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copied!<p>The short answer is "templates are not string substitution". <code>void f(void)</code> has meaning only so far as it is an alias for <code>void f()</code> in C++, in order to be backwards compatible with C.</p> <p>The first step is to use variadics, as noted elsewhere.</p> <p>The second step is figuring out how to map <code>void</code> returning functions to ... well, maybe something like <code>std::function<void()></code>, or maybe something else. I say maybe something else because unlike the other cases, you cannot call <code>std::function<void()> foo; foo( []()->void {} );</code> -- it isn't a true continuation.</p> <p>Something like this maybe:</p> <pre><code>template<typename T> struct Continuation { typedef std::function<void(T)> type; }; template<> struct Continuation<void> { typedef std::function<void()> type; }; </code></pre> <p>then use it like this:</p> <pre><code>auto someFunc = []()->void {}; Continuation<decltype(someFunc())>::type c; </code></pre> <p>which gives you the type you want. You could even add in an apply to continuation:</p> <pre><code>template<typename T> struct Continuation { typedef std::function<void(T)> type; template<typename func, typename... Args> static void Apply( type const& cont, func&& f, Args... args) { cont( f(args...) ); } }; template<> struct Continuation<void> { typedef std::function<void()> type; template<typename func, typename... Args> static void Apply( type const& cont, func&& f, Args... args) { f(args...); cont(); } }; </code></pre> <p>which lets you apply a continuation to an execution of a function uniformly if the incoming type is a void or if it is a non-void type.</p> <p>However, I would ask "why would you want to do this"?</p>

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