Enthuware

How come the compiler knows about the value of the counter, before the program has been run? or how will it evaluates if condition before the program being executed?

It doesn't know the value of counter but it knows that there is an if part and there is an else part. The if part contains break and the else part contains continue, so no matter what is the value of counter, nothing after the if/else statement can be executed.

Is it really one of the OCA exam objectives to know when the compiler can identify unreachable code?

Yes, you may get questions that require you know when the code will not compile because of this reason.

in the syntax (value<4) in the question does it mean that the program is trying to access the whole value of the array & check if its less than 4 ?

AnneLy wrote:in the syntax (value<4) in the question does it mean that the program is trying to access the whole value of the array & check if its less than 4 ?

No, the code is inside the forEach loop i.e. forEach(int value : arr){ ... }. So value > 4 compares a value in the array with 4.

and remember break and continue only works for loops.

that makes no sense.
the following run well , but the code below continue is also unreachable,which is never executed.


int[] arr = {1,2,3,4,5};
for (int value :arr
) {
if (value ==0){
System.out.println("0000");
}else {
continue;
}
if (true){
System.out.println(true);
}
System.out.println("last line");
}

if(true) is an exception to the rule. See this (read the part towards the end about conditional compilation): https://docs.oracle.com/javase/specs/jl ... #jls-14.21
This point is covered in other questions.