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copied!<p>This is really late, but I was trying to figure out how to do this and ran into this question. The environment I'm using can't use C++11 (aka C++0x) or Boost, although both of those are awesome, so I figured I'd post how I figured out to do this without either for the sake of posterity.</p> <p>As UncleBens alluded to, the functional header in the STL has some useful features if you aren't using C++11 or Boost: <a href="http://www.cplusplus.com/reference/std/functional/" rel="nofollow">http://www.cplusplus.com/reference/std/functional/</a></p> <p>This problem is a bit more general than just not wanting to call a second template function. For example, one might want to build a vector of the return type of a functor, in which case calling a second template function might not work.</p> <p>By using some function overloading (to operate on both function pointers and functors) and stl's , we can make this work. Here's an example that prints out the result of a one-argument functor/function argument after declaring the variable explicitly:</p> <pre><code>#include <iostream> #include <functional> using namespace std; // Simple function [pointer] that adds one to its argument int addOne(int n) { return n + 1; } // Simple functor that multiplies its argument by two class timesTwo { public: int operator()(int n) const { return n * 2; } }; // Simple higher-order function: takes a functor f and calls f on n, returning the result // This is your template function in which you want to know the return type of f template <typename Functor> void printResultImpl(Functor f, typename Functor::argument_type n) { typename Functor::result_type r = f(n); cout << r << endl; } // Wrapper function for function pointer template <typename Arg, typename Result> void printResult(Result (*f)(Arg), Arg n) { printResultImpl(ptr_fun(f), n); } // Wrapper function for functor (function object) template <typename Functor, typename Arg> void printResult(Functor f, Arg n) { printResultImpl(bind1st(mem_fun(&Functor::operator()), &f), n); } // Prints out 8 then 14 int main() { printResult(addOne, 7); printResult(timesTwo(), 7); } </code></pre> <p>There are a couple limitations to this method: 1. You can't have your function return the result type of the functor (since the wrapper function doesn't know the result type) 2. It relies upon unary_function or binary_function in the stl. As UncleBens demonstrated, it's possible to extend to other types - simply follow the pattern of the declarations in : <a href="http://www.cplusplus.com/reference/std/functional/" rel="nofollow">http://www.cplusplus.com/reference/std/functional/</a></p> <p>But it worked for what I needed; maybe it'll work for someone else.</p>

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