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Found 9321 Articles for Object Oriented Programming
![Sreemaha](https://www.tutorialspoint.com/assets/profiles/13570/profile/60_64977-1512709068.jpg)
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You can simply iterate through list and fill the array as shown below −import java.util.ArrayList; import java.util.List; public class Tester { public static void main(String[] args) { List list = new ArrayList(); list.add(new Integer(1)); list.add(new Integer(2)); list.add(new Integer(3)); list.add(new Integer(4)); int[] array = new int[list.size()]; for(int i=0;i
![Prabhas](https://www.tutorialspoint.com/assets/profiles/13554/profile/60_90506-1512543588.jpg)
4K+ Views
Yes. From Java 8 onwards, we can do so using method references.Method references help to point to methods by their names. A method reference is described using "::" symbol. A method reference can be used to point the following types of methods −Static methodsInstance methodsConstructors using new operator (TreeSet::new)Method Reference ExampleCreate the following Java program using any editor of your choice in, say, C:\> JAVA.Java8Tester.java Live Demo import java.util.List; import java.util.ArrayList; public class Java8Tester { public static void main(String args[]) { List names = new ArrayList(); names.add("Mahesh"); names.add("Suresh"); names.add("Ramesh"); ... Read More
![seetha](https://www.tutorialspoint.com/assets/profiles/13549/profile/60_42938-1512640859.jpg)
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A method can give multiple values if we pass an object to the method and then modifies its values. See the example below −Examplepublic class Tester { public static void main(String[] args) { Model model = new Model(); model.data1 = 1; model.data2 = 2; System.out.println(model.data1 + ", " + model.data2); changeValues(model); System.out.println(model.data1 + ", " + model.data2); } public static void changeValues(Model model) { model.data1 = 100; model.data2 = 200; } } class Model { int data1; int data2; }Output1, 2 100, 200
![varun](https://www.tutorialspoint.com/assets/profiles/13559/profile/60_29674-1512637943.jpg)
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Call by Value means calling a method with a parameter as value. Through this, the argument value is passed to the parameter.While Call by Reference means calling a method with a parameter as reference. Through this, the argument reference is passed to the parameter.In call by value, the modification done to the parameter passed does not reflect in the caller's scope while in call by reference, the the modification done to the parameter passed are persistent and changes are reflected in the caller's scope.But Java uses only call by value. It creates a copy of references and pass them as ... Read More
![usharani](https://www.tutorialspoint.com/assets/profiles/13564/profile/60_32161-1512638316.jpg)
350 Views
Yes, break statement can be used in nested for loops. It works on the for loop in which it is applied. See the example below.Examplepublic class Tester { public static void main(String[] args) { for(int i = 0; i< 2; i++) { for(int j = 0; j < 2; j++){ if(i == 1) { break; } System.out.println("i = " + i+",j = " + j); } } } }
![Rahul Sharma](https://www.tutorialspoint.com/assets/profiles/14297/profile/60_102977-1513054044.jpg)
4K+ Views
Use List.clear() method to empty an array.Exampleimport java.util.ArrayList; import java.util.List; public class Tester { public static void main(String[] args) { List list = new ArrayList(); list.add("A"); list.add("B"); list.add("C"); list.add("D"); list.add("E"); list.add("F"); list.add("G"); list.add("H"); System.out.println("Original List " + list); list.clear(); System.out.println("Cleared List " + list); } }OutputOriginal List [A, B, C, D, E, F, G, H] Cleared List []
![Johar Ali](https://www.tutorialspoint.com/assets/profiles/14293/profile/60_101691-1513053093.jpg)
520 Views
Array of Object class can be created which can accept any type of object. During operation on such array, instanceof operator can be used.Examplepublic class Tester { public static void main(String[] args) { Object[] dataArray = new Object[3]; dataArray[0] = new Integer(0); dataArray[1] = new String("1"); dataArray[2] = new Boolean(false); for(Object data: dataArray){ if(data instanceof Integer){ System.out.println(((Integer) data).intValue()); } if(data instanceof String){ System.out.println(data); } if(data instanceof Boolean){ System.out.println(((Boolean) data).booleanValue()); } } } }Output0 1 false
![Ali](https://www.tutorialspoint.com/assets/profiles/14292/profile/60_101553-1513052445.jpg)
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Following example uses Contains method to search a String in the Array.Exampleimport java.util.ArrayList; public class Main { public static void main(String[] args) { ArrayList objArray = new ArrayList(); ArrayList objArray2 = new ArrayList(); objArray2.add(0, "common1"); objArray2.add(1, "common2"); objArray2.add(2, "notcommon"); objArray2.add(3, "notcommon1"); objArray.add(0, "common1"); objArray.add(1, "common2"); System.out.println("Array elements of array1"+objArray); System.out.println("Array elements of array2"+objArray2); System.out.println("Array 1 contains String common2?? " +objArray.contains("common1")); System.out.println("Array 2 contains Array1?? " +objArray2.contains(objArray) ); } }OutputThe above code sample will produce the following result.Array elements of array1[common1, common2] Array elements of array2[common1, common2, notcommon, notcommon1] Array ... Read More
![Every ; Thing](https://www.tutorialspoint.com/assets/profiles/27891/profile/60_61166-1522731475.jpg)
110 Views
Exampleclass Shape{ private String shapeName; private int numSides; Shape(String shapeName, int numSides){ this.shapeName = shapeName; this.numSides = numSides; } public String toString(){ return shapeName + " has " + numSides + " sides."; } } class ObjectList{ private Object[] list = new Object[10]; private int numElement = 0; public void add(Object next){ list[numElement] = next; numElement++; } @Override public String toString(){ String str=""; int i=0; while(list[i] != null){ str += list[i]+""; i++; } return str; } } public class Driver{ public static void main(String[] args){ ObjectList list = new ObjectList(); Shape square = new Shape("square", 4); Shape hex = new Shape("hexagon", 6); list.add(hex); list.add(square); System.out.println(list); } }