call by reference

Hi friend,

callByvalue

When you call a method, the method sees a copy of any primitives passed to it. Thus any changes it makes to those values have no effect on the caller?s variables. This also applies to references passed as parameters. The caller cannot change the caller?s reference variables, i.e. make them point to different objects, but it can change the fields in the caller?s objects they point to. In Java, you cannot pass objects as parameters, only references to objects.

This is code of callByvalue

public class CallByValue {

public static void main(String[] args) {
int i= 10;
double d= 50.0;
System.out.println("This is call by value example");
System.out.println("Massage 1: i= " + i + ", d= " + d);
Double(i, d); // No return value
System.out.println("Massage 2: i= " + i + ", d= " + d);
Double(i, i); //Java converts int to double
System.out.println("Massage 3: i= " + i);
}
public static void Double(int ii, double dd){
System.out.println("Triple 1: ii= " + ii + ", dd= " + dd);
ii *= 3; // ii= ii*3;
dd *= 3.0;
System.out.println("Triple 2: ii= " + ii + ", dd= " + dd);
}
}

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call By Reference

When you call a method by reference, the callee sees the caller?s original variables passed as parameters, not copies. References to the callee?s objects are treated the same way. Thus any changes the callee makes to the caller?s variables affect the caller?s original variables. Java never uses call by reference. Java always uses call by value.

import java.awt.*;

public class callByReference {
public static void main(String[] args) {
Rectangle rec = new Rectangle(20,30,200,50);
System.out.println("Before move up massage");
System.out.println("r1 is now\n" + r1);
System.out.println("After move up massage display");
moveUp(rec);
System.out.println("r1 is now\n" + rec);
}
static void moveUp(Rectangle rect) {
// move up ten units
rect.translate(0,-20);
}
}

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