Question

The following code gives an unreachable statement compiler error

public static void main(String[] args) {
    return;
    System.out.println("unreachable");
}

Sometimes for testing purposes a want to prevent a method from being called, so a quick way to do it (instead of commenting it out everywhere it’s used) is to return immediately from the method so that the method does nothing. What I then always do to get arround the compiler error is this

public static void main(String[] args) {
    if (true) {
        return;
    }
    System.out.println("unreachable");
}

I’m just curious, why is it a compiler error?? Will it break the Java bytecode somehow, is it to protect the programmer or is it something else?

Also (and this to me is more interesting), if compiling java to bytecode does any kind of optimization (or even if it doesn’t) then why won’t it detect the blatant unreachable code in the second example? What would the compiler pseudo code be for checking if a statement is unreachable?

Answer 1

Unreachable code is meaningless, so the compile-time error is helpful. The reason why it won’t be detected at the second example is, like you expect, for testing / debugging purposes. It’s explained in The Specification:

if (false) { x=3; }

does not result in a compile-time error. An optimizing compiler may
realize that the statement x=3; will never be executed and may choose
to omit the code for that statement from the generated class file, but
the statement x=3; is not regarded as “unreachable” in the technical
sense specified here.

The rationale for this differing treatment is to allow programmers to
define “flag variables” such as:

static final boolean DEBUG = false;

and then write code such as:

if (DEBUG) { x=3; }

The idea is that it should be possible to change the value of DEBUG
from false to true or from true to false and then compile the code
correctly with no other changes to the program text.

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