Exception propagation in Java - Mini Project: Build & Apply

Exception Propagation in Java

📖 Scenario: Imagine you are building a simple banking application. You want to check if a withdrawal amount is valid. If the amount is negative, an exception should be thrown and passed up the call chain until it is handled.

🎯 Goal: Build a Java program that demonstrates how exceptions can be thrown in one method and caught in another method higher up the call stack.

📋 What You'll Learn

Create a method that throws an exception when a negative withdrawal amount is passed

Create a method that calls the first method and propagates the exception

Handle the exception in the main method

Print a message when the exception is caught

💡 Why This Matters

🌍 Real World

Exception propagation is used in real-world applications to handle errors like invalid inputs or failed operations without crashing the program immediately.

💼 Career

Understanding exception propagation is essential for writing robust Java applications and is a common topic in software development interviews.

1

Create the withdraw method that throws an exception

Create a method called withdraw that takes an int amount parameter. If amount is less than 0, throw a new IllegalArgumentException with the message "Negative amount not allowed".

Java

public class Bank {

    public static void main(String[] args) {

    }

    // Create the withdraw method below
    
    // Your code here
}

Hint

Use throw new IllegalArgumentException("Negative amount not allowed") inside an if statement checking if amount < 0.

2

Create the processWithdrawal method that calls withdraw

Create a method called processWithdrawal that takes an int amount parameter and calls the withdraw method with this amount. Do not handle the exception here; let it propagate.

Java

public class Bank {

    public static void main(String[] args) {

    }

    public static void withdraw(int amount) {
        if (amount < 0) {
            throw new IllegalArgumentException("Negative amount not allowed");
        }
    }

    // Create the processWithdrawal method below
    
    // Your code here
}

Hint

Simply call withdraw(amount) inside processWithdrawal without try-catch.

3

Call processWithdrawal from main and handle the exception

In the main method, call processWithdrawal with the value -100. Use a try-catch block to catch IllegalArgumentException and print "Exception caught: " followed by the exception message.

Java

public class Bank {

    public static void main(String[] args) {
        // Call processWithdrawal with -100 and handle exception
        
        // Your code here
    }

    public static void withdraw(int amount) {
        if (amount < 0) {
            throw new IllegalArgumentException("Negative amount not allowed");
        }
    }

    public static void processWithdrawal(int amount) {
        withdraw(amount);
    }
}

Hint

Use try and catch (IllegalArgumentException e) around the call to processWithdrawal(-100). Print the message with System.out.println.

4

Print the final output

Run the program and print the output that shows the exception message caught in main.

Java

public class Bank {

    public static void main(String[] args) {
        try {
            processWithdrawal(-100);
        } catch (IllegalArgumentException e) {
            System.out.println("Exception caught: " + e.getMessage());
        }
    }

    public static void withdraw(int amount) {
        if (amount < 0) {
            throw new IllegalArgumentException("Negative amount not allowed");
        }
    }

    public static void processWithdrawal(int amount) {
        withdraw(amount);
    }

    // Program ends here
}

Hint

Run the program to see the exception message printed from the catch block.

Practice

1. What does exception propagation mean in Java?

easy

A. An exception is passed up the call stack until caught or program ends

B. An exception is ignored and the program continues normally

C. An exception is automatically fixed by the JVM

D. An exception is converted into a warning message

Solution

1. Step 1: Understand exception propagation concept

   When an exception occurs, Java looks for a matching catch block in the current method. If none is found, it passes the exception to the caller method.

2. Step 2: Follow the exception up the call stack

   This passing continues up the call stack until a catch block handles it or the program terminates if uncaught.

