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copied!<p>Until the auto keyword makes it really trivial, one can rely on the convention that function objects provide certain typedef's, including result_type. One just needs to wrap a plain function into a function object. The standard "functional" header provides some (pointer_to_unary_function/pointer_to_binary_function in addition to a couple of member function wrappers). When these are not enough, boost library provides more powerful wrappers. </p> <pre><code>#include <iostream> #include <boost/function.hpp> #include <boost/bind.hpp> int answer() { return 42; } template <class T> T unknown() { return T(); } template <class Function> void foobar(Function unknown) { typename Function::result_type x = unknown(); std::cout << x << '\n'; } int main() { foobar(boost::function<int()>(answer)); foobar(boost::bind(unknown<int>)); } </code></pre> <p>And below's an example, how you might add a pointer_to_zeronary_function. (I suppose the helper function that helps you create one, ptr_fun, might be added to the standard namespace as well as an overload(?)</p> <pre><code>template <class T> class pointer_to_zeronary_function { typedef T(*zeronary_func)(); zeronary_func func; public: typedef T result_type; pointer_to_zeronary_function(zeronary_func f): func(f) {} T operator()() const { return func(); } }; template <class T> pointer_to_zeronary_function<T> ptr_fun(T(*f)()) { return pointer_to_zeronary_function<T>(f); } ... //usage: foobar(ptr_fun(answer)); </code></pre>

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