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    <p><strong>Edit:</strong> this is a correct answer to a different question.</p> <p>Using the typename <code>T</code> twice confuses the issue a little, because they are compiled separately and are not connected in any way.</p> <p>You can overload the method to take a templated parameter:</p> <pre><code>template &lt;typename T&gt; class MyTemplateClass { void DoSomething(T&amp; t) { } template &lt;typename U,typename V&gt; void DoSomething(SomeRandomClass&lt;&lt;U,V&gt;&amp; r) { } }; </code></pre> <p>This maps <code>U</code> and <code>V</code> in the new method to <code>T'</code> and <code>S'</code> in <code>SomeRandomClass</code>. In this setup, either <code>U</code> or <code>V</code> could be the same type as <code>T</code>, but they don't have to be. Depending on your compiler, you ought to be able to do</p> <pre><code>MyTemplateClass&lt;string&gt; mine; SomeRandomClass&lt;int,double&gt; random; // note: nevermind the non-const ref on the string literal here... mine.DoSomething("hello world"); mine.DoSomething(random); </code></pre> <p>and the templated call will be selected as the matching overload without having to respecify the types explicitly.</p> <p><strong>Edit:</strong></p> <p>To do with with template specialization makes no difference to the overload of <code>DoSomething</code>. If you specialize the class as follows</p> <pre><code>template &lt;&gt; class SomeRandomClass &lt;int,double&gt; { // something here... }; </code></pre> <p>then the overload above will eat up this specialized implementation gladly. Just be sure the interfaces of the specialized template and the default template match.</p> <p>If what you're wanting is to specialize <code>DoSomething</code> to take a specific pair of types for <code>SomeRandomClass</code> then you've already lost generality...that's what specialization is.</p>
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    1. POTemplated class specialization where template argument is a template
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    1. USRyan Calhoun
    1. USRyan Calhoun
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    1. COThat's not quite it. That looks like it's just overloading of the DoSomething member, rather than a specialization of the class itself.
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    2. COYou're right, but it makes no difference to the method overload whether you're passing a specialized class or a regular template.
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      1. USRyan Calhoun
    3. COI finally see now what you were asking. This answer was totally backward.
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      1. USRyan Calhoun
 

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