Why does `instanceof` error rather than return `false` when used for 2 incompatible classes? [duplicate]

Question

I'm reading this:
http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.20.2

They say:

Consider the example program:
class Point { int x, y; }
class Element { int atomicNumber; }
class Test {
        public static void main(String[] args) {
                Point p = new Point();
                Element e = new Element();
                if (e instanceof Point) {       // compile-time error
                        System.out.println("I get your point!");
                        p = (Point)e;           // compile-time error
                }
        }
}
The instanceof expression is incorrect because no instance of Element or any of its possible subclasses (none are shown here) could possibly be an instance of any subclass of Point.

Why does this result in an error, rather than simply in instanceof returning false?

It sounds like you're previous experience might be with languages which were mostly loosely typed and mostly interpreted (ex: most BASICs) as opposed to strongly typed and compiled. With a language that is mostly interpreted, almost all type-like validation occurs at run-time. However, with compiled languages, the compiler is able to do quite a bit of "type and execution flow analysis". As a result, if it finds pathways that can never be executed, it can remove the code in that pathway (i.e. dead code removal). And it generates an error if it finds an operation that cannot ever succeed. — chaotic3quilibrium
– chaotic3quilibrium, Commented Dec 16, 2010 at 16:36

AlexW · Accepted Answer · 2016-07-11 10:45:35Z

14

instanceof check is a runtime check. The compiler is able to discover that this condition is incorrect at compile time (much earlier), so it tells you that it is wrong. Always remember, that failing fast is a good practice, it will save you a lot of time and nerves.

edited Jul 11, 2016 at 10:45

AlexW

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answered Dec 16, 2010 at 11:42

Vladimir Ivanov

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1 Comment

JDelage Over a year ago

Thank you - this explains the rational for this decision. It completely makes sense, indeed, to error at compile time rather than have a false result at run time.

Spiny Norman · Accepted Answer · 2010-12-16 11:40:03Z

11

I'd say, because you know at compile-time that it will never be true. Therefore, it's safe to assume this is not what the programmer meant :)

However, there probably is a more java-technical explanation.

answered Dec 16, 2010 at 11:40

Spiny Norman

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Comments

pmariano · Accepted Answer · 2010-12-16 11:40:17Z

4

Because the compiler knows that is impossible to an Element be a Point, so you get an compilation error.

answered Dec 16, 2010 at 11:40

pmariano

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user467871 · Accepted Answer · 2010-12-16 11:41:05Z

1

Because of the inheritance tree. if A inherited from B then you can write A instance of B

Integer i = 3;

System.out.println(i instanceof String); // compile time error

System.out.println(i instanceof Number); // true

System.out.println(i instanceof Object); // true

answered Dec 16, 2010 at 11:41

user467871

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