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 ofElement
or any of its possible subclasses (none are shown here) could possibly be an instance of any subclass ofPoint
.
Why does this result in an error, rather than simply in instanceof
returning false?
Thanks,
JDelage
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.
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.
Because the compiler knows that is impossible to an Element be a Point, so you get an compilation error.
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
来源:https://stackoverflow.com/questions/4460355/why-does-instanceof-error-rather-than-return-false-when-used-for-2-incompati