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copied!<pre><code>#include <iostream> void MyFunction(float a){std::cout << "float\n";} void MyFunction(int a){std::cout << "int\n";} template<typename Func, typename T> void Do( Func f, T t ) { f(t); } template<typename T> void DoMyFunction( T t ) { Do(static_cast<void(*)(T)>(MyFunction), t); } template<typename T> void DoSomeFunction( T t, void(*func)(T) ) { Do(func, t); } int main() { float a; MyFunction(a); void (*manualFunctionPointer)(float) = MyFunction; manualFunctionPointer(a); // Do(MyFunction, a); -- does not compile Do(static_cast<void(*)(float)>(MyFunction), a); DoMyFunction(a); DoSomeFunction(a, MyFunction); } </code></pre> <p>The above works. I picked a MyFunction in 4 different ways.</p> <p>If you are willing to do some boilerlate, and want to solve the "what if a is a <code>char</code>" problem, then this might help:</p> <pre><code>// wrap the idea of calling MyFunction in a type: struct MyFunctionFunctor { template<typename T> static auto Do( T&& t )->decltype(MyFunction(std::forward(t))) { return MyFunction(std::forward(t)); } }; // Calling MyFunctionFunctor::Do( x ) will basically do static dispatch on all of // the overloads of MyFunction // wrap the idea of dispatching a variable to a functor: template<typename T, typename Functor> struct Dispatch { static auto Do( T t )->decltype( Functor::Do( t ) ) { return Functor::Do( t ); } } int main() { char a; auto func_ptr = &Dispatch<decltype(a), MyFunctionFunctor>::Do; func_ptr(a); } </code></pre> <p>but, as noted, this requires we wrap our <code>MyFunction</code> up so that it is described by a type. I'm not aware of a way to do this without boilerplate.</p>

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