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    <p>I don't think you actually need to specialize for void return type. A void function is allowed to return the "result" of another void function for exactly this scenario.</p> <pre><code>void foo() { } void bar() { return foo(); } //this is OK int main() { bar(); } </code></pre> <p>So your only problem would be how to determine the return type.</p> <p>It appears that <code>boost::function</code> has a typedef for <code>result_type</code> (see <a href="http://beta.boost.org/doc/libs/1_37_0/doc/html/boost/functionN.html" rel="nofollow noreferrer">http://beta.boost.org/doc/libs/1_37_0/doc/html/boost/functionN.html</a>)</p> <pre><code>#include &lt;boost/function.hpp&gt; template&lt;class T&gt; class Container { public: typedef typename boost::function&lt;T&gt;::result_type result_type; private: boost::function&lt;T&gt; f; }; Container&lt;bool(int, int)&gt;::result_type r = true; </code></pre> <hr> <p>Edit: Now that you know what the <code>result_type</code> is, and you do need to distinguish between void/non-void results, you can employ <code>enable_if</code> and <code>disable_if</code>. The only complication is that those only work with function templates, so a non-template <code>foo</code> calls a templated <code>do_foo</code>.</p> <pre><code>#include &lt;boost/function.hpp&gt; #include &lt;boost/utility/enable_if.hpp&gt; #include &lt;boost/type_traits.hpp&gt; #include &lt;cstdio&gt; template&lt;class T&gt; class Container { public: typedef typename boost::function&lt;T&gt;::result_type result_type; result_type foo() { return do_foo&lt;result_type&gt;(); //note that this still works because you can return the void result! :) } private: //use this if the result_type is void template &lt;class U&gt; typename boost::enable_if&lt;boost::is_same&lt;U, void&gt;, U &gt;::type do_foo() { std::puts("for void"); } //else template &lt;class U&gt; typename boost::disable_if&lt;boost::is_same&lt;U, void&gt;, U&gt;::type do_foo() { std::puts("other"); return U(); } private: boost::function&lt;T&gt; f; }; int main() { Container&lt;void()&gt; a; a.foo(); Container&lt;int()&gt; b; b.foo(); } </code></pre>
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    1. POAccess result type of a function template parameter in the template?
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    1. COFor void return type, it's true you don't need to specialise in the case I gave above. But suppose I have a collection of boost::function<f> and that I want to store the result of each call to f() [0..n] into a collection, assuming that DoSomething returns a collection now rather than ::result_type. In the void case a collection won't compile (std::list<void>) and, indeed, wouldn't make any sense even if it did. That's why I wanted something that would discriminate between the two cases and not try to instantiate the non-void function if the return type of T is void.
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    2. CO@robinsont: Edited answer to show how `enable_if` could be used to distinguish between the two cases.
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    3. COThanks very much all. This is extremely helpful.
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