib/CoolFramework.jar into your classpath. The VM will look inside the jar file for the org/javaguy/coolframework part, and find your class.

So, classpaths contain:

• JAR files, and
• Paths to the top of package hierarchies.

How do you set your classpath?

The first way everyone seems to learn is with environment variables. On a unix machine, you can say something like:

export CLASSPATH=/home/myaccount/myproject/lib/CoolFramework.jar:/home/myaccount/myproject/output/

On a Windows machine you have to go to your environment settings and either add or modify the value that is already there.

The second way is to use the -cp parameter when starting Java, like this:

java -cp "/home/myaccount/myproject/lib/CoolFramework.jar:/home/myaccount/myproject/output/"  MyMainClass

A variant of this is the third way which is often done with a .sh or .bat file that calculates the classpath and passes it to Java via the -cp parameter.

There is a "gotcha" with all of the above. On most systems (Linux, Mac OS, UNIX, etc) the colon character (':') is the classpath separator. In windowsm the separator is the semicolon (';')

So what's the best way to do it?

Setting stuff globally via environment variables is bad, generally for the same kinds of reasons that global variables are bad. You change the CLASSPATH environment variable so one program works, and you end up breaking another program.

The -cp is the way to go. I generally make sure my CLASSPATH environment variable is an empty string where I develop, whenever possible, so that I avoid global classpath issues (some tools aren't happy when the global classpath is empty though - I know of two common, mega-thousand dollar licensed J2EE and Java servers that have this kind of issue with their command-line tools).

Answer 2

将其视为 Java 对 PATH 环境变量的回答——操作系统在 PATH 上搜索 EXE，Java 在类路径上搜索类和包。

Answer 3

The classpath is the path where the Java Virtual Machine look for user-defined classes, packages and resources in Java programs.

In this context, the format() method load a template file from this path.

Answer 4

The classpath in this context is exactly what it is in the general context: anywhere the VM knows it can find classes to be loaded, and resources as well (such as output.vm in your case).

I'd understand Velocity expects to find a file named output.vm anywhere in "no package". This can be a JAR, regular folder, ... The root of any of the locations in the application's classpath.

Answer 5

Setting the CLASSPATH System Variable

To display the current CLASSPATH variable, use these commands in Windows and UNIX (Bourne shell): In Windows: C:\\> set CLASSPATH In UNIX: % echo $CLASSPATH

To delete the current contents of the CLASSPATH variable, use these commands: In Windows: C:\\> set CLASSPATH= In UNIX: % unset CLASSPATH; export CLASSPATH % unset CLASSPATH; export CLASSPATH

To set the CLASSPATH variable, use these commands (for example): In Windows: C:\\> set CLASSPATH=C:\\users\\george\\java\\classes In UNIX: % CLASSPATH=/home/george/java/classes; export CLASSPATH % CLASSPATH=/home/george/java/classes; export CLASSPATH

Answer 6

Classpath is an environment variable of system. The setting of this variable is used to provide the root of any package hierarchy to java compiler.

Answer 7

CLASSPATH is an environment variable (ie, global variables of the operating system available to all the processes) needed for the Java compiler and runtime to locate the Java packages used in a Java program. (Why not call PACKAGEPATH?) This is similar to another environment variable PATH, which is used by the CMD shell to find the executable programs.

CLASSPATH can be set in one of the following ways:

CLASSPATH can be set permanently in the environment: In Windows, choose control panel ⇒ System ⇒ Advanced ⇒ Environment Variables ⇒ choose "System Variables" (for all the users) or "User Variables" (only the currently login user) ⇒ choose "Edit" (if CLASSPATH already exists) or "New" ⇒ Enter "CLASSPATH" as the variable name ⇒ Enter the required directories and JAR files (separated by semicolons) as the value (e.g., ".;c:\javaproject\classes;d:\tomcat\lib\servlet-api.jar"). Take note that you need to include the current working directory (denoted by '.') in the CLASSPATH.

To check the current setting of the CLASSPATH, issue the following command:

> SET CLASSPATH

CLASSPATH can be set temporarily for that particular CMD shell session by issuing the following command:

> SET CLASSPATH=.;c:\javaproject\classes;d:\tomcat\lib\servlet-api.jar

Instead of using the CLASSPATH environment variable, you can also use the command-line option -classpath or -cp of the javac and java commands, for example,

> java –classpath c:\javaproject\classes com.abc.project1.subproject2.MyClass3

Answer 8

Environment variable is a global system variable accessible by all the processes running under the Operating System (OS).

classpath in Java is an environment variable used by Java Virtual machine to locate or find class files( .class ) in Java during class loading by ClassLoader .

Java Compiler ( javac ), Java Runtime ( java ) and other Java tools searches for classes used in your program in this order:

1. Java platform (bootstrap) classes
2. Java Extension Directories

3. User classes search path (in short, classpath ): determined in the following order:

   a. Defaulted to the current working directory (.) .

   b. Entries in the classpath environment variable , which overrides the default.

   c. Entries in the -cp (or -classpath ) command-line option, which overrides the classpath environment variable.

   d. The runtime command-line option -jar , which override all the above.

The classpath environment variable could include directories (containing many class files) and JAR files (a single-file archive of class files). If classpath is not set, it is defaulted to the current directory . If you set the classpath , it is important to include the current working directory (.) . Otherwise, the current directory will not be searched.

