I must be doing something stupid, because this seems like such a easy operation. For some reason, I just cannot get this command to work for me. I've installed JDK, and I go into the
/bin
folder. I type:
./javap -classpath /home/Matt/workspace/VariablesTestProject/src/ -s VariablesTest.Variable
My actual filepath for the class is
/home/Matt/workspace/VariablesTestProject/src/VariablesTest/Variable.java
The error I get back is:
ERROR:Could not find VariablesTest.Variable
A command like this works however:
./javap -s java.lang.String
These are not the only commands I've tried. I've literally tried every variation I can think of, and none of them work. My javac and java commands both work fine.
Any suggestions?
Javap works against the .class bytecode. So point the classpath at the VariablesTestProject/out or whatever.
If you want perform operations using the javap command, first go to the jar location in cmd prompt.
Eg:
cd /usr/lib/jvm/java-6-openjdk-i386/jre/lib (Linux)
cd D:\javasoftware\Java\jdk1.6.0_43\jre\lib (Windows)
The above locations are the rt.jar locations. java.lang.String is the one of the classes in this jar.
Using javap you can get the information from this class.
Eg:
cd /usr/lib/jvm/java-6-openjdk-i386/jre/lib
javap java.lang.String
One recent use of javap which I've found is to check which version of JDk compiled a particular class file on Windows:
javap -v path to .class file | findstr major
This will give you major version number :
50 - JDK6, 51 - JDK7, 52 - JDK8, 53 - JDK9.
I will make it simple words to understand the use of Javap Command in Java .
javap command is a tool or you can say it dissembler which can read class files .
I have sample java program like below.
class Bank {
private int code = 210;
class SBI {
void display() {
System.out.println("Code is "+code);
}
}
public static void main(String args[]){
Bank b = new Bank();
Bank.SBI sbi = b.new SBI();
sbi.display();
}
}
After compilation I got Bank.class file.Next my friends Subbu and Narasimha asked me compiled version of source file.Later on they want to see what vikram is actually used then they will use the following command to see the basic details like what methods used in this Bank.class file.
C:\ >javap Bank.class
Compiled from "Bank.java"
class Bank {
Bank();
public static void main(java.lang.String[]);
static int access$000(Bank);
}
java has pretty cool tools to check out many features few of them like javaws.jconsole,javaws,jmap,jhat and jinfo
javap is the command used to see the java profile/method declaration of a Inbuilt class/User Defined Class/Interfaces/Abstract Class.
-> If you want to see the methods of a pre defined class then in this path C:\Program Files\Java\jre1.8.0_171\lib you have rt.jar if you installed jdk.
-> Extract that rt.jar file in some folder D:\rt and then go to this folder and type
javap java.util.collection or javap java.lang.* or any other predefined package to see the methods.
-> If you want to see the user defined class methods then go to the workspace\javaproject\specificpackage\bin and then type javap classname and you can see all the methods declaration.
Related
I am trying to learn Java and I've made my first program and compiled it into a class file (the file is called aye.java and when compiled I have aye.class, I think the compilation worked). However when I use the java command in the folder where the class is located it just returns below error -
Could not find or load main class aye.class.
I have tried including the package name (com.java24hours) but it still doesn't work.. please help!
Commands I have tried:
java aye.class
java com.java24hours.aye.class
java aye
java com.java24hours.aye
program code:
package com.java24hours;
class aye {
public static void main(String[] args) {
//java code yeet
String aye = "Hello World!";
System.out.println(aye);
}
}
(I am running Linux on a Chromebook and have installed Java via the ppa:webupd8team/java)
Thanks.
I suppose you wanna put binaries to ./bin folder.
Compile aye.java:
javac -d ./bin aye.java
Then cd to ./bin directory and run the program:
cd bin
java com.java24hours.aye
well im stupid
since im new to java, i didn't know anything about packages and such. turns out all i had to do was put the class file in a folder named "ya" (that's the name of the package - i updated the program) and run the command
java -cp /home/ramsey/Documents/ya aye
(-cp stands for classpath, and you use it when you want to specify where you class is located MAKE SURE TO PUT IT IN A FOLDER NAMED AFTER YOUR PACKAGE!!!)
the wiki page is helpful: https://en.wikipedia.org/wiki/Classpath_(Java)
its under the section setting the path to execute java programs
thanks for the help everyone!
