Question

use the following class and program to answer the question.
hint: does the value of original variable change if we pass primitive data type as the argument?

program:
public static void main(string args)
{
int age = 42;
add5(age);
system.out.println(\after add5 age = \ + age);
}

method:
public static void add5(int n)
{
n = n + 5;
}

what output is generated when the program is executed?

• after add5 age = 47
• after add5 age = 5
• after add5 age = 42

Explanation:

🆕 New Concept Discovered: Pass-by-Value in Java
How values are copied when passed to methods.

Step 1: Analyze the main method execution

In the main method, an integer variable named age is declared and initialized:

int age = 42;

This creates a variable in memory holding the primitive integer value 42.

Step 2: Analyze the method call behavior

The program then calls the method add5(age).

In Java, all arguments are passed by value. This means that when you pass a primitive variable (like an int) to a method, Java makes a copy of the value (42) and passes that copy to the method.

Inside the add5 method:

public static void add5(int n) {
    n = n + 5;
}

• The parameter n is a local variable that receives the copied value 42.
• The operation n = n + 5 changes the value of the local variable n to 47.
• This change only affects the local variable n inside the scope of add5. It has absolutely no effect on the original age variable in the main method.

Step 3: Determine the final printed output

After the add5(age) method finishes executing, control returns to the main method. The original variable age remains unchanged:

System.out.println("After add5 age = " + age);

Since age is still 42, the printed output is:
After add5 age = 42

Answer:

After add5 age = 42