Bitwise Copy

When u check the binary value of it using
Long.toBinaryString(longVal) and
Integer.toBinaryString(intVal), you will notice the diffrence

Yes, you are quite right. I apologize. The reason for the behaviour you see is that integers (and longs) are signed quantites. The value 0xffffffff represents -1 in an integer. When it is converted to a long, the sign bit is extended leftward (i.e. if the topmost bit is 0 the top 4 bytes of the long are set to 0, but if the top bit is 1 the top 4 bytes of the long are set to 1). Thus the long becomes 0xffffffffffffffff which also represents -1.

To force the long to contain 0xffffffff you will need to mask the top 4 bytes to 0:

long l=i&0xffffffffL;

Make sure you add the L to the constant which will cause the constant to be long (i.e. 8 bytes). This will cause the variable i to be widened (and sign extended) (in order to do the bitwise operation), then top 4 bytes will by masked off.

Without the trailing L, the bitwise operation would occur first (having no effect) and then the result would be widened (leaving you with the sign extension again.

Thanks very much, I was not getting the exact way to do a bitwise operation, I was doing something like this

private static long getBinaryLongValue(int val)
{
String s = Integer.toBinaryString(val);
long l = Long.parseLong(s,2);
return l;  
             
}