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    <p>We first need some helper templates. The first one checks if two types are the same:</p> <pre class="lang-cpp prettyprint-override"><code>template &lt;typename T, typename U&gt; struct is_same { // Default case: T and U are not the same type static const bool value = false; }; template &lt;typename T&gt; struct is_same&lt;T, T&gt; { // Specialization: both template arguments are of the same type static const bool value = true; }; </code></pre> <p>We can now use the boolean <code>is_same&lt;T, U&gt;::value</code> to determine if the types <code>T</code> and <code>U</code> are equivalent.</p> <p>Using <code>is_same</code> we can now write a template which checks if a specific type occurs in a type list:</p> <pre><code>template &lt;typename TList, typename T&gt; struct contains { static const bool value = is_same&lt;typename TList::head, T&gt;::value // Base case || contains&lt;typename TList::tail, T&gt;::value; // Recursion }; template &lt;typename T&gt; struct contains&lt;nulltype, T&gt; { // Termination condition static const bool value = false; }; </code></pre> <p>This is a recursive template, using a specialization on <code>nulltype</code> to terminate the recursion.</p> <p>One final helper template we need is <code>enable_if</code>:</p> <pre><code>template &lt;bool Cond, typename T=void&gt; struct enable_if { // Default case: Cond assumed to be false, no typedef }; template &lt;typename T&gt; struct enable_if&lt;true, T&gt; { // Specialization: Cond is true, so typedef typedef T type; }; </code></pre> <p><code>enable_if&lt;true, T&gt;::type</code> yields <code>T</code>, whereas <code>enable_if&lt;false, T&gt;::type</code> is undefined. We exploit the <a href="http://en.wikipedia.org/wiki/Substitution_failure_is_not_an_error" rel="nofollow">SFINAE</a> rule the enable or disable a function depending on its template argument, like so:</p> <pre><code>template &lt;typename TList&gt; class OutputClass { public: // Only enable the function if T is in TList (by SFINAE) template &lt;typename T&gt; typename enable_if&lt;contains&lt;TList, T&gt;::value&gt;::type output(T t) { std::cout &lt;&lt; t &lt;&lt; std::endl; } }; </code></pre> <p>Quite a trip, but if you understand all of this you're well on your way to master template metaprogramming. For an in-depth discussion of template metaprogramming I recommend you pick up a copy of <a href="https://www.pearsonhighered.com/program/Abrahams-C-Template-Metaprogramming-Concepts-Tools-and-Techniques-from-Boost-and-Beyond/PGM93893.html" rel="nofollow">C++ Template Metaprogramming</a> (ISBN-13: 9780321227256).</p>
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    1. COworked like a charm. I find these type lists really fascinating, I will definitely have to check out that book. Thanks again! You've really helped me understand type lists much better now
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    2. COIf the standard form for a function template is `template <typename T> [return type] [name]([arguments])`, then does that mean that the return type of `OutputClass::output()` is the result of instantiating the template `typename enable_if<contains<TList, T>::value>::type` ?
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    3. CO@Chris `enable_if<contains<TList, T>::value>::type` is technically not a template, it is a `typedef` which happens to be defined in a template. But yes, it is the return type of `OutputClass<TList>::output<T>()`. Did that answer your question?
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