CODEWITHSUNDEEP
javacmdjavac
parsecer
parsecer

Reputation: 5106

Can't compile and execute java program

I have the following simplified file structure:

C:/red/green/black/yellow/white/pink. I am now in cmd in folder 'black'.

The Hello.javafile has package yellow.white.pink in it. In folder 'pink' there is a Hello.java source file.

First I try to compile the file:

javac -classpath . -encoding ISO-8859-1 yellow.white.pink.Hello.java

javac -cp . -encoding ISO-8859-1 yellow.white.pink.Hello.java

These two give me an error:

File not found yellow.white.pink.Hello.java.

Then I try

javac -encoding ISO-8859-1 C:/red/green/black/yellow/white/pink/Hello.java

It compiles just fine.

To run it I do

java -classpath . yellow.white.pink.Hello And it runs just fine. But in this case

java C:/red/green/black/yellow/white/pink/Hello.class

doesn't work - gives Could not find or load main class error.

Why is that? Why can't I compile .java file when being in the root folder and giving a full package name and it works only with the whole path to the source, while the reverse is true for executing the program?

Upvotes: 1

Views: 202

Answers (1)

Sasha Shpota
Sasha Shpota

Reputation: 10300

java command takes as an argument the fully qualified name of the class (which is package name plus class name).

In both cases the fully qualified name must be yellow.white.pink.Hello and it should not be changed.

However in your second run you pass C:/red/green/black/yellow/white/pink/Hello.class which is path to the compiled file but not the fully qualified name of the class.

What is also different in this two runs is classpass declaration. In the first run it is path of the current directory (can be passed as .). In the second run it was not specified at all. Java treat this as current directory path.

To make the second example work you have to specify classpath as well as class name.

java -classpath C:/red/green/black yellow.white.pink.Hello

Upvotes: 3

Related Questions