4 ms·
Consider the following pseudo-Java: Object foo = ... if (foo instanceof String) { System.out.println(foo.toLowerCase()) } The above is not
by gizmo686 4y ago
Consider the following pseudo-Java:
Object foo = ...
if (foo instanceof String) {
System.out.println(foo.toLowerCase())
}
The above is not valid Java, because foo has type Object, which does not have a toLowerCase() method, so the compiler refuses to compile it. What you need to do is something like:
Object foo = ...
if (foo instanceof String) {
String sfoo = (String) foo
System.out.println(sfoo.toLowerCase())
}
Granted, you don't necessarily need to explicitly redefine in, as Java allows anonymous casting:
Object foo = ...
if (foo instanceof String) {
System.out.println(((String) foo).toLowerCase())
}
Regardless, the point still stands. Java (and most strongly typed languages with an inheritance based type system) requires an explicit cast, even if a straight forward static analysis would show that the object in question is already the desired type.
- GavinBierman 4y agoThis is all true, but with pattern matching for instanceof you don't need the explicit cast: Object foo = ... if (foo instanceof String s) { System.out.println(s.toLowerCase()) } In this case, the scope of the pattern variable `s` is just the then block, so are free to reuse it elsewhere in the block.