executing a dynamically created file

Question

import java.io.*;

public class Demo{

    public static void main(String[] args){
     File f = new File("abc.txt") ;

     try{
      System.setOut(new PrintStream( new FileOutputStream(f) ) ) ;
     }
     catch(FileNotFoundException fnfe){
      System.out.println(fnfe.getMessage()) ;
     }

     System.out.println("Hello\n") ;

     try{
      //throwing exception,
      //is there any method to close the f File,
      //before we try to open the file referred by f.
      Process p = Runtime.getRuntime().exec(f.getPath()) ;
     }
     catch(IOException io){
      System.out.println(io.getMessage()) ;
     }
    }

}

and the content of abc.txt after executing Demo is:-

Hello

Cannot run program "abc.txt": CreateProcess error=32, The process cannot access the file because it is being used by another process

how to avoid the exception.....

as many people here suggested, i have tried the following code, but sadly, even that is also throwing excption....:-(

import java.io.*;

class Demo{

    public static void main(String[] args){
     File f = new File("abc.txt") ;

     FileOutputStream fos = null ;
     try{
      fos = new FileOutputStream(f) ; 
     }
     catch(FileNotFoundException fnfe){
      System.out.println(fnfe.getMessage()) ;
     }

     PrintStream ps = new PrintStream(fos) ;
     ps.println("Hello") ;

     try{
      fos.close() ;

      //throwing exception again
      Process p = Runtime.getRuntime().exec(f.getAbsolutePath()) ;
     }
     catch(IOException io){
      System.out.println(io.getMessage()) ;
     } 
    }
}

??????????

Answer 1

Close the FileOutputStream (or PrintStream) before you try to execute it.

Answer 2

Close the file before executing it (and don't redirect System.out):

f = new File("abc.txt");
FileOutputStream fos = new FileOutputStream(f);

// You would likely use fos.write instead, but here we go
PrintStream ps = new PrintStream(fos);
ps.println("Hello\n");

fos.close();

Process p = Runtime.getRuntime().exec(f.getPath());

Answer 3

1. What your program has created is a text file, not an executable file. An executable file is a binary file (or in some cases a script with a special header) which the operating system knows how to execute. I'm not sure what you expect should happen when you execute a text file whose contents are "Hello\n".

2. You are redirecting stdout (standard out) to your file. Later, you printed the exception stack trace to stdout, which is why the trace appears in your file. It would probably make more sense for you to write to the file directly, instead of redirecting stdout.

Answer 4

I am assuming that the reason for calling Runtime.getRuntime().exec(f.getPath()); is to open up the abc.txt file in a text editor. It would be better to provide complete command for opening the text editor along with the file path. I tried this with notepad.exe (windows) and it worked.

import java.io.*;

public class Demo{

    public static void main(String[] args){
        File f = new File("abc.txt") ;

        try{
                System.setOut(new PrintStream( new FileOutputStream(f) ) ) ;
        }
        catch(FileNotFoundException fnfe){
                System.out.println(fnfe.getMessage()) ;
        }

        System.out.println("Hello\n") ;

        try{
                Process p = Runtime.getRuntime().exec("notepad.exe " + f.getPath()) ;
        }
        catch(IOException io){
                System.out.println(io.getMessage()) ;
        }
    }

}

Following code dynamically generates java code and uses javac to compile it

import java.io.*;

public class Demo{

    public static void main(String[] args){

        File f = new File("Abc.java") ;
        PrintWriter writer = null;
        BufferedReader reader = null;
        InputStream pStream = null;

        try{
                // Open File Stream and write code
                writer = new PrintWriter( new FileOutputStream(f) );
                String javaCode = "public class Abc { \r\n" +
                                  "public static void main(String[] args) {\r\n" +
                                  "System.out.println(\"Hello World!\");\r\n" +
                                  "}\r\n" +
                                  "}\r\n";

                writer.println(javaCode) ;
                writer.close();

                // Run Javac to compile the code
                Process p = Runtime.getRuntime().exec("javac " + f.getPath()) ;
                p.waitFor();

                // status = 0 => Process executed without errors
                //        = 1 => Process executed with errors
                int status = p.exitValue();
                if( status ==  0 )
                {
                    pStream = p.getInputStream();
                }
                else
                {
                    pStream = p.getErrorStream();
                }

                // Display the output from the process
                reader = new BufferedReader(new InputStreamReader(pStream));
                String ln = null;
                while( (ln = reader.readLine()) != null )
                {
                    System.out.println(ln);
                }
        }
        catch(Exception ex){
                System.out.println(ex.getMessage()) ;
        }
        finally{
            try{
               if( writer != null ){writer.close();}
                if( pStream != null ){pStream.close();}
                if( reader != null ){reader.close();}
            }
            catch(Exception ex){
                System.out.println(ex.getMessage()) ;
            }
        }
    }
}

Answer 5

assuming you are running on windows, instead of

Process p = Runtime.getRuntime().exec(f.getPath());

you could use

Process p = Runtime.getRuntime().exec("start " + f.getPath());

which should choose whatever application you have associated with .txt files.