Why does the following code cause a compiler error?
Try
{
Number = Integer.parseInt(str);
}
Catch (IllegalArgumentException e)
{
System.out.println("Bad number format.");
}
Catch (NumberFormatException e)
{
System.out.println(str + " is not a number.");
}
A) Because you can have only one catch clause in a try statement
B) Because NumberFormatException inherits from IllegalArgumentException. The code should handle NumberFormatException before IllegalArgumentException
C) Because the Integer.parseInt method does not throw a NumberFormatException
D) Because the Integer.parseInt method does not throw an IllegalArgumentException

Question

Quizplus · Accepted Answer

NumberFormatException is a subclass of IllegalArgumentException, therefore the catch clause for NumberFormatException should be before the catch clause for IllegalArgumentException.

Why Does the Following Code Cause a Compiler Error