Java - fill in specific indices for a byte array

Question

Suppose I have a byte array with 65 indices, and I want to fill the first 62 indices (or bytes) with data since index 63-65 are reserved. How can I move a byte array to the first 62 indices of the byte array?

String message = "Hello to the client. This is the message that you will receive"; //62 bytes message
byte[] b = message.getBytes();

byte[] sendData = new byte[65];
//how can I transfer byte[] b to 0-62?

There is no such thing as a 62 bytes message, because getBytes depends on an encoding. Can you be sure that 62 bytes will always be enough? — Paul Boddington
– Paul Boddington, Commented Nov 26, 2015 at 21:36
I know, but that result is platform-dependent. See the docs for String.getBytes. — Paul Boddington
– Paul Boddington, Commented Nov 26, 2015 at 21:39

user2390182 · Accepted Answer · 2015-11-26 21:54:26Z

1

Loop through the 62 indeces (0-61 btw) and copy the cells.

for (int i = 0; i < b.length; i++) {
    sendData[i] = b[i];
}

edited Nov 26, 2015 at 21:54

answered Nov 26, 2015 at 21:41

user2390182

73.7k6 gold badges71 silver badges95 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Dylan Wheeler · Accepted Answer · 2015-11-26 21:42:11Z

0

I'm not entirely sure what you're getting at, but I think the Arrays.copy method will help you. With this method, you can copy certain parts of an array to other arrays, saving the sections you're concerned about.

answered Nov 26, 2015 at 21:42

Dylan Wheeler

7,11414 gold badges58 silver badges81 bronze badges

Collectives™ on Stack Overflow

Java - fill in specific indices for a byte array

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related