java: find program name, parse integer argument

Question

A simple test case to demonstrate my 2 problems:

public class Numbers {

    private static void usage() {
        System.err.println("Usage: java " + getClass().getName() + " range");
        System.exit(1);
    }

    public static void main(String[] args) throws IOException {
        try {
            int range = Integer.parseInt(args[0]);
        } catch (Exception e) {
            usage();
        }
    }
}

Can't call getClass() from a static method
If no arguments have been supplied at the command line, I'll get ArrayIndexOutOfBoundsException message instead of the usage() output. Why doesn't catch (Exception e) catch it?

score 6 · Answer 1 · answered Jul 15 '11 at 07:29

6

1) getClass is a method on the Object type. In static methods there is no object to call the getClass on

2) The exception is caught in your example - I just tested it.

answered Jul 15 '11 at 07:29

Petar Ivanov

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score 3 · Accepted Answer · answered Jul 15 '11 at 07:39

Works for me, exception is caught.

Getting the class name from a static method without referencing the Numbers.class.getName() is difficult.

But I found this

String className = Thread.currentThread().getStackTrace()[2].getClassName(); 
System.err.println("Usage: java " + className + " range");

score 1 · Answer 3 · answered Jul 15 '11 at 07:33

How about this in your static method to get the class name: get the name of the class at the top of the current stack trace.

StackTraceElement[] stackTraceElements= new Exception().getStackTrace();
String className = stackTraceElements[0].getClassName();

score 1 · Answer 4 · edited May 23 '17 at 12:08

1

Your first question can be answered by looking at this question: Getting the class name from a static method in Java.

edited May 23 '17 at 12:08

Community

1
1

answered Jul 15 '11 at 07:34

Ray Toal

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Kowser · Answer 5 · 2011-07-15T07:57:20.417

1

You can't use getClass() method without an reference object.

Try this

System.err.println("Usage: java " + Numbers.class.getName() + " range");

It is not possible to use member variable/method without object reference from a static method.

int range = Integer.parseInt(args[0]);

Above will return ArrayIndexOutOfBoundException, not IOException.

So your code won't compile.

edited Jul 15 '11 at 07:57

answered Jul 15 '11 at 07:37

Kowser

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`Numbers.class.getClass()` returns Class> not Class. He should use `Numbers.class.getName()` – helios Jul 15 '11 at 07:45
I'm sorry, but this is useless then. I can just hardcode the String "Numbers" in my program. – Alexander Farber Jul 15 '11 at 08:29

score 0 · Answer 6 · answered Jul 15 '11 at 07:34

0

If you create a new instance of Numbers you can call getClass() on that.

(new Numbers()).getClass().getName()

and as @Petar Ivanov has already said, the Exception is caught as expected.

answered Jul 15 '11 at 07:34

andyb

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score 0 · Answer 7 · answered Jul 15 '11 at 07:36

To get the class you can do 1 of two things:

Class cl = null;
try
{
    cl = Class.forName("Numbers");
}
catch(ClassNotFoundException ex)
{
}

or:

Numbers n = new Numbers();
Class cl = n.getClass();

The first is obviously better because you don't waste memory allocating. Though if you're just planning on returning right away, then this probably doesn't matter much in this case.

java: find program name, parse integer argument

7 Answers7