CODEWITHSUNDEEP
java
Whiteox
Whiteox

Reputation: 39

Since java performs both compilation and interpretation for the source code does that mean javac compiles the code and java interprets\runs code

I wrote this code in file named Hello.java

public class Hello{
    System.out.println("Hello there");
}

then I wrote command javac hello.java and java hello and prints successfully

However if I make changes to the code and write

public class Demo{
     System.out.println("New hello");
}

and run command java hello.java This also works and prints successfully. However I am not following any Java parameters(file name and class name should be same) for defining class name so why does this works?

I know I am doing the wrong way but if I didn't follow the pattern it shouldn't execute, right ?! and if I knowingly make semicolon mistake and run the command java hello.java it says compilation failed, but isn't the task of compilation complete ?

Upvotes: 0

Views: 62

Answers (1)

the Hutt
the Hutt

Reputation: 18418

What you've encountered here is a "source-file" mode of java launcher: Java launcher has 4 launching modes:

  • launching a class file
  • launching the main class of a JAR file
  • launching the main class of a module.
  • launching a class declared in a source file.

We are talking about fourth option here:

In source-file mode, the effect is as if the source file is compiled into memory, and the first class found in the source file is executed. For example, if a file called HelloWorld.java contains a class called hello.World, then the command
java HelloWorld.java
is informally equivalent to
javac -d <memory> HelloWorld.java
java -cp <memory> hello.World

The compiler does not enforce the optional restriction defined at the end of JLS §7.6, that a type in a named package should exist in a file whose name is composed from the type name followed by the .java extension.

For more info read the specification Jep 330.

Upvotes: 2

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