. confuse that how actually reference is passed to a method ?

[naved]

i know this code is example of pass by value , so
how to make the same code as pass by reference ?
also explain how and what happens when you alter the same code as i new bie to java and confuse that how actually reference is passed to a method ?

class sample
{
	 int  value = 60;
	void changeVal(int val)
	{
		val=19;
		System.out.println("inside method" +val);
	}
	

	public static void main(String[] args)
	{
		int val = 10;
		sample s ;
		s = new sample(50);
		
		
		//System.out.println(" in main method " + s.value);
		//System.out.println(" in main method " + s1.value);
		s.changeVal(val);
		System.out.println(" in main method " + val);
		
		
	}
}

[Iron Lion]

You can't pass primitive values by reference in Java. You'll have to use objects instead.

public class A {

	public static void main(String[] args) {
		IntWrapper i = new IntWrapper();
		i.val = 1;
		addFive(i);
		System.out.println(i.val);
	}
	public static void addFive(IntWrapper i) {
		i.val += 5;
	}
}

class IntWrapper {
	int val;
}

Note that this won't work with the built-in Integer class because of auto-unboxing. If you were to use Integer instead of IntWrapper here, the expression i += 5 would create a new Integer with value equal to i+5, and then assign a reference to the new Integer to the local variable i. The original Integer still exists at its original location with an unchanged value.

[DarrylBurke]

You can't pass primitive values by reference in Java.

You can't pass any values by reference in Java. All passing is by value. Primitives are passed by value. References are passed by value.

There's a very good anecdotal explanation here:
JavaRanch Campfire - Cup Size: a Story About Variables
JavaRanch Campfire - Pass By Value, Please

Make sure to read both articles, in the correct order.

db

[Iron Lion]

Thanks for the correction.