function-object

The key reason this works is that for_each () doesn’t actually assume its third argument to be a function. It simply assumes that its third argument is something that can be called with an appropriate argument. A suitably defined object serves as well as – and often better than – a function. For example, it is easier to inline the application operator of a class than to inline a function passed as a pointer to function. Consequently, function objects often execute faster than do ordinary functions. An object of a class with an application operator (§11.9) is called a functionlike object, a

The C++ standard defines function objects as: A function object type is an object type that can be the type of the postfix-expression in a function call. ( link ) First I was thinking that function objects were functors, but then I realized that for a function pointer ptr of type P (not a function, but a function pointer), std::is_object_v<P> is true and can be called with the ptr(Args...) syntax. I am right that function pointers are considered as function objects by the standard? And if they are not what part of the definition is not satisfied by function pointers? Yes, they are. The term

I try to pass a function object to a thread. I am confused when I found the copy constructor is called two times in the 'main' thread. Why not simply copy once instead of twice? The second copy is useless. Code Snippet: #include <iostream> #include <thread> using namespace std; struct A { A() { cout << "constructor this=" << this << " thread_id=" << this_thread::get_id() << endl; } A(const A &other) { cout << "Copy constructor this=" << this << " thread_id=" << this_thread::get_id() << endl; } void operator()() { cout << "operator() this=" << this << " thread_id=" << this_thread::get_id() <<

In the wikipedia article about function objects it says such objects have performance advantages when used with for_each because the compiler can "inline" them. I'm a bit foggy on exactly what this means in this context... or any context I'm embarrassed to say. Thanks for any help! The last parameter of for_each template is a functor . Functor is something that can be "called" using the () operator (possibly with arguments). By defintion, there are two distinctive kinds of functors: Ordinary non-member functions are functors. Objects of class type with overloaded () operator (so called

In C++17, it is trivial to implement an overload(fs...) function that, given any number of arguments fs... satisfying FunctionObject , returns a new function object that behaves like an overload of fs... . Example: template <typename... Ts> struct overloader : Ts... { template <typename... TArgs> overloader(TArgs&&... xs) : Ts{forward<TArgs>(xs)}... { } using Ts::operator()...; }; template <typename... Ts> auto overload(Ts&&... xs) { return overloader<decay_t<Ts>...>{forward<Ts>(xs)...}; } int main() { auto o = overload([](char){ cout << "CHAR"; }, [](int) { cout << "INT"; }); o('a'); //

In N3421 - Making Operator Functors greater<> , the new specialization for the std function objects is: template <> struct plus<void> { template <class T, class U> auto operator()(T&& t, U&& u) const -> decltype(std::forward<T>(t) + std::forward<U>(u)); }; instead of template <> struct plus<void> { template <class T, class U> auto operator()(T&& t, U&& u) const noexcept(noexcept(decltype(std::forward<T>(t) + std::forward<U>(u)) (std::move(std::forward<T>(t) + std::forward<U>(u))))) -> decltype(std::forward<T>(t) + std::forward<U>(u)); }; Is there a reason for that? Does the omission of