AP Computer Science Exam Question

// Consider the following Java code:

public class VariableExample {
  
  public static void main(String[] args) {
    int num1 = 5;
    
    for(int i = 0; i < 3; i++) {
      int num2 = num1 + i;
      System.out.println("Value of num2: " + num2);
    }
    
    System.out.println("Value of num1: " + num1);
    System.out.println("Value of num2: " + num2);
  }
}

// What is the output of the above code and explain why?

a) Compilation error
b) It prints:
   Value of num2: 5
   Value of num2: 6
   Value of num2: 7
   Value of num1: 5
   Compilation error

c) It prints:
   Value of num2: 5
   Value of num2: 6
   Value of num2: 7
   Value of num1: 5
   Value of num2: 7

d) It prints:
   Value of num2: 5
   Value of num2: 6
   Value of num2: 7
   Value of num1: 5
   Compilation error

Answer

The correct answer is d) It prints:

Value of num2: 5
Value of num2: 6
Value of num2: 7
Value of num1: 5
Compilation error

Explanation: The variable num1 has a scope within the main method, so it can be accessed throughout the method. Its initial value is 5.

Inside the for loop, a new variable num2 is declared and assigned the value of num1 + i. The scope of num2 is limited to the for loop block. It exists only within the loop and cannot be accessed outside of it.

During each iteration of the loop, num2 is assigned a new value based on num1 + i. It is then printed using the System.out.println statement inside the loop. Thus, the output will be:

Value of num2: 5
Value of num2: 6
Value of num2: 7

After the for loop, num1 is printed using System.out.println. Since num1 is still in scope, its value remains the same and will be printed as 5:

Value of num1: 5

However, when we attempt to print num2 after the loop, a compilation error occurs. This is because num2 is only accessible within the for loop block, and cannot be accessed outside of it. Hence, the code will not compile, and the output will be:

Compilation error

Therefore, option d) is the correct answer.