I have a set of instructions to create a Java application that takes multiple arguments when running the application from a CMD line.
The instructions state:
Thus, using the example set of above, assuming the main() method of your program is in a class called JavaClassName, the output should be:
$ java JavaClassName 4 7 file.csv
program output here
My question is:
Isn't this skipping the compile process?
Would they assume that loading the Java classes onto a computer that has never run this application before (or a directory with purely the .java files needed to run); running the cmd
$ java JavaClassName 4 7 file.csv
would output something?
Sidenote: Currently, running that CMD outputs
Error: Could not find or load main class JavaClassName
Have ran through multiple SO questions and online tutorials trying to get this to even run but I have yet to get it to work.
You ask:
Isn't this skipping the compile process?
Absolutely yes. A command line like java JavaClassName 4 7 file.csv assumes that there is a compiled class file "JavaClassName.class" in the current directory (or in some other directory or Zip/Jar file found in the CLASSPATH environment variable). And yes, to produce that "JavaClassName.class" class file, you have to use a java compiler first.
from Java 10 it is possible to run java programs that fit a single file without manually run the compiler first.
It will be compiled on the fly before execution. Nice and useful for scripting.
e.g. writing HelloWorld.java file
public class HelloWorld{
public static void main(String[] args) {
System.out.println("Hello World");
}
}
we can run it as
java HelloWorld.java
We can also add more classes in the single file.
from Java 11 we can also use shebang files
we have to specify which version of the java language you want to use
we have to save the file without .java extension
remember to give executable permissions chmod +x HelloWorld
writing HelloWorld file
#!/path/to/java --source 11
public class HelloWorld{
public static void main(String[] args) {
System.out.println("Hello World");
}
}
and we can run it as
./HelloWorld
I am very new to linux environment.
I am trying to run an simple hello world java class in linux environment.
Hello .java
package com.util;
public class Hello {
/**
* #param args
*/
public static void main(String[] args) {
System.out.println("hi");
}
}
I have compiled java class in windows environment and uploaded the .class file to linux system into /home/scripts path.
my command is as follows,
java -cp /home/scripts com.util.Hello
when i am executing this command from this same /home/scripts where Hello.class is there i am getting,
Error: Could not find or load main class com.util.Hello and not able to proceed further.
can some one help me in this issue?
navigate to /home/scripts using terminal
javac com/util/Hello.java
then
cd /home/scripts
java -cp . com.util.Hello
Or,
java -cp "/home/scripts" com.util.Hello
At first you must generate your .class file :
javac ./hello.java
This command has generated hello.class file
And after you can run your class file ! :)
java hello
We first know javac command work well.
I also met this error,and i have resolved this.Let me share this.
First we need to find the parent path of your package in your java codes.
Then cd to that path using java package + fileName should work well at that moment.
I had the exact same issue on windows, and I solved it by adding path "." to both CLASSPATH and PATH, maybe you can try this on Linux as well.
Your .class file should not reside in /home/scripts/, but in /home/scripts/com/util/. Take a look at this document that explains the relation between classpath, packages and directories.
Before Specifying the path,ensure you follow these three things meticulously,
1. Close the command prompt window, before specifying the path.
2. When adding path, add bin and semi- colon at the end and
3. If JAVAC command has worked properly, try java -cp class name.
if you want to run program in current working directory where your class reside.
java gives three options.
first option
java -cp Tester
Second option for current working directory
java -cp . Tester
Third option export CLASSPATH variable
export CLASSPATH=$CLASSPATH:. (this is the best one if your directory changes) or
export CLASSPATH=$PWD
or
export CLASSPATH=
after that you must sorce the bashrc or bashprofile.
Running basic java programs from commands line is a 3 steps process:
Write code:
public class HelloWorld {
public static void main(String[] args) {
System.out.println("Hello, World");
}
}
Compile by javac HellWorld.java which would check for errors & generate HelloWorld.class file.
run code by giving the class name --> java HelloWorld
Now,
I am curious to know why:
java HelloWorld works but when we give fullpath of the classfile, it throws an error
$ java HelloWorld.class
Error: Could not find or load main class HelloWorld.class
What does it make a difference if we give just the classname Vs classname with file-extension?
