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    <p>While both statements are saying essentially the same thing, the minor semantic difference does have at least one maintenance effect.</p> <pre><code>class A&lt;T&gt; where T : IFoo {} </code></pre> <p>This says that A will work on any type, so long as it has IFoo implemented.</p> <pre><code>class B&lt;IFoo&gt; {} </code></pre> <p>This says that B will only work on types that implement IFoo.</p> <p>The distinction is that you can still expand A to also require other things, like that IBar also be implemented, so you can do more things in A.</p> <p>B on the other hand works better as a base class, <code>B&lt;T&gt;</code>, and then you can extend <code>C&lt;IFoo&gt;</code> from it. C is now a B that works on IFoo. Later on, you could extend class <code>D&lt;IBar&gt;</code> to have a B that works on only IBars (not both IBar and IFoo). </p> <p>So you'd use the first form if you're writing one class that always needs to work on IFoo. This use case is pretty common, maybe writing a helper class for some interface types you wrote. You'd use the second form if you're writing a generic base class that can work on anything that you specify the interface for. </p> <p>I've used this form a fair bit, for example I just wrote a generic abstract serializer to handle the basic operations of serialization, but leaving abstract methods (like <code>int GetKeyField(T record)</code>) in it (the template method pattern) to allow specialization. So then I can have <code>IFooSerialize : Serializer&lt;IFoo&gt;</code> that serializes IFoos according to what is specific about IFoo.</p>
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    1. POWhat is the purpose of constraining a type to an interface?
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    1. USCodexArcanum
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    1. CO+1 for getting what I was trying to ask, despite the fact that I completely botched the explanation at first.
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