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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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