abstract class ReferencePipeline<P_IN, P_OUT> extends AbstractPipeline<P_IN, P_OUT, Stream<P_OUT>> implements Stream<P_OUT> … It’s ReferencePipeline that implements them. For example: @Override public final boolean anyMatch(Predicate<? super P_OUT> predicate) { return evaluate(MatchOps.makeRef(predicate, MatchOps.MatchKind.ANY)); } 2 solved How does “Stream” in java8 work?
Interfaces let you define methods that will be common to a group of classes. The implementation of the interface can be different for each class, and it’s up to those classes to independently implement what that methods do. interface Animal { // Define the interface Boolean canBite(); } class Dog implements Animal { // Define … Read more
You are not permitted to mutate a list while you are iterating over it. Your l.remove(j) causes the list l to change, but you’re inside a for (Integer j : l) loop. 2 solved java+ ConcurrentModificationException forEach(enhanced) loop single thread [duplicate]
public static long weightedSum(int[] a, int[] b, int n) { if (n == 0) return 0; else return a[n – 1] * b[n – 1] + weightedSum(a, b, n – 1); } Output : int[] arr1 = { 1, 2, 3, 4, 5 }; int[] arr2 = { 6, 7, 8, 9, 10 }; System.out.println(weightedSum(arr1, … Read more
An HTTP POST request cannot be redirected. 10.3 Redirection 3xx This class of status code indicates that further action needs to be taken by the user agent in order to fulfill the request. The action required MAY be carried out by the user agent without interaction with the user if and only if the method … Read more
As much as i understand it should be something like this System.out.println(“Write year between 1950-2050: “); int keyboard = input.nextInt(); int OL = (keyboard); int WC = (keyboard); int nothingspec = (keyboard); int instru = (keyboard); boolean blOL = false; boolean blWC = false; //this occurs whenever the number can be divided by 4 if(keyboard>=1950&&keyboard<=2050){ … Read more
In display method acdyear= sharedPreferencess1.getString(“acamadic”,”0″); 0 solved NumberFormatException: Invalid int: “”
| is a bitwise or, || is a logical or. You should know the difference. Work this way though: if(!Character.isLetter(c) && !Character.isDigit(c)) return false; => If the character is not a letter nor a digit return false 2 solved Java password checking, code modification
This is one simple example for your singleton. Feel free to add setters and getters as you need it for the fields. I’m not sure what the set-method in your diagram should do but maybe you don’t need it anyway. public class LibraryInfo { private static final LibraryInfo instance = new LibraryInfo(); public static LibraryInfo … Read more
You need to declare hexText as a String in main. You need to declare character as a char in hexStr(String). import java.util.Scanner; public class charToText { public static void main(String[] args) { Scanner input = new Scanner(System.in); String text, hexStr; System.out.print(“Enter some text: “); text = input.nextLine(); System.out.println(); System.out.println(“Hex value”); String hexText = hexStr(text); System.out.println(hexText); … Read more
The results of pow are often irrational, fractional or too large to store as a long, If you want to use powers of integers you can use BigInteger bi = BigInteger.valueOf(100); BigInteger bi2 = bi.pow(20); // 10^40 Another reason maybe that Math has many function which are taken from C, and also from common CPU … Read more
This code doesn’t use the enhanced for loop properly: for (int i : myInts) { if (myInts[i] != 0) { newInts[i] = myInts[i]; } } It tries to read element i from myInts, but i is the content of an element, not its index. So, as soon as some element contains a value > array’s … Read more
Your String result = “”; is null check it Your are returning null string change it You should return sb.toString(); instead of return result; EDIT public class PreDefinedAttributes { private Context mContext; private String mobile_os, mobile_model, mobile_brand, mobile_version, mobile_manufacturer; private String sdk_version, src, appname, appversion; private String lat = “”, lng = “”, device_id; private … Read more
this is what you try to parse: “+91 72072 21721 +91 79 8105 5662 ” step one, perform a .split(“\\+”); on your String, it will actually split it in the seperate numbers. Next, remove all the spaces of the number you try to parse, so you’ll end up parsing “917207221721” and “917981055662” This will bring … Read more
Not that this is the best idea to verify names. But assuming you’ve settled your requirements, then you can use next example: if(input.matches(“[a-zA-Z]+(\\-[a-zA-Z]+)?”)) { //OK } else { //Invalid } Several examples I’ve tested using this page: String matches? qwe Yes qwe- No qwe-qwe Yes qwe-qwe- No qw2e-qwe2 No qwe-qwe-qwe No solved How to create … Read more