Java Reference
In-Depth Information
The program results in a compile-time error, because the override of method
move
in
class
CheckedPoint
declares that it will throw a checked exception that the
move
in class
Point
has not declared. If this were not considered an error, an invoker of the method
move
on a reference of type
Point
could find the contract between it and
Point
broken if
this exception were thrown.
Removing the
throws
clause does not help:
class CheckedPoint extends Point {
void move(int dx, int dy) {
if ((x + dx) < 0 || (y + dy) < 0)
throw new BadPointException();
x += dx; y += dy;
}
}
A different compile-time error now occurs, because the body of the method
move
can-
not throw a checked exception, namely
BadPointException
, that does not appear in the
throws
clause for
move
.
Example 8.4.8.3-4. Erasure Affects Overriding
A class cannot have two member methods with the same name and type erasure:
class C<T> {
T id (T x) {...}
}
class D extends C<String> {
Object id(Object x) {...}
}
This is illegal since
D.id(Object)
is a member of
D
,
C<String>.id(String)
is declared in a
supertype of
D
, and:
• The two methods have the same name,
id
•
C<String>.id(String)
is accessible to
D
• The signature of
D.id(Object)
is not a subsignature of that of
C<String>.id(String)
• The two methods have the same erasure
Two different methods of a class may not override methods with the same erasure:
class C<T> {
