About Question enthuware.ocajp.i.v7.2.964 :

[ETS User]

System.out.println("c") is always unreachable , so why not it cause compiler error.
It the answer is: The compiler cannot decide during compile time that the statement is unreachable, then how do we decide if the compiler will be able to decide or not during compile time?

[admin]

The detailed rules are given here: Section 14.21 of JLS

It may seem complicated but the general idea is to look for compile time constants i.e. final variables and literals. For example,

Code: Select all

int x = 5;
while (x > 100){ System.out.println(x); }

will work. But this will not:

Code: Select all

final int x = 5;
while (x > 100){ System.out.println(x); }

because compiler knows that x cannot change from 5.

On top of this, there are some exceptions such as the if statement. It is explained in the above mentioned section.

HTH,
Paul.

[jeremy_dd]

I pasted the code in my eclipse, it fails to compile because m1 is supposed to throw an exception, which is not caught...

[admin]

I pasted the code in my eclipse, it fails to compile because m1 is supposed to throw an exception, which is not caught...

The question and the explanation are correct. Please make sure you paste the code exactly as given.

HTH,
Paul.

[KKreator]

I pasted the code in my eclipse, it fails to compile because m1 is supposed to throw an exception, which is not caught...

I think you forgot to add "...throws Exception" to the main function.

[JavaSoftware]

Hi! But if we catch the exception in catch block and print some message say "message" , then the statement System.out.println("C"); will be executed ?

[admin]

What happened when you tried it out?
Paul.

[JavaSoftware]

Thank you for your promotion to write and test code by myself and sorry for my hastiness.
And the statement System.out.println("C"); will be printed after we catch the exception

[admin]

Very good
This exam is all about writing short code examples to understand the basic concepts.