Invoke c# là gì

Admin 04/04/2022 53

The concept of callable is defined in http://en.cppreference.com/w/cpp/concept/Callable.

Bạn sẽ xem: In C++ 11

Suppose I have a callable object f that has one argument of type T* & return type void. f can be any callable type (a function object, a pointer to member function, a pointer lớn data member, etc). How can I invoke f?

Simply calling f(x) fails since f can be a pointer to member function or data member. Is there a simple way to hotline f? One possible solution is std::bind(f, x)(), but this solution becomes more complex when f has more arguments.

This is exactly what std::invoke does, but it won"t be standard until C++17. You can make your own version, but it can be pretty complicated if it is fully general.

Here"s the basic idea for two cases (code taken from cppreference.com):

template inline tự động INVOKE(F&& f, Args&&... Args) -> decltype(std::forward(f)(std::forward(args)...)) return std::forward(f)(std::forward(args)...);template inline tự động hóa INVOKE(T Base::*pmd, Derived&& ref) -> decltype(std::forward(ref).*pmd) return std::forward(ref).*pmd;

Rather than implementing INVOKE yourself, use one of the library features that uses it. In particular, std::reference_wrapper works. Thus you can have the effect of std::invoke(f, args...) with std::ref(f)(args...):

templateauto invoke(F f, Args&&... Args) -> decltype(std::ref(f)(std::forward(args)...)) return std::ref(f)(std::forward(args)...);I didn"t forward f because std::reference_wrapper requires that the object passed in is not an rvalue. Using std::bind instead of std::ref doesn"t fix the problem. What this means is that for a function object lượt thích this:

I found the technique from this paper

Use boost::hof::apply:

#include // ...boost::hof::apply(f, args...);boost::hof::apply performs the same operation that INVOKE does.

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Alternatively, use boost::hana::apply, which does the same thing

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