Question

Java try and catch problem

public void method1(){
int[] array = new int[1];

try{ array[2] = 0;}

catch(ArithmeticException e){array[2] = 0} // ?

finally{ return 0;}}

Could you please tell me why the line with the question mark will not report an error?

Answer 1

Detailed explanation :-

try{

//The code that might generate an exception is placed here .

// In the given code , array[2]=0 generates the exception . It is because the we try to store value 0 in a unallocated memory.

// When such exception occurs, it is caught by the catch block that immediately follows it.

}

catch{

//catch block handles such exceptions .

// Catch block is executed only when their is an exception in try block.Otherwise , catch block is skipped .

// catch block is used to handle the Exception by declaring the type of exception within the parameter.

// In the above code , ArithmeticException class is mentioned in the parameter .Hence , all the arithemtic Exceptions like array overflow are handled .Therefore , the error array[2]=0 is handled by the catch block and the code executes without any errors and the rest of program continues .

}

finally{

// finally block is always executed whether exception is found or not Java finally block follows try or catch block.

}

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Thank you.