Question

i have to write in a file 4bytes representing an integer in little endian (java use big endian) because an external c++ application have to read this file. My code don’t write anything in te file but de buffer has data inside. why?
my funcion:

public static void copy(String fileOutName, boolean append){
    File fileOut = new File (fileOutName);

    try {
         FileChannel wChannel = new FileOutputStream(fileOut, append).getChannel();

         int i = 5;
         ByteBuffer bb = ByteBuffer.allocate(4);
         bb.order(ByteOrder.LITTLE_ENDIAN);
         bb.putInt(i);

         bb.flip();

         int written = wChannel.write(bb);
         System.out.println(written);    

         wChannel.close();
     } catch (IOException e) {
     }
}

my call:

copy("prueba.bin", false);

Answer 1

When you don’t know why something failed, it is a bad idea to ignore exceptions in an empty try-catch block.

Odds are excellent that you are running the program in an environment where the file cannot be created; however, the instructions you gave to handle such an exceptional situation is to do nothing. So, odds are you have a program that attempted to run, but failed with some reason, which was handled by not even displaying the reason to you.

try this

public static void copy(String fileOutName, boolean append){
    File fileOut = new File (fileOutName);

    try {
         FileChannel wChannel = new FileOutputStream(fileOut, append).getChannel();

         int i = 5;
         ByteBuffer bb = ByteBuffer.allocate(4);
         bb.order(ByteOrder.LITTLE_ENDIAN);
         bb.putInt(i);

         bb.flip();

         int written = wChannel.write(bb);
         System.out.println(written);    

         wChannel.close();
     } catch (IOException e) {
// this is the new line of code
         e.printStackTrace();
     }
}

And I’ll bet you find out why it doesn’t work right away.