You can override classpath in Java, defined by environment variable by providing option -cp or -classpath while running Java program and this is the best way to have different classpath for different Java application running on same Unix or Windows machine.

java -cp "<path/1><separator><path/2>" <path.to.your.MainClass>

separators are semi-colon ; - Windows, colon : - Linux

For example command for MacOS

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

Read more here and here

Answer 9

The classpath is one of the fundamental concepts in the Java world and it's often misunderstood or not understood at all by java programmes, especially beginners.

Simply put, the classpath is just a set of paths where the java compiler and the JVM must find needed classes to compile or execute other classes.

Let's start with an example, suppose we have a Main.java file thats under C:\\Users\\HP\\Desktop\\org\\example ,

package org.example;

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

And Now, suppose we are under C:\\ directory and we want to compile our class, Its easy right, just run:

javac .\Users\HP\Desktop\org\example\Main.java

Now for the hard question, we are in the same folder C:\\ and we want to run the compiled class.

Despite of what you might think of to be the answer, the right one is:

java -cp .\Users\HP\Desktop org.example.Main 

I'll explain why, first of all, the name of the class that we want ro tun is org.exmaple.Main not Main , or Main.class or .\\users\\hp\\desktop\\org\\example\\Main.class ! This is how things works with classes declared under packages.

Now, we provided the name of the class to the JVM (java command in this case), But how it (JVM) will know where to find the .class file for the Main class? Thats where the classpath comes into picture. Using -cp flag (shortcut for -classpath), we tell the JVM that our Main.class file will be located at C:\\users\\hp\\Desktop .. In fact, not really, we tell it to just go to the Desktop directory, and, because of the name of the class org.example.Main, the JVM is smart and it will go from Desktop to org directory, and from org to example directory, searching for Main.class file , and it will find it and it will kill it, I mean, it will run it :D .

Now lets suppose that inside the Main class we want to work with another class named org.apache.commons.lang3.StringUtils and the latter is located in a jar file named commons-lang3-3.10.jar thats inside C:\\Users\\HP\\Downloads . So Main.java will look like this now:

package org.example;

import org.apache.commons.lang3.StringUtils;

public class Main {
    public static void main(String[] args) {
        System.out.println("Hello world");
        System.out.println(StringUtils.equals("java", "java")); //true
    }   
}

How to compile the Main.java if we are always inside C:\\ ? The answer is:

javac -cp .\Users\HP\Downloads\commons-lang3-3.10.jar .\Users\HP\Desktop\org\example\Main.java

• .\\Users\\HP\\Desktop\\org\\example\\Main.java is because our .java file is there in the filesystem.

• -cp .\\Users\\HP\\Downloads\\commons-lang3-3.10.jar is because the java compiler (javac int this case) need to know the location of the class org.apache.commons.lang3.StringUtils , so we provided the path of the jar file, and the compiler will then go inside the jar file and try to find a file StringUtils.class inside a directory org\\apache\\commons\\lang3 .

And if we want to run the Main.class file, we will execute:

java -cp ".\Users\HP\Desktop\;.\Users\HP\Downloads\commons-lang3-3.10.jar" org.example.Main

• org.example.Main is the name of the class.

• ".\\Users\\HP\\Desktop\\;.\\Users\\HP\\Downloads\\commons-lang3-3.10.jar" are the paths (separated by ; in Windows) to the Main and StringUtils classes.

Answer 10

Static member of a class can be called directly without creating object instance. Since the main method is static Java virtual Machine can call it without creating any instance of a class which contains the main method, which is start point of program.

Answer 11

For linux users, and to sum up and add to what others have said here, you should know the following:

1. $CLASSPATH is what Java uses to look through multiple directories to find all the different classes it needs for your script (unless you explicitly tell it otherwise with the -cp override). Using -cp requires that you keep track of all the directories manually and copy-paste that line every time you run the program (not preferable IMO).

2. The colon (":") character separates the different directories. There is only one $CLASSPATH and it has all the directories in it. So, when you run "export CLASSPATH=...." you want to include the current value "$CLASSPATH" in order to append to it. For example:

    export CLASSPATH=. export CLASSPATH=$CLASSPATH:/usr/share/java/mysql-connector-java-5.1.12.jar

   In the first line above, you start CLASSPATH out with just a simple 'dot' which is the path to your current working directory. With that, whenever you run java it will look in the current working directory (the one you're in) for classes. In the second line above, $CLASSPATH grabs the value that you previously entered (.) and appends the path to a mysql dirver. Now, java will look for the driver AND for your classes.

3.  echo $CLASSPATH

   is super handy, and what it returns should read like a colon-separated list of all the directories, and .jar files, you want java looking in for the classes it needs.

4. Tomcat does not use CLASSPATH. Read what to do about that here: https://tomcat.apache.org/tomcat-8.0-doc/class-loader-howto.html

Answer 12

classpath and path are the evironment variables . usually , you have to put the jdk/bin to path so that u could use java compiler everywhere , classpath is the path of your .class files . the classpath has a default path a period(.) which means the current directory. but when u used the packages . u would either specify the full path of the .class file or put the .class file path in the classpath which would save a lots of works !