Remove some bytes from byte buffer and returning same byte buffer

Question

I have a byte buffer from which i need to remove some bytes of carriage return /r and return the same byte buffer after removal. With help i was able to remove the the /r using stream as below but that return a int[], is there any way where i do not need to create another byte buffer and use the same one after removing the /r? Below is the code i used

IntStream.range(bb.position(), bb.limit())
          .filter(i -> bb.get(i) != 13)
          .map(i -> bb.get(i)) 
          .toArray();

Let me know any other way to do this?

What type of buffer do you mean, could you provide the decleration of your buffer, the ´bb´? — user9069309
– user9069309, Commented Jun 25, 2020 at 6:32
"is there any way where i do not need to create another byte buffer and use the same one after removing the /r?" – I don't understand. My answer to your previous question tells you exactly how to do this (similar to Lino's answer here), but in the comments you said you want a new buffer instance. — Slaw
– Slaw, Commented Jun 25, 2020 at 6:44
@Slaw as you've provided the solution in your answer to the other question, I think we can close this as a dupe? — Lino
– Lino, Commented Jun 25, 2020 at 6:52
@Lino Not sure. My answer may solve this question but the two questions themselves are different (the solution to this question in my answer was more of a "bonus"). — Slaw
– Slaw, Commented Jun 25, 2020 at 6:59

Lino · Accepted Answer · 2020-06-25 07:04:07Z

1

You could use this helper method:

public static int removeAll(ByteBuffer buf, int b) {
    int removed = 0;
    for (int i = 0, start = 0, cap = buf.capacity(); i < cap; i++) {
        byte read = buf.get(i);
        buf.put(i, (byte) 0);
        if (read != b) {
            buf.put(start++, read);
        } else {
            removed++;
        }
    }
    return removed;
}

Which you can then call like this:

removeAll(buf, '\r');

It simply iterates the buffer, and removes the bytes that are equal to the provided argument, leaving 0 at the end of the buffer to account for the missing elements.

edited Jun 25, 2020 at 7:04

answered Jun 25, 2020 at 6:39

Lino

20k6 gold badges55 silver badges73 bronze badges

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

2 Comments

user4910279 Over a year ago

removeAll(ByteBuffer.wrap(new byte[] {1, 2, 3, 13, 4, 5, 13}, 13) -> {0, 0, 0, 4, 5, 0, 13} 　　Do you need buf.put(i, (byte) 0);?

Lino Over a year ago

@saka1029 good catch, fixed it, by always putting a 0 at index i into the buffer, and yes it is needed, else we would have trailing bytes at the end which we have moved earlier to another position

Collectives™ on Stack Overflow

Remove some bytes from byte buffer and returning same byte buffer

1 Answer 1

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related