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Deducing return type of map generator helper (#1576)
* Deduce map return type implicitly Giving the first template argument to map generator function to deduce return type is now optional even if the return type is different from the type generated by mapped generator.
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committed by
Martin Hořeňovský

parent
296d447452
commit
54089c4c8c
@@ -169,16 +169,28 @@ namespace Generators {
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}
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};
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template <typename T, typename U, typename Func>
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#if defined(__cpp_lib_is_invocable) && __cpp_lib_is_invocable >= 201703
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// std::result_of is deprecated in C++17 and removed in C++20. Hence, it is
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// replaced with std::invoke_result here. Also *_t format is preferred over
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// typename *::type format.
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template <typename Func, typename U>
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using MapFunctionReturnType = std::remove_reference_t<std::remove_cv_t<std::invoke_result_t<Func, U>>>;
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#else
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template <typename Func, typename U>
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using MapFunctionReturnType = typename std::remove_reference<typename std::remove_cv<typename std::result_of<Func(U)>::type>::type>::type;
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#endif
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template <typename Func, typename U, typename T = MapFunctionReturnType<Func, U>>
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GeneratorWrapper<T> map(Func&& function, GeneratorWrapper<U>&& generator) {
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return GeneratorWrapper<T>(
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pf::make_unique<MapGenerator<T, U, Func>>(std::forward<Func>(function), std::move(generator))
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);
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}
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template <typename T, typename Func>
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GeneratorWrapper<T> map(Func&& function, GeneratorWrapper<T>&& generator) {
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template <typename T, typename U, typename Func>
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GeneratorWrapper<T> map(Func&& function, GeneratorWrapper<U>&& generator) {
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return GeneratorWrapper<T>(
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pf::make_unique<MapGenerator<T, T, Func>>(std::forward<Func>(function), std::move(generator))
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pf::make_unique<MapGenerator<T, U, Func>>(std::forward<Func>(function), std::move(generator))
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);
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}
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