You can try the below code to convert string array to string public class JavaApplication20 { public static void main(String[] args) { // TODO code application logic here String test1 = “”; String[] test = {“this”,”this2″}; int length = test.length; //System.out.println(“Length :”+length); for(int i=0; i<length; i++){ //System.out.println(test[i]); test1 += test[i]+”,”; } int len = test1.length(); … Read more
Use Comparable and Comparator as shown below, you can also visit https://www.journaldev.com/780/comparable-and-comparator-in-java-example for further details. import java.util.Comparator; class Employee implements Comparable<Employee> { private int id; private String name; private int age; private long salary; public int getId() { return id; } public String getName() { return name; } public int getAge() { return age; } … Read more
Your first port of call should be the documentation which explains it reasonably clearly: Thrown to indicate that an array has been accessed with an illegal index. The index is either negative or greater than or equal to the size of the array. So for example: int[] array = new int[5]; int boom = array[10]; … Read more
Use it like this – String [] myArray = myList.toArray(new String[myList.size()]); solved JAVA casting list of strings to string array [duplicate]
Use this String[] params = data.split(” “); String firstString = params[0]; split() function will split your whole string using space as delimiter, resulting in array of strings. First element of the array is the string you are looking for. 2 solved Extract specific string from String [closed]
for (int i = 0; i < userInput; i++) { System.out.print(” ” +(i%7)+1); } Assuming it is java code that you are talking about. 2 solved using for loop in repeating number of day [closed]
It seems that you are looking implementation of thread pool. Fortunately JDK has one built-in. Take a look on Exectutors framework. solved Open source Java Thread server [closed]
No this is not valid how ever you can do following: String[] name=new String[]{“h”, “e”, “l”, “l”, “o”}; or String name=new String(new char[]{‘h’, ‘e’, ‘l’, ‘l’, ‘o’}); solved Can we declare the reference variable as String for the char array? [closed]
You do it by actually using the i parameter. private static final String ALPHABET = “abcdefghijklmnopqrstuvwxyz”; public static void binary(int n, String str, int i) { if (i == ALPHABET.length()) return; if (n == 0) { System.out.println(str); return; } binary(n – 1, str + ALPHABET.charAt(i), 0); // next letter starts at beginning of alphabet binary(n, … Read more
A simple test proves the loop indeed works as intended: public static void main(String[] args) { int enemysno = 5; for (int i = 0; i < enemysno; i++) { System.out.println(“lalala ” + i); } } this works fine producing lalala 0 lalala 1 lalala 2 lalala 3 lalala 4 It was kind of obvious, … Read more