unreachable code?

[jon ruane]

can someone please explain why the System.out statement at line //1 didnt throw an 'unreachable code error'. i felt for sure that it would, as the call above it redirects it to another method.

are 'unreachable code' errors usually found in loops?

thank you.



public class MyProgram {
public static void throwIt() {
throw new RuntimeException();
}

public static void main(String[] args) {
try {
System.out.println("hello world");
throwIt();
System.out.println("done with block"); //1
}
catch (RuntimeException e) {
System.out.println("catch");
}
finally {
System.out.println("finally");
}
}
}

[Chandra Bhatt]

Hi Jon,

originally posted by jon
 
public class MyProgram {
public static void throwIt() {
throw new RuntimeException();
}
 
public static void main(String[] args) {
try {
System.out.println("hello world");
throwIt();
System.out.println("done with block"); //1
}catch (RuntimeException e) {
System.out.println("catch");
}
finally {
System.out.println("finally");
}
}
}

Output:
hello world
catch
finally


You code compiles and runs fine.

You did convert a line 1, you should have written

throw new RuntimeException();

instead of a method call that throws the RuntimeException.
Then only you are right that "next line of code is not reachable"

Regards,
cmbhatt
[ April 11, 2007: Message edited by: Chandra Bhatt ]

[Peter Chase]

The compiler doesn't go as far as looking into the throwIt() method and determining that it always throws.

[Jesper de Jong]

The 'unreachable code' error has its limits. The compiler checks for some cases of unreachable code, but it will not go and trace into the throwIt() method to see if there's something there that might make some code in main() unreachable. If the compiler would have to check all possible kinds of unreachable code, the compiler would have to be incredibly complicated - it would have to trace all kinds of strange code paths. What if your code looked like this:

public class MyProgram {
    public static void method1() {
        throw new RuntimeException();
    }
 
    public static void method2() {
        method1();
    }
 
    public static void main(String[] args) {
        method2();
        System.out.println("Hello World!");  // Unreachable code?
    }
}

Would you expect the compiler to check if method2 never returns normally, and to do that it have to check method1, etc. What if method1 was in a different class or somewhere else, probably in a JAR file in a third-party library? Would the compiler have to examine the compiled classes in that library to check if your println line was unreachable?

You can make up other arbitrarily complex examples:

import java.util.Calendar;
 
public class MyProgram {
    public static void method1() {
        if (Calendar.getInstance().get(Calendar.MONTH) % 2 == 0) {
            throw new RuntimeException("Try again next month");
        }
        else {
            while (true) { }    // Never return
        }
    }
 
    public static void main(String[] args) {
        method1();
        System.out.println("Hello World!");  // Unreachable code?
    }
}

[ April 11, 2007: Message edited by: Jesper Young ]

[Chandra Bhatt]

Thanks Jesper Young,

By the way I clicked to this link; I got quite nicely given
explanation. What are limitations of the compiler while
recognizing the non-reachable code.

It can only recognize like:

while(true) {
}
System.out.println("Can you get me!"); //Not reachable code

try{
throw new ParseException();
System.out.println("Sure! You can't get me!"); //Unreachable code
}catch(Exception e) {}

Can you count more cases? please give here!!!


Thanks,
cmbhatt