long in a byte array

[Santoshkumar Jeevan Pawar]

I want to store a long in a byte array, transfer it over a network, then convert it back to a long.And In Java, a long is 8 bytes not 4.
But i want it a 4 bytes,is there any way to convert .

[W. Joe Smith]

EDIT:

I totally misread the OP. My response was not relevent.

[Jesper de Jong]

Well, 4 bytes is obviously not enough to store all possible values that a long can have in Java, so no, there is no way you're ever going to fit an 8-byte large long into 4 bytes.

[Rob Spoor]

Force the long into an int, then use ByteBuffer for the conversion of int to long:

int i = (int)myLong;
ByteBuffer buffer = ByteBuffer.allocate(4);
buffer.putInt(i);
byte[] b = buffer.array();

[naresh voota]

putting long in int will reduce the precision of the value stored.

long l = 9;
String s= Long.toString(l);
byte[] b = s.getBytes();

hopefully this will help. Thanks.

[naresh voota]

putting long in int will reduce the precision of the value stored.

long l = 9;
String s= Long.toString(l);
byte[] b = s.getBytes();

once you pass this byte array over network,

String s = new String(b);
Long l = Long.valueOf(s);

hopefully this will help. Thanks.

[Steve Luke]

naresh voota wrote:putting long in int will reduce the precision of the value stored.

long l = 9;
String s= Long.toString(l);
byte[] b = s.getBytes();

once you pass this byte array over network,

String s = new String(b);
Long l = Long.valueOf(s);

hopefully this will help. Thanks.

That works for values less than 10,000. Once you get to 10,000 you need 5 bytes to hold the String, so, you loose even more than you do when you convert to int. Oh, and any value less than 1,000 would take 3 or fewer bytes, so then you would have to test if < 1000, pad byte array, else if > 9999, throw bytes away.

[Bauke Scholtz]

Steve Luke wrote:
That works for values less than 10,000.

Correction: less than Integer.MAX_VALUE.
One byte is 8 bits, you know.

[Mike Simmons]

No, if you look at the code Steve is referring to, one byte ends up representing one character in a String representation. So the largest number that can be represented this way is 9999, at four digits. Of course this isn't a very good way to try to fit a long into 4 bytes. Simply casting to int would be easier, and covers a wiser range of possible values.

[Steve Luke]

Bauke Scholtz wrote:

Steve Luke wrote:
That works for values less than 10,000.

Correction: less than Integer.MAX_VALUE.
One byte is 8 bits, you know.

That is true for the answers suggesting to use a cast to int. Naresh, whom I was responding to, suggested converting the long to a String, then the String to byte[]. When you do this and have more than 4 characters (digits) you get more than 4 bytes. See:

public class LongToByteArray 
{
	
	static byte[] toByteArray_STR(long l)
	{
		String s= Long.toString(l); 
		byte[] b = s.getBytes();
		return b;
	}
	
	static long toLong_STR(byte[] b)
	{
		String s = new String(b); 
		Long l = Long.valueOf(s);
		return l;
	}

	public static void main(String[] args) 
	{
		long l = 10000L;
		byte[] buffer = toByteArray_STR(l);
		System.out.println("Length of byte[] for "+l+" is: "+buffer.length);
	}

}

output:

Length of byte[] for 10000 is: 5

[Santoshkumar Jeevan Pawar]

Hello all,

Here we lose the actual value,please check below code when we typecast from long to int.


long l = new Date().getTime(); //1238583275781
int i = (int)myLong; // value becomes 1633321080

System.out.println(new Date(i));
System.out.println(new Date(myLong));

But at the same time i dont want to lose actual value......

Thank You.
Have a Nice Time.

-Santosh

[Rob Spoor]

Santoshkumar Jeevan Pawar wrote:But at the same time i dont want to lose actual value......

Then you have no choice but to increase your array size to 8. A long simply requires 64 bits (= 8 bytes), nothing more and nothing less. By trying to squeeze it into an int you lose information. You can also convert the long to a float (4 bytes), but then you lose precision on the lower end:

long l = 1238583275781L;
System.out.println(l);
float f = l;
l = (long)f;
System.out.println(l);

Output:
1238583275781
1238583214080

So the only way to keep your actual value is: make sure you send the actual complete long.