On Monday, March 21, 2016 at 8:34:16 AM UTC-5, Louis Dionne wrote:
On Sunday, March 20, 2016 at 10:44:45 AM UTC-5, Louis Dionne wrote:
[...] How so? The implementations are very similar, except Hana also allows for non-const function objects. I'd be grateful if you could provide a benchmark, perhaps I could improve Hana's implementation.
Actually looking at your implementation closer, I realized thats not the case. I originally had an implementation where I ranked the overloads, and your implementation looked similar to that when using `which`. However, after looking at it closer, it won't ever instantiate the second function if
Paul Fultz II wrote the
first function is callable. Sorry for the confusion.
I'm not sure I understand what you mean. I mean, it is expected that the function is never called when the first one could be called, since it's overload _linearly_, right?
Yes, but another way to achieve that is to use a ranked tag-dispatching. Like this: template<int N> struct rank : rank<N-1> {}; template<> struct rank<0> {}; // Call F template<class... Ts> auto foo_impl(Ts&&... xs, rank<1>) -> decltype(std::declval<F>()(std::forward<Ts>(xs)...)); // Call G template<class... Ts> auto foo_impl(Ts&&... xs, rank<0>) -> decltype(std::declval<G>()(std::forward<Ts>(xs)...)); template<class... Ts> auto foo_impl(Ts&&... xs) -> decltype(foo_impl(std::forward<Ts>(xs)..., rank<1>{})); So `G` will not be called if `F` is callable, however, the compiler will instantiate both `F` and `G` with this implementation. I used to use an implementation like this, but it is more costly at compile time. Furthermore, lazy instantiation is nice feature as well, especially when using constexpr, as the body of a constexpr function is always instantiated during substitution.
Louis
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