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    <p>I think it is possible to specialize traits for lambdas and do pattern matching on the signature of the unnamed functor. Here is the code that works on g++ 4.5. Although it works, the pattern matching on lambda appears to be working contrary to the intuition. I've comments inline.</p> <pre><code>struct X { float operator () (float i) { return i*2; } // If the following is enabled, program fails to compile // mostly because of ambiguity reasons. //double operator () (float i, double d) { return d*f; } }; template &lt;typename T&gt; struct function_traits // matches when T=X or T=lambda // As expected, lambda creates a "unique, unnamed, non-union class type" // so it matches here { // Here is what you are looking for. The type of the member operator() // of the lambda is taken and mapped again on function_traits. typedef typename function_traits&lt;decltype(&amp;T::operator())&gt;::return_type return_type; }; // matches for X::operator() but not of lambda::operator() template &lt;typename R, typename C, typename... A&gt; struct function_traits&lt;R (C::*)(A...)&gt; { typedef R return_type; }; // I initially thought the above defined member function specialization of // the trait will match lambdas::operator() because a lambda is a functor. // It does not, however. Instead, it matches the one below. // I wonder why? implementation defined? template &lt;typename R, typename... A&gt; struct function_traits&lt;R (*)(A...)&gt; // matches for lambda::operator() { typedef R return_type; }; template &lt;typename F&gt; typename function_traits&lt;F&gt;::return_type foo(F f) { return f(10); } template &lt;typename F&gt; typename function_traits&lt;F&gt;::return_type bar(F f) { return f(5.0f, 100, 0.34); } int f(int x) { return x + x; } int main(void) { foo(f); foo(X()); bar([](float f, int l, double d){ return f+l+d; }); } </code></pre>
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    1. COThanks, that did the trick. It didn't occur to me to just chain the non-specialized template onto the specialized version like that. Once I saw it, it looks incredibly elegant and obvious.
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    2. COI'm glad it helped. It seems like even though the syntax of getting at the operator() of a lambda is like class's member function, lambda is afterall an anonymous free standing function and matches R (*)(A...) but not R (C::*)(A...).
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    3. COI suppose the reasoning behind it might be that "C" in this case would be an implementation-defined type, and providing access to it through a template argument would allow us to do nonsensical things like define typedef aliases for it. Since we can't do anything with the lambda type except call it like a function, it makes more sense from a language design perspective to force that to be the specialization that applies.
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