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POHow do I make a class, which I can't change, implement an interface?
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  1. POHow do I make a class, which I can't change, implement an interface?
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    copied!<p>I have a class from another library that is closed-source, but I want to be able to use an interface for it. The reason being that I don't want to do <code>instanceof</code> checks or <code>null</code>-checks everywhere, but I also don't want to extend the existing class.</p> <p>For example, let's say I have this code:</p> <pre><code>public class Example { // QuietFoo is from another library that I can't change private static QuietFoo quietFoo; // LoudFoo is my own code and is meant to replace QuietFoo private static LoudFoo loudFoo; public static void main(String[] args) { handle(foo); } private static void handle(Object foo) { if (foo instanceof QuietFoo) ((QuietFoo) foo).bar(); else if (foo instanceof LoudFoo) ((LoudFoo) foo).bar(); } } </code></pre> <p>I can't change <code>QuietFoo</code>:</p> <pre><code>public class QuietFoo { public void bar() { System.out.println("bar"); } } </code></pre> <p>But I <em>can</em> change <code>LoudFoo</code>:</p> <pre><code>public class LoudFoo { public void bar() { System.out.println("BAR!!"); } } </code></pre> <p>The problem is, there may be many other implementations of <code>bar</code> in many classes, and there may be more methods than just <code>bar</code>, so not only would my <code>handle</code> method get slow and ugly with lots of <code>instanceof</code> statements, but I would have to write one of these handle methods for each method on <code>QuietFoo</code> and <code>LoudFoo</code>. Extending isn't a viable solution because it violates the whole <em>is-a</em> contract since <code>LoudFoo</code> is not a <code>QuietFoo</code>.</p> <p>Basically, given <code>Foo</code>:</p> <pre><code>public interface Foo { void bar(); } </code></pre> <p>How can I make <code>QuietFoo</code> implement <code>Foo</code> without changing its source so I don't have to do casting and <code>instanceof</code> calls everywhere in my code?</p>
 

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