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POCan I use boost::enable_if on a member function?
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  1. POCan I use boost::enable_if on a member function?
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    copied!<p>I'm writing a template class, and I want to allow an additional method to exist only for a certain template type. Currently the method exists for all template types, but causes a compilation error for all other types.</p> <p>Complicating this is that it's an overloaded operator(). Not sure if what I want to do is actually possible here.</p> <p>Here's what I have now:</p> <pre><code>template&lt;typename T, typename BASE&gt; class MyClass : public BASE { public: typename T&amp; operator() (const Utility1&lt;BASE&gt;&amp; foo); typename T const&amp; operator() (const Utility2&lt;BASE&gt;&amp; foo) const; }; </code></pre> <p>I want the <code>T&amp;</code> version always available, but the <code>T const&amp;</code> version only available if <code>Utility2&lt;BASE&gt;</code> is valid. Right now, both methods exist, but attempting to use the const version gives a weird compilation error if <code>Utility2&lt;BASE&gt;</code> is invalid. I'd rather have a sensible error, or even a "no such member function" error.</p> <p>Is this possible?</p> <p><strong>EDIT</strong>: After reading through the boost docs, here's what I've come up with, and it seems to work:</p> <pre><code>template&lt;typename T, typename BASE&gt; class MyClass : public BASE { public: typename T&amp; operator() (const Utility1&lt;BASE&gt;&amp; foo); template&lt;typename U&gt; typename boost::enable_if&lt;boost::is_same&lt;Utility2&lt;BASE&gt;, U&gt;, T&gt;::type const &amp; operator() (const U&amp; foo) const; }; </code></pre> <p>So that method doesn't exist unless someone tries to use it with Utility2, and they can only create a Utility2 if it's valid for that BASE type. But when it's not valid for that BASE type, MyClass will not waste time creating the accessor method.</p>
 

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