Question

parameters reference vs primitive
what output is generated when the program is executed?

program:
public static void main(string args)
{
anotherclass n = new anotherclass(8, \yes\);
int a = 1;
int b = n.taska(a);
system.out.println(\b = \ + b);
}

class:
public class anotherclass
{
private int mynum;
private string myname;
public anotherclass(int x, string s)
{
mynum = x;
myname = s;
}
public int taska(int y)
{
y = y + 3;
return y;
}
public int getmynum()
{
return mynum;
}
public string getmyname()
{
return myname;
}
}

options:

• b = 1
• b = 9
• b = 4

Explanation:

🆕 New Concept Discovered: Java Parameter Passing (Value vs. Reference)
How Java passes variables to methods

Step 1: Analyze the main program execution

We start inside the main method:

1. An object n of type AnotherClass is instantiated:

   AnotherClass n = new AnotherClass(8, "YES");

This sets the instance variable myNum to 8 and myName to "YES".

1. A primitive integer variable a is declared and initialized to 1:

   int a = 1;

1. The method taskA is called on object n with a passed as an argument:

   int b = n.taskA(a);

Step 2: Trace the method execution

Let's look at the definition of taskA in AnotherClass:

public int taskA(int y)
{
    y = y + 3;
    return y;
}

1. When n.taskA(a) is called, the value of a (which is 1) is copied into the local parameter y.
2. Inside the method, y is updated:

$$ y = 1 + 3 = 4 $$

1. The method returns the value of y, which is 4.

Step 3: Determine the final output

1. The returned value 4 is assigned to the variable b in the main method:

   int b = n.taskA(a); // b becomes 4

1. The program prints the value of b:

   System.out.println("b = " + b);

This outputs: b = 4.

Answer:

b = 4