Let’s declare two interfaces and a class that implements them both: interface I1 { } interface I2 { } class C implements I1, I2 { } objects can have multiple types In the following code, it can be seen that a C instance has the type of C as well as I1 and I2: C …
You could use remote debugging by running your jar like this java -Xdebug -Xrunjdwp:transport=dt_socket,server=y,suspend=n,address=5005 -jar yourJar.jar And then connecting from your IDE to that port
Java 8 and later: JDBCType & SQLType With improvements in the API’s, as of Java 8 and JDBC 4.2, we have JDBCType and SQLType, and in the same spirit as some of the other examples can be simply used as follows: String typeName = JDBCType.valueOf(-5).getName(); But of course, why using the numeric types to begin …
You need to get the parent directory from the WatchKey to resolve the full path WatchKey key; WatchEvent<Path> event; Path dir = (Path)key.watchable(); Path fullPath = dir.resolve(event.context()); This piece of code reads like it needs accompanying documentation to be grasped, it makes little sense on its own. What were their intentions with this particular API …
The standard answer is to take two iterators at the beginning, increment the first one once, and the second one twice. Check to see if they point to the same object. Then repeat until the one that is incrementing twice either hits the first one or reaches the end. This algorithm finds any circular link …
I found the cause of my problem. This problem occurs because JAX-WS generates a class for each method and the class name is constructed by concatenating methodName + “Response”. In my case, the newly generated class by JAX-WS will have the same name as my response object. Example: @Stateless @WebService() public class AccountWS { @WebMethod() …
Check out this article: Boxed values and equality When comparing wrapper types such as Integers, Longs or Booleans using == or !=, you’re comparing them as references, not as values. If two variables point at different objects, they will not == each other, even if the objects represent the same value. Example: Comparing different Integer …
String ip; try { Enumeration<NetworkInterface> interfaces = NetworkInterface.getNetworkInterfaces(); while (interfaces.hasMoreElements()) { NetworkInterface iface = interfaces.nextElement(); // filters out 127.0.0.1 and inactive interfaces if (iface.isLoopback() || !iface.isUp()) continue; Enumeration<InetAddress> addresses = iface.getInetAddresses(); while(addresses.hasMoreElements()) { InetAddress addr = addresses.nextElement(); ip = addr.getHostAddress(); System.out.println(iface.getDisplayName() + ” ” + ip); } } } catch (SocketException e) { throw new …
This is a nice way of doing it for any two-dimensional array, assuming you want them in the following order: [[array[0]-elems], [array[1]elems]…] public <T> List<T> twoDArrayToList(T[][] twoDArray) { List<T> list = new ArrayList<T>(); for (T[] array : twoDArray) { list.addAll(Arrays.asList(array)); } return list; }
You must use: String [] temp = s.split(“\\|”); This is because the split method takes a regular expression, and | is one of the special characters. It means ‘or’. That means you are splitting by ” or ”, which is just ”. Therefore it will split between every character. You need two slashes because the …