Unreachable code compiler error [duplicate]

Answer 1

http://docs.oracle.com/javase/specs/jls/se8/html/jls-14.html#jls-14.21

says:

14.21. Unreachable Statements

It is a compile-time error if a statement cannot be executed because it is unreachable.

Answer 2

This error is mainly there to prevent programmer errors (a swap of 2 lines or more). In the second snippet, you make it clear that you don't care about the system.out.println().

Answer 3

Will it break the Java bytecode somehow, is it to protect the programmer or is it something else?

This is not required as far as Java/JVM is concerned. The sole purpose of this compilation error is to avoid silly programmer mistakes. Consider the following JavaScript code:

function f() {
    return 
        {
            answer: 42
        }
}

This function returns undefined as the JavaScript engine adds semicolon at the end of the line and ignores dead-code (as it thinks). Java compiler is more clever and when it discoveres you are doing something clearly and obviously wrong, it won't let you do this. There is no way on earth you intended to have dead-code. This somehow fits into the Java premise of being a safe language.

Answer 4

It's because it's a waste of resources for it to even be there. Also, the compiler designers don't want to assume what they can strip out, but would rather force you to remove either the code that makes it unreachable or the unreachable code itself. They don't know what is supposed to be there. There's a difference between the optimizations where they tweak your code to be a bit more efficient when it's compiled down to machine code and blatantly just removing code "you didn't need."

Answer 5

Its because the compiler writer assumed that the human at the controls is dumb, and probably didn't mean to add code that would never be executed - so by throwing an error, it attempts to prevent you from inadvertently creating a code path that cannot be executed - instead forcing you to make a decision about it (even though, as you have proven, you still can work around it).

Answer 6

Example 1:

In this case you are return before any statement because of it compiler never going to execute that code.

public static void main(String[] args) {

    return;
    System.out.println("unreachable");
}

In second code I have put the statement above of return and its work now :)

Example 2:

    public static void main(String[] args) {

    System.out.println("unreachable"); // Put the statement before return
    return;
}

The reason behind this is that if you return sometime then code after it never going to execute because you already return the function data and as so it is shown unreachable code.