Convert long to byte array and add it to another array

Question

I want to change a values in byte array to put a long timestamp value in in the MSBs. Can someone tell me whats the best way to do it. I do not want to insert values bit-by-bit which I believe is very inefficient.

long time = System.currentTimeMillis();
Long timeStamp = new Long(time);
byte[] bArray = new byte[128];

What I want is something like:

byte[0-63] = timeStamp.byteValue(); 

Is something like this possible . What is the best way to edit/insert values in this byte array. since byte is a primitive I dont think there are some direct implementations I can make use of?

Edit:
It seems that System.currentTimeMillis() is faster than Calendar.getTimeInMillis(), so replacing the above code by it.Please correct me if wrong.

Solution 1

There are multiple ways to do it:

• Use a ByteBuffer (best option - concise and easy to read):

  byte[] bytes = ByteBuffer.allocate(Long.SIZE / Byte.SIZE).putLong(someLong).array();
  

• You can also use DataOutputStream (more verbose):

  ByteArrayOutputStream baos = new ByteArrayOutputStream();
  DataOutputStream dos = new DataOutputStream(baos);
  dos.writeLong(someLong);
  dos.close();
  byte[] longBytes = baos.toByteArray();
  

• Finally, you can do this manually (taken from the LongSerializer in Hector's code) (harder to read):

  byte[] b = new byte[8];
  for (int i = 0; i < size; ++i) {
    b[i] = (byte) (l >> (size - i - 1 << 3));
  }
  

Then you can append these bytes to your existing array by a simple loop:

// change this, if you want your long to start from 
// a different position in the array
int start = 0; 
for (int i = 0; i < longBytes.length; i ++) {
   bytes[start + i] = longBytes[i];
}

Solution 2

If you want to really get under the hood...

public byte[] longToByteArray(long value) {
    return new byte[] {
        (byte) (value >> 56),
        (byte) (value >> 48),
        (byte) (value >> 40),
        (byte) (value >> 32),
        (byte) (value >> 24),
        (byte) (value >> 16),
        (byte) (value >> 8),
        (byte) value
    };
}

Solution 3

For me ByteBuffer and other utils are expensive from time perspective. Here are 2 methods that you can use:

// first method (that is using the second method), it return the array allocated and fulfilled

public byte[] longToByteArray(long value) 
{
        byte[] array = new byte[8];

        longToByteArray(value,array,0);
        return array;
}

// this method is useful if you have already allocated the buffer and you want to write the long a specific location in the array.

public void longToByteArray(long value, byte[] array, int startFrom) 
{
    for (int i=7; i>=0; i--)
    {
        array[startFrom+7-i] = (byte) (value >> i*8);
    }
}

Comments

[codeObserver]

thanks Bozho. I think this is what I was looking for . I think the first method would be faster than the later one. i am just concerned about the performance impact of using a ByteBuffer instead of byte array. This link talks more about the performance in the ned bytebuffer lib: jroller.com/cpurdy/date/20040405#raw_nio_performance ; Also if the impact is not huge, can I just use ByteBuffer instead of a byte array ?

[Bozho]

Use ByteBuffer. I doubt it will cause any problems.

[mR_fr0g]

I found that I had to set the order on the ByteByffer like so, Otherwise the bytes were the in an order reverse to what i was expecting byte[] bytes = ByteBuffer.allocate(8).order(ByteOrder.LITTLE_ENDIAN).putLon‌​g(bits).array();

[white_gecko]

I assume Java is generally big endian, while x86 is little endian.

[mike jones]

To avoid hardcoding the size of a long I would use allocate(Long.SIZE)

[overthink]

@mikejones Note that Long.SIZE will give you the number of bits used to represent Long (docs.oracle.com/javase/6/docs/api/java/lang/Long.html#SIZE)‌​. You'd need (Long.SIZE / Byte.SIZE) to match the example above.

[Max]

@overthink Java 8 introduced a BYTES constant for this (docs.oracle.com/javase/8/docs/api/java/lang/Long.html#BYTES‌​).