Byte array with padding of null bytes at the end: how to efficiently copy to smaller byte array

Question

Have:

[46][111][36][11][101][55][87][30][122][75][66][32][49][55][67][77][88][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0]

Want:

[46][111][36][11][101][55][87][30][122][75][66][32][49][55][67][77][88]

I have an array of bytes size 8192 to start, and starting at some index in that first array until the end of the array the bytes are all null bytes. So there might be 6000 bytes with values and 2196 null bytes at the end in the array. How do I efficiently create a new array of size (6000) and copy those bytes over? Note: I do not know how many null bytes, or bytes with values there will be.

What are the remaining bytes in your new array going to equal? What you're looking to do doesn't make much sense. — user142162
– user142162, Commented Jun 8, 2013 at 19:44
A good old for loop? It will be simpler if you can't have zeros in the data part. — Pragmateek
– Pragmateek, Commented Jun 8, 2013 at 19:44
A new byte[] will have 0 in all elements, so you won't actually accomplish anything if you want an 8192 element array at the end. — kdgregory
– kdgregory, Commented Jun 8, 2013 at 19:45

Pragmateek · Accepted Answer · 2013-06-08 22:26:42Z

34

Here is my try:

static byte[] trim(byte[] bytes) { int i = bytes.length - 1; while (i >= 0 && bytes[i] == 0) { --i; } return Arrays.copyOf(bytes, i + 1); } public static void main(String[] args) { byte[] bytes = { 0, 1, 2, 0, 3, 4, 5, 0, 6, 0, 0, 7, 8, 9, 10, 0, 0, 0, 0 }; byte[] trimmed = trim(bytes); return; }

answered Jun 8, 2013 at 22:26

Pragmateek

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3 Comments

Sagar Trehan Over a year ago

Is it safe to remove all bytes from end which are 0. This won't anything?

Sagar Kharab Over a year ago

@Pragmateek : I think in some cases it's going to make it lose the data, I have an xlsx file which has last octet of hex as D5000000 however if I use this as leave only D5, file corrupts.

Pragmateek Over a year ago

@SagarKharab Indeed tweaking a binary file always calls for caution, but this is most probably another use-case as the one of the PO. :)

Rich Benner · Accepted Answer · 2016-06-23 08:46:49Z

8

I think we can do in this way also

byte []array={0, 69, 0, 71, 0, 72}; byte ar[]=new String(array).replaceAll("\0", "").getBytes();

edited Jun 23, 2016 at 8:46

Rich Benner

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answered Jun 23, 2016 at 8:45

Dattatray Satpute

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1 Comment

Aaron Over a year ago

I like the use of chaining. This is clear and makes it a one liner!

Mohammed Falha · Accepted Answer · 2013-06-08 20:22:15Z

why not try the static method array copy in the system class just give the source array src start position , destination array , destination start position and the length

 System.arraycopy(src, srcPos, dest, destPos, length); byte [] dest= new byte [6000]; System.arraycopy(src, 0, dest, 0, 6000);

The asker specified that he does not know how many non-zero bytes there are. This approach would only work if he knew beforehand.

Uri Loya · Accepted Answer · 2022-11-14 09:04:07Z

Another option using java string trim():

byte[] array={65, 66, 67, 0, 0, 0}; byte[] trimmedArray = new String(array).trim().getBytes();

Nilesh Deokar · Accepted Answer · 2018-10-23 13:34:32Z

With kotlin :

val strWithoutZeros = String(byteArray,StandardCharsets.UTF_8).replace(0.toChar().toString(), "") val byteArrayWithoutZeros = strWithoutZeros.toByteArray(StandardCharsets.UTF_8)

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