2
\$\begingroup\$

I have this code that duplicates a byte array 5 times.

class Multiple {
    public static void main(String[] args) {
        byte[] result = new byte[] {0x41,0x42,0x43,0x44};
        int len = result.length;
        byte[] multipled = new byte[len*5];
        for (int i = 0; i < 5; i++) {
            for (int j = 0; j < len; j++) {
                multipled[i*len + j] = result[j];
            }
        } 
        System.out.println(new String(multipled));
        System.out.println(new String(result));
    }
}

Example:

ABCDABCDABCDABCDABCD
ABCD

The code uses multiple loops and assignment, can it be better or shorter?

\$\endgroup\$
0
4
\$\begingroup\$

It can be made shorter:

public static void main(String[] args) {
    byte[] result = new byte[] {0x41, 0x42, 0x43, 0x44};
    byte[] multipled = new byte[result.length * 5];
    for (int i = 0; i < multipled.length; i++)
        multipled[i] = result[i % result.length];
    ...
}
\$\endgroup\$
2
  • \$\begingroup\$ @NexusDuck using new String in this case is necessary, because an array's toString just prints the object address, not the content. \$\endgroup\$ Apr 11 '15 at 20:03
  • \$\begingroup\$ @DavidWallace I had no idea. Disregard my comment then \$\endgroup\$
    – NexusDuck
    Apr 11 '15 at 20:05
2
\$\begingroup\$

This operation is worth defining as a function for clarity and reuse.

To copy an array, use System.arraycopy().

public static byte[] repeat(byte[] array, int times) {
    byte[] repeated = new byte[times * array.length];
    for (int dest = 0; dest < repeated.length; dest += array.length) {
        System.arraycopy(array, 0, repeated, dest, array.length);
    }
    return repeated;
}
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.