Java is pass-by-value.
Pass by value: make a copy in memory of the actual parameter"s value that is passed in.
Pass by reference: pass a copy of the address of the actual parameter.
This code will not swap anything:
void swap(Type a1, Type a2) { Type temp = a1; a1 = a2; a2 = temp; }
For this code, since the original and copied reference refer the same object, the member value gets changed:
class Apple { public String color = "red"; } public class main { public static void main(String[] args) { Apple a1 = new Apple(); System.out.println(a1.color); //print "red" changeColor(a1); System.out.println(a1.color); //print "green" } public static void changeColor(Apple apple) { apple.color = "green"; } }
Java does manipulate objects by reference, and all object variables are references. However, Java doesn"t pass method arguments by reference; it passes them by value.
Some more examplespublic class Main { public static void main(String[] args) { Student s = new Student("John"); changeName(s); System.out.printf(s); // will print "John" modifyName(s); System.out.printf(s); // will print "Dave" } public static void changeName(Student a) { Student b = new Student("Mary"); a = b; } public static void modifyName(Student c) { c.setAttribute("Dave"); } }
public static void changeContent(int[] arr) { // If we change the content of arr. arr[0] = 10; // Will change the content of array in main() } public static void changeRef(int[] arr) { arr = new int[2]; // If we change the reference arr[0] = 15; // Will not change the array in main() } public static void main(String[] args) { int [] arr = new int[2]; arr[0] = 4; arr[1] = 5; changeContent(arr); System.out.println(arr[0]); // Will print 10.. changeRef(arr); System.out.println(arr[0]); // Will still print 10.. }
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