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copied!<p><code>IEnumerable<object></code> is not a supertype of <code>IEnumerable<T></code>, so it is not a supertype of <code>List<T></code> either. See <a href="https://stackoverflow.com/questions/2575363/generics-list-extends-animal-is-same-as-listanimal">question 2575363</a> for a brief overview of why this is the case (it's about Java, but the concepts are the same). This problem has been solved in C# 4.0, by the way, which <a href="https://stackoverflow.com/questions/245607/how-is-generic-covariance-contra-variance-implemented-in-c-4-0">supports covariant generics</a>.</p> <p>The reason why you didn't find this error is because you used <code>x as T</code>, where you should have been using a normal cast (<code>(T)x</code>), see <a href="https://stackoverflow.com/questions/2139798/why-is-the-c-as-operator-so-popular">question 2139798</a>. The resulting <code>InvalidCastException</code> would have pointed you at your error. (In fact, if the type relationship were correct (i.e. if <code>IEnumerable<object></code> were a supertype of <code>List<T></code>), you wouldn't need a cast at all.)</p> <p>To solve your problem, make your method generic, so that it accepts an <code>IEnumerable<T></code> instead of an <code>IEnumerable<object></code>, and skip the cast completely.</p> <pre><code> public static void VerifyNotNullOrEmpty<T>(IEnumerable<T> theIEnumerable, string theIEnumerableName, string theVerifyingPosition) { ... } </code></pre>

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