3. Final Answer:

   An exception is passed up the call stack until caught or program ends -> Option A

4. Quick Check:

   Exception moves up call stack = A [OK]

Hint: Exception moves up call stack until caught [OK]

Common Mistakes:

• Thinking exceptions are ignored automatically
• Believing JVM fixes exceptions silently
• Confusing exceptions with warnings

2. Which of the following is the correct way to declare a method that might throw an exception?

easy

A. public void readFile() throws IOException {}

B. public void readFile() throw IOException {}

C. public void readFile() throws IOException() {}

D. public void readFile() throws-IOException {}

Solution

1. Step 1: Recall correct syntax for throws clause

   The keyword throws is used followed by the exception class name without parentheses.

2. Step 2: Check each option syntax

   public void readFile() throws IOException {} uses correct syntax: throws IOException. The other options have syntax errors.

3. Final Answer:

   public void readFile() throws IOException {} -> Option A

4. Quick Check:

   Correct throws syntax = B [OK]

Hint: Use 'throws' keyword followed by exception class name [OK]

Common Mistakes:

• Writing 'throw' instead of 'throws'
• Adding parentheses after exception name
• Using invalid symbols like '-'

3. What will be the output of the following code?

public class Test {
  static void method() throws Exception {
    throw new Exception("Error occurred");
  }
  public static void main(String[] args) {
    try {
      method();
    } catch (Exception e) {
      System.out.println(e.getMessage());
    }
  }
}

medium

A. Compilation error

B. Exception in thread "main" java.lang.Exception

C. No output

D. Error occurred

Solution

1. Step 1: Analyze method throwing exception

   The method explicitly throws a new Exception with message "Error occurred".

2. Step 2: Check exception handling in main

   The main method calls method() inside try-catch. The catch block prints the exception message.

3. Final Answer:

   Error occurred -> Option D

4. Quick Check:

   Exception caught and message printed = C [OK]

Hint: Exception message prints if caught in try-catch [OK]

Common Mistakes:

• Assuming uncaught exception causes crash
• Confusing exception message with full stack trace
• Thinking code won't compile without throws in main

4. Identify the error in this code snippet:

public class Demo {
  static void risky() {
    throw new IOException("IO error");
  }
  public static void main(String[] args) {
    risky();
  }
}

medium

A. main method must catch IOException

B. IOException cannot be thrown directly

C. Method risky() must declare 'throws IOException'

D. No error, code is correct

Solution

1. Step 1: Check exception type thrown

   IOException is a checked exception and must be declared or caught.

2. Step 2: Verify method declaration

   Method risky() throws IOException but does not declare it with 'throws' keyword, causing a compile error.

3. Final Answer:

   Method risky() must declare 'throws IOException' -> Option C

4. Quick Check:

   Checked exceptions require throws declaration = A [OK]

Hint: Checked exceptions need 'throws' or try-catch [OK]

Common Mistakes:

• Ignoring throws declaration for checked exceptions
• Thinking IOException is unchecked
• Assuming main must catch exception always

5. Consider this code:

class A {
  void process() throws Exception {
    throw new Exception("Error in A");
  }
}
class B extends A {
  @Override
  void process() throws Exception {
    super.process();
  }
}
public class Main {
  public static void main(String[] args) {
    A obj = new B();
    try {
      obj.process();
    } catch (Exception e) {
      System.out.println(e.getMessage());
    }
  }
}

What will be the output and why?

hard

A. Runtime error due to invalid override

B. Error in A, because exception propagates from superclass method

C. Compilation error due to throws mismatch

D. No output, exception is swallowed silently

Solution

1. Step 1: Understand method overriding with exceptions

   Subclass B overrides process() and calls super.process(), which throws Exception.

2. Step 2: Exception propagates to main and is caught

   Main calls obj.process() on B instance, exception thrown by A.process() propagates and is caught in main's try-catch, printing the message.

3. Final Answer:

   Error in A, because exception propagates from superclass method -> Option B

4. Quick Check:

   Exception propagates through override and caught = D [OK]

Hint: Overridden method can throw same exceptions up [OK]

Common Mistakes:

• Thinking override cannot throw exceptions declared in superclass
• Assuming exception is lost in subclass
• Confusing compile error with runtime behavior