What does it make a difference if we give just the classname Vs classname with file-extension?
The argument you give to the java binary isn't meant to be a filename. It's meant to be a class name. So in particular, if you're trying to start a class called Baz in package foo.bar you would run:
java foo.bar.Baz
So similarly, if you try to run java HelloWorld.class it's as if you're trying to run a class called class in a package HelloWorld, which is incorrect.
Basically, you shouldn't view the argument as a filename - you should view it as a fully-qualified class name. Heck there may not even be a simple Baz.class file on the file system - it may be hidden away within a jar file.
What does it make a difference if we give just the classname Vs classname with file-extension?
It is because that is the way it is. Sun / Oracle have implemented the java command to work that way since Java 1.0, and changing it would be massively disruptive.
As Jon says, the argument to the command is a fully qualified class name, not a filename. In fact, it is quite possible that a file with the name HelloWorld.class does not exist. It could be a member of a JAR file ... or in some circumstances, just about anything.
In Java 11 and later it is also possible to compile and run a single Java source file with a single command as follows:
java HelloWorld.java
(This possible because Oracle no longer supports Java distributions without a Java bytecode compiler.)
In the Java programming language, source files (.java files) are compiled into (virtual) machine-readable class files which have a .class extension.
When you run java class file after compile then run the following command:
java HelloWorld
Note: Need to setup java classpath
I have just installed JDK on Windows Vista. After that I set proper values for the 4 environment variables: classpath, include, lib, path. After that I was able to compile my HelloWorld-program (I got a *.class file). But when I try to execute the compiled program (I type java HelloWorldApp) it does not work. The Java write a lot of stuff and in the end it is written that it "could not find the main class: HelloWorldApp". Can anybody, pleas, help me with this problem?
Just for clarity; you are saying that you have a class in the default package, that is you have not included a package specifier in the Java file, and your class is called HelloWorldApp. When you compiled this, you got a classfile HelloWorldApp.class in the current directory.
Assuming the above to be true then try:
java -cp . HelloWorldApp
For example, the following works on a unix box:
$ echo 'class HelloWorldApp { public static void main(String []argv) { System.out.println("Hello World!"); } }' > HelloWorldApp.java
$ javac HelloWorldApp.java
$ java -cp . HelloWorldApp
Hello World!
Of course, you should indent your code a little nicer than just shoving the whole thing onto one line ;-)
Edit: To answer the comment:
Normally, the default classpath is the runtime libraries and the current directory. However, if you have the CLASSPATH variable set, then this will override the default, and you need to explicitly set the classpath back to its "default value". To verify if the CLASSPATH environment variable is set, you can do (again assuming unix):
set | grep CLASSPATH
If it is set, that is why you need to manually include . on your classpath.
create a file called HelloWorld.java;
paste the code posted below inside HelloWorld.java:
compile it by executing the command: javac HelloWorld.java in the same folder as HelloWorld.java is in;
execute the code by doing: java -cp . HelloWorld in the same folder as HelloWorld.java is in.
public class HelloWorld {
public static void main(String[] args) {
System.out.println("HelloWorld works!");
}
}
How the classpath works, can be read here: http://en.wikipedia.org/wiki/Classpath_%28Java%29
Have you included . and .. in your path? Just for clarification . represents your current directory and .. represents your parent directory. You are telling that the java has to search the current directory and the parent directory to find the class. Add the same to your classpath too.
What happens if you use:
java -cp {path to directory with HelloWorldApp in it} HelloWorldApp
That path should be contained within your CLASSPATH environment variable. Is that exported to your command shell ? Do you need to start a new command shell to get the most recent version of CLASSPATH ?
Post your code. I believe the problem is that your main class is not defined properly. I did this the other day.
public static void main(String[] args){
//code
}
The class path concept and the logical difference between Java source code and compiled byte code is notoriously hard to get right.
I would strongly recommend you familiarize yourself with the Sun Java Tutorial. The relevant section is
http://java.sun.com/docs/books/tutorial/getStarted/cupojava/win32.html