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copied!<p>An <code>interface</code> in Java is nothing more but a contract; at the compiler level, the only contract is that if you <code>implements</code> an interface, you must have an implementation of all methods (for classes which are not <code>abstract</code>):</p> <pre><code>public interface Foo { /* * Any invocation of this method sends the instance to the moon */ void foo(); } </code></pre> <p>If you have a concrete class which you can instantiate and which implements <code>Foo</code>, such as:</p> <pre><code>public class MyClass implements Foo { // must implement foo() @Override public void foo() { stayOnEarth(); } } </code></pre> <p>then you can do:</p> <pre><code>final Foo foo = new MyClass(); </code></pre> <p>As you have a reference to an instance of type <code>Foo</code>, it means you can invoke <code>foo()</code> on it:</p> <pre><code>foo.foo(); </code></pre> <p>The JVM will look up method <code>foo()</code> and find it.</p> <p>Now: <strong>the fact that you have implemented <code>foo()</code> is one thing; the fact that you obey the contract of this interface is another. The compiler does not give a <beep> if you stay on Earth or go right to Antares: it cannot enforce that</strong>.</p> <p>This is a question of trust first and foremost: if you implement an interface, you are expected to understand what this interface means; which is why there is documentation. The compiler cannot help any further but to enforce that you <em>at least</em> implement the needed methods.</p>

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