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5117641
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0
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0
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3
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2011-02-25T13:12:31.443
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2011-02-25T13:12:31.443
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632030
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20
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The presence of a non-templated T::operator() for a given type T can be detected by: <pre><code>template<typename C> // detect regular operator() static char test(decltype(&C::operator())); template<typename C> // worst match static char (&test(...))[2]; static const bool value = (sizeof( test<T>(0) ) </code></pre> The presence of a templated operator can be detected by: <pre><code>template<typename F, typename A> // detect 1-arg operator() static char test(int, decltype( (*(F*)0)( (*(A*)0) ) ) = 0); template<typename F, typename A, typename B> // detect 2-arg operator() static char test(int, decltype( (*(F*)0)( (*(A*)0), (*(B*)0) ) ) = 0); // ... detect N-arg operator() template<typename F, typename ...Args> // worst match static char (&test(...))[2]; static const bool value = (sizeof( test<T, int>(0) ) == 1) || (sizeof( test<T, int, int>(0) ) == 1); // etc... </code></pre> However, these two do not play nicely together, as decltype(&C::operator()) will produce an error if C has a templated function call operator. The solution is to run the sequence of checks against a templated operator first, and check for a regular operator() if and only if a templated one can not be found. This is done by specializing the non-templated check to a no-op if a templated one was found. <pre><code>template<bool, typename T> struct has_regular_call_operator { template<typename C> // detect regular operator() static char test(decltype(&C::operator())); template<typename C> // worst match static char (&test(...))[2]; static const bool value = (sizeof( test<T>(0) ) == 1); }; template<typename T> struct has_regular_call_operator<true,T> { static const bool value = true; }; template<typename T> struct has_call_operator { template<typename F, typename A> // detect 1-arg operator() static char test(int, decltype( (*(F*)0)( (*(A*)0) ) ) = 0); template<typename F, typename A, typename B> // detect 2-arg operator() static char test(int, decltype( (*(F*)0)( (*(A*)0), (*(B*)0) ) ) = 0); template<typename F, typename A, typename B, typename C> // detect 3-arg operator() static char test(int, decltype( (*(F*)0)( (*(A*)0), (*(B*)0), (*(C*)0) ) ) = 0); template<typename F, typename ...Args> // worst match static char (&test(...))[2]; static const bool OneArg = (sizeof( test<T, int>(0) ) == 1); static const bool TwoArg = (sizeof( test<T, int, int>(0) ) == 1); static const bool ThreeArg = (sizeof( test<T, int, int, int>(0) ) == 1); static const bool HasTemplatedOperator = OneArg || TwoArg || ThreeArg; static const bool value = has_regular_call_operator<HasTemplatedOperator, T>::value; }; </code></pre> If the arity is always one, as discussed above, then the check should be simpler. I do not see the need for any additional type traits or library facilities for this to work.
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1. POC++ metafunction to determine whether a type is callable
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1. COHmm.. In retrospect, I suppose those templated checks could simply use int instead...
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2. COThat works very well, thank you!
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