[Solved] What is wrong with this Java code snippet?try { int result = 10 / 0; } catch (Exception e) { System.out.println("Exception caught"); } catch (ArithmeticException e) { System.out.... | Java

Java - Exception Handling

What is wrong with this Java code snippet?

try {
  int result = 10 / 0;
} catch (Exception e) {
  System.out.println("Exception caught");
} catch (ArithmeticException e) {
  System.out.println("Arithmetic Exception caught");
}

AThe catch blocks are ordered incorrectly; specific exceptions must come before general ones

BThe try block does not throw any exception

CArithmeticException cannot be caught

DThere is no error; code is correct

Step-by-Step Solution

Solution:

Step 1: Understand catch block order
In Java, catch blocks must be ordered from most specific to most general exceptions.
Step 2: Analyze code
The general Exception catch block comes before ArithmeticException, causing a compile-time error.
Final Answer:
The catch blocks are ordered incorrectly; specific exceptions must come before general ones -> Option A
Quick Check:
Specific exceptions first, then general [OK]

Quick Trick: Order catch blocks from specific to general [OK]

Common Mistakes:

Placing general Exception catch block before specific exceptions
Assuming catch block order does not matter
Ignoring compile-time errors from incorrect order

Master "Exception Handling" in Java

9 interactive learning modes - each teaches the same concept differently

Learn Why Deep Visual Try Challenge Project Recall Time

More Java Quizzes

What is wrong with this Java code snippet?

Step 1: Understand catch block order

Step 2: Analyze code

Final Answer:

Quick Check:

Want More Practice?