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    <p>This is a similar problem to loading JDBC drivers. If the drivers don't exist you don't want to crash out your entire app.</p> <p>The basic principle of the solution is to use <a href="http://developer.android.com/reference/java/lang/Class.html#forName%28java.lang.String%29" rel="nofollow">Class.forName()</a></p> <p>This lets you check if the class exists and throws an exception if it doesn't.</p> <p><strong>Edit:</strong></p> <p>To be clear - the reason why this works and why this is <em>necessary</em> is that if Java is unable to execute code because it doesn't have a class then Java will treat this as a catastrophic failure (an <a href="http://developer.android.com/reference/java/lang/Error.html" rel="nofollow">Error</a> - specificity a <a href="http://developer.android.com/reference/java/lang/NoClassDefFoundError.html" rel="nofollow">NoClassDefFoundError</a>). Therefore executing any direct reference to a missing class will throw this Error. According to the <a href="http://developer.android.com/reference/java/lang/Error.html" rel="nofollow">documentation</a>:</p> <blockquote> <p>When errors are thrown, they should not be caught by application code.</p> </blockquote> <p>However when <code>Class.forName()</code> is unable to find a class (or more direct calls to a <a href="http://developer.android.com/reference/java/lang/ClassLoader.html" rel="nofollow">ClassLoader</a>) an <a href="http://developer.android.com/reference/java/lang/ClassNotFoundException.html" rel="nofollow">Exception</a> is thrown. Because this is an exception and not an Error it can be caught. And because the function takes a String as an argument, you don't need to directly reference the class (just give it's name in the string) so the check can be performed without directly referencing the class and thus without crashing out your program.</p>
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    1. POHow can i check if android device supports a library?
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    1. COdid it work? could using a try catch on an initialization function also work instead ?
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    2. COno, i didn't mean it , as this is what the post says. i asked if what the post says works, and if as an alternative, trying to add a try-catch for the initialization function would work too (again, instead of using the Class.ForName()) .
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    3. CONo. you get an Error not an Exception. As per the documentation you really should not catch errors, you get into very dangerous territory when you do that.
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