Question
How can I efficiently convert a long timestamp value into a byte array and insert it into an existing byte array without using bit-wise manipulation?
long time = System.currentTimeMillis();
byte[] bArray = new byte[128];
Answer
To convert a long timestamp to a byte array and insert it into a specific section of another byte array, we can use the ByteBuffer class from Java's NIO package. This approach avoids inefficient bit-by-bit insertion and provides a cleaner, more efficient way to handle byte arrays.
import java.nio.ByteBuffer;
long time = System.currentTimeMillis();
byte[] bArray = new byte[128];
// Create a ByteBuffer with the specified order and insert the long value
ByteBuffer byteBuffer = ByteBuffer.wrap(bArray);
byteBuffer.putLong(0, time); // Insert timestamp at the beginning (MSBs) of the array
// Now bArray contains the timestamp in the first 8 bytes.
Causes
- Attempting to insert data bit-by-bit can be slow and inefficient.
- Inadequate understanding of Java's primitive and wrapper types might lead to confusion about manipulating byte arrays.
Solutions
- Use the ByteBuffer class to convert the long timestamp to a byte array efficiently.
- Utilize the `putLong` method to insert the timestamp into the specific indices of the target byte array.
Common Mistakes
Mistake: Not allocating enough space in the target byte array for the long value.
Solution: Ensure that the byte array has at least 8 bytes available for a long value.
Mistake: Using incorrect indexing when inserting the long into the byte array.
Solution: Always start inserting at 0 for the MSBs if that is your intention.
Helpers
- convert long to byte array
- insert long into byte array
- byte array manipulation
- Java byte array
- ByteBuffer example
- System.currentTimeMillis()
