1. How is this 13 bytes long? stackoverflow.comTwo quotes:
All of the remaining messages in the protocol take the form of |
2. Assigning a byte to a long coderanch.comHi, Originally posted by Mikael Jonasson: Try casting it. /Mike some things I have tried: public class AssignByteLong { public static void main(String[] args) { byte b = (byte)0xFF; long l = 0xFF; System.out.println("b = 0xFF: b = " + b); // -1 System.out.println("l = 0xFF: l = " + l); // 255 l = 0; l = b; System.out.println("l ... |
3. Byte to long coderanch.com |
4. Memory usage of byte and long variable. coderanch.comLet's assume arrays have no overhead. Each byte[]/long[] has 10*1024*128 = 1,310,720 different array values. In scenario 2 each element takes 1 byte, therefore the total is 1,310,720 bytes or 1.25MB. In scenario 1 each element takes not 1 but 8 bytes, therefore the total is 8 * 1,310,720 = 10,485,760 bytes or 10MB. Now alone this shouldn't cause an OutOfMemoryError. ... |
5. What happens internally when a long is typecaseted to a byte coderanch.comlong a = 12345678L; byte[] bytes = new byte[8]; for (int i = 0; i < 8; i++) { // Get the lowest 8 bits of the long bytes[8 - i] = (byte) a; // Shift the long 8 bits to the right a >>= 8; } for (byte b : bytes) { System.out.println(b + "\t"); } |
6. How to get the bytes of a long? forums.oracle.comHi, I need an 8-byte number for my application. First I create an int and then cast it to long. After that this long number must be represented by 8 bytes, or 64 bits, in memory. As far as I know the leading zeros in its binary representation are present, even though they mean nothing. However, I need to get those ... |
7. sign problems byte[6] -> long forums.oracle.com |
8. From byte to long forums.oracle.comThis solution works but this is not what I am looking for. Imagine that you do not know the value of the byte. This byte is only a variable. And the purpose is to set this byte at the end of the long bit String, that is, the first 8 bits starting from the right. Any ideas? |