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While looking at the implementation of std::reference_wrapper here

The constructors and operators are obvious to me but I didn't understand this part

template< class... ArgTypes >
typename std::result_of<T&(ArgTypes&&...)>::type
  operator() ( ArgTypes&&... args ) const {
  return std::invoke(get(), std::forward<ArgTypes>(args)...);
}

Could someone simplify it for me ... would be appreciated

Edit: and would be great to give useful example for operator() of std::reference_wrapper

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Laith
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1 Answers1

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This defines the operator() member function, which is applicable for an std::reference_wrapper wrapping a reference to a Callable. The purpose of it is to call the underlying Callable.

  • The template parameter class ... Args is to make it generic in terms of the parameters that can be passed to the underlying Callable.
  • The return type of the operator has to be the return type produced by invoking the Callable, which is obtained by the typename std::result_of<T&(ArgTypes&&...)>::type part
  • It uses the invoke call as a general-purpose way of calling the Callable, which works irrespective of what type of Callable it is (Functor, function pointer, member function pointer etc).
  • It uses std::forward in passing the argument list to achieve perfect forwarding - so for example lvalue and rvalues passed in to the original call retain their l/rvalue-ness in the underlying call.
Smeeheey
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