Question

parameters reference vs primitive

what output is generated when the program is executed?

program:
public static void main(string args)
{
classa a = new classa(5, 4);
classa b = new classa(-1, 3);
a.methodb(b);
system.out.println(\b.mynum2 = \ + b.getmynum2());
}

class:
public class classa
{
private int mynum1;
private int mynum2;
public classa(int x, int y)
{
mynum1 = x;
mynum2 = y;
}

public void methodb(classa c)
{
mynum2 = c.getmynum1();
c.setmynum2(5 + 1);
mynum1 += 3;
}

public int getmynum1()
{
return mynum1;
}

public int getmynum2()
{
return mynum2;
}

public void setmynum2(int v)
{
mynum2 = v;
}
}

options:

• b.mynum2 = 6
• b.mynum2 = 1
• b.mynum2 = 5

Explanation:

🆕 New Concept Discovered: Object Reference Pass-by-Value
How Java passes object references to methods.

Step 1: Analyze Object Initialization

In the main method, two objects of type ClassA are instantiated:

• a is initialized as new ClassA(5, 4). This means for object a:
• myNum1 = 5
• myNum2 = 4
• b is initialized as new ClassA(-1, 3). This means for object b:
• myNum1 = -1
• myNum2 = 3

Step 2: Trace the Method Call a.methodB(b)

The method methodB is called on object a with object b passed as the argument c.

Inside methodB(ClassA c):

• The calling object is a (so this refers to a).
• The parameter c is a reference pointing to the same object as b.

Let's execute the statements inside methodB step-by-step:

1. myNum2 = c.getMyNum1();

• c.getMyNum1() retrieves myNum1 of object b, which is -1.
• This value is assigned to this.myNum2 (which is a.myNum2).
• Thus, a.myNum2 becomes -1.

1. c.setMyNum2(5 + 1);

• This calls the setter method on object c (which is object b) with the argument 6.
• This changes b.myNum2 to 6.

1. myNum1 += 3;

• This adds 3 to this.myNum1 (which is a.myNum1).
• a.myNum1 becomes 5 + 3 = 8.

Step 3: Determine the Output

After a.methodB(b) finishes executing, control returns to the main method:

• System.out.println("b.myNum2 = " + b.getMyNum2());
• Since b.myNum2 was updated to 6 in Step 2, b.getMyNum2() returns 6.

The printed output is:
b.myNum2 = 6

Answer:

b.myNum2 = 6