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PO
 

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    <p>If you only want to call methods from the interface (good!), then you usually don't need to now the name of the implementor.</p> <pre><code>getRunnableFromSomewhere().run(); </code></pre> <p>always works and calls the <code>run()</code> method on the instance that is returned by that method.</p> <p>If you want to now the class name at runtime, simpy call <code>getClass().getName()</code> on the instance:</p> <pre><code>System.out.println(getRunnableFromSomewhere().getClass().getName()); </code></pre> <hr> <p>A simple example with the <code>Number</code> interface:</p> <pre><code>public class NumberExample { public static void main(String[] args) { MagicNumber magic = MagicNumberProvider.get(); // a random implementation System.out.println(magic.getMagicNumber().doubleValue()); // We know nothing about the implementations } } class MagicNumberProvider { public static MagicNumber get() { return Math.random() &gt; 0.5d ? new ItsMagicOne() : new ItsMagicTwo(); } } interface MagicNumber { public Number getMagicNumber(); } class ItsMagicOne implements MagicNumber { @Override public Number getMagicNumber() {return new Long(1);} } class ItsMagicTwo implements MagicNumber { @Override public Number getMagicNumber() {return new Double(2.5);} } </code></pre> <p>It only calls interface methods and we have, from the perspective of the main method, no idea, which implementation of <code>MagicNumber</code> is used (it's random) and on which implementation of <code>Number</code> we actually call the <code>doubleValue()</code> method.</p>
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    1. POHow can you call a class that is implementing a certain interface without knowing its name?
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    1. COI'm not sure I totaly understand the getRunnableFromSomwhere(), can you explain?
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    2. COThis is a dummy method / example that returns just some `Runnable`, assume it has a signature like `public Runnable getRunnableFromSomewhere() { ... }`
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    3. COYes but to instanciate a class that implements an interface you need the class name, wich i don't have
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