Unreachable Code

Why does this code compile?
finally block is unreachable code, right?

Hello, and welcome to the Ranch!

Do you think the Java compiler considers unreachable code an error?

I believe so

If you want to run the code in finally clause, use "return" or throw some exception instead of exit().

Agree Petr, the intention is not to run the code but to understand about unreachable code for exam preparation (OCP Java21)

I think Java cannot declare a method as "noreturn". At least System.exit(int) is not declared so (see System.exit(int)). So the compiler does not know that System.exit() never returns.

Look Oracle system class. Method exit() terminates JVM. The compiler doesn’t treat the finally block as unreachable, because it cannot predict runtime behavior. Compilation only checks syntax, not what actually executes.

JLS 14.22 Unreachable Statements gives precise rules for how Java determines whether code is considered "unreachable" in a way that causes a compiler error.  Unfortunately, it's not easy reading.  But the most relevant part here is that when determining if code is unreachable, if the compiler sees any method call, it will never use any knowledge of what that method does internally, in the body of the method declaration.  In fact I think the only thing that it will look at from a method declaration is the throws list - it can use knowledge of checked exceptions thrown by methods to determine whether a given catch block is reachable or not.  But in general, for reachability, the compiler does not know or care what a method call does.  It essentially assumes that it could complete normally, or that it could throw an unchecked exception, or a checked exception if it's declared to be able to do so.  That's all that it knows about any method call, when determining if code is reachable.

The JLS even permits if (false) foo(); to be regarded as reachable; that is exp;ained in the section MS linked.

Mike Simmons wrote:. . . the compiler . . . can use knowledge of checked exceptions . . . to determine whether a given catch block is reachable or not. . . .

The following little example shows this, compiled on JShell:-

jshell> void foo()
  ...> {
  ...>     try
  ...>     {
  ...>         System.out.println("CodeRanch is brilliant.");
  ...>     }
  ...>     catch (java.io.IOException ex)
  ...>     {
  ...>         ex.printStackTrace();
  ...>     }
  ...> }
|  Error:
|  exception java.io.IOException is never thrown in body of corresponding try statement
|      catch (java.io.IOException ex)
|      ^-----------------------------...

jshell>

I think you can get catch (Exception ex) ... to compile, because Exception has unchecked subtypes. Of course, I avoid catch (Exception ex) ... and throwing and catching the same exception in the same method as far as possible.