Java: Can Vector be called as Vector?

Question

I've got a class foo, a class that derives from food called bar, and I've got a method in foo that takes another foo

boolean baz(foo c)
{
return (condition)?true:false;
}

I want to write an overload for baz that takes a Vector and calls baz on all of them -- something like

boolean baz(Vector<foo> v)
{
    for(int i=0;i<v.size();++i)
    {
    if baz(v.get(i))
        return true;
    }
    return false;
}

and I want to use call this method on a Vector of bar. I tried writing this in the way I just outlined, and I get compiler errors when I try to call this method on a vector of bar.

What am I missing?

Can you post the exact compiler error? This will help people identify the problem. — maerics, Feb 06 '11 at 20:48
possible duplicate of [Why aren't Java's generics implicitly polymorphic?](http://stackoverflow.com/questions/2745265/why-arent-javas-generics-implicitly-polymorphic) — Bert F, Feb 06 '11 at 20:55

score 6 · Accepted Answer · answered Feb 06 '11 at 20:48

This is expected, because generics are invariant.

List<Derived> is not List<Base>. Imagine if you call that method, but inside it you call list.add(anotherDerived).

You can "fix" this by using List<? extends Base>. Thus you won't be able to add elements, and hence won't be able to violate the generics contract.

(I'm using List instead of Vector, because Vector is replaced (in most cases) by ArrayList)

score 6 · Answer 2 · answered Feb 06 '11 at 20:49

6

Just declare the method like this:

boolean baz(Vector<? extends foo> v)

Two additional points:

Java has an extremely strong convention for class names beginning with an uppercase letter. Anyone reading your code will be irritated by lowercase class names.
Vector is an obsolete class that should not be used anymore unless you're dealing with an API that does and which you don't control (like some parts of AWT). Use ArrayList instead.

answered Feb 06 '11 at 20:49

Michael Borgwardt

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I know about the uppercase class name convention and use it in my own code. I just haven't internalized it to the degree that I automatically used it in my example :). Thanks for the hint about vector, though! – Alexander Questioning Bresee Feb 08 '11 at 04:08

score 1 · Answer 3 · answered Feb 06 '11 at 20:47

1

instead of :

boolean baz(Vector<foo> v)

try

boolean baz(Vector<? extends foo> v)

answered Feb 06 '11 at 20:47

JohnQApple

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4

Java: Can Vector be called as Vector?

3 Answers3