About shared_ptr and pointer to member operator `->*` and `std::bind`

问题

Recently I discovered that shared_ptr does not have pointer to member operator ->*. I created simple example:

template <typename Pointer, typename Function, typename... Args>
auto invoke1(Pointer p, Function f, Args... args) -> decltype((p->*f)(args...))
{
  return (p->*f)(args...);
}
struct A { 
    void g() { std::cout << "A::g()\n"; } 
};
int main() {
  A a;
  invoke1(&a, &A::g); // works!!
  std::shared_ptr<A> sa = std::make_shared<A>();
  invoke1(sa, &A::g); // compile error!!
}

Q1: Why is so? Why shared_ptr does not have this operator?

I added such operator for shared_ptr and the example started to work:

template <typename T, typename Result>
auto operator ->* (std::shared_ptr<T> pointer, Result (T::*function)()) ->decltype(std::bind(function, pointer))
{
    return std::bind(function, pointer);
}
template <typename T, typename Result, typename Arg1>
auto operator ->* (std::shared_ptr<T> pointer, Result (T::*function)(Arg1 arg1)) ->decltype(std::bind(function, pointer, std::placeholders::_1))
{
    return std::bind(function, pointer, std::placeholders::_1);
}

Q2: Is this right implementation for this operator? Are there somewhere any "gold" rules how to implement such operator, probably either I reinvented the wheel or go in completely wrong direction, what do you think? Is there a way to have a single function implementing this operator instead of as many function as there are placeholders in std...

After that I came to conclusion that std::bind can be used in my invoke method.

template <typename Pointer, typename Function, typename... Args>
auto invoke2(Pointer p, Function f, Args... args) 
                     -> decltype(std::bind(f, p, args...)())
{
   return std::bind(f, p, args...)();
}

In this way my example also works without need to add operator ->* to shared_ptr.

Q3: So, is std::bind now considered as a replacement for operator->*?

回答1:

In a nut shell: yes std::bind is a replacement for member function pointers.

Why? because member function pointers are awful, and their only purposes is to implement delegates, which is why std::bind and std::function do

For reference on how member function pointers are implemented, see my previous answer here. In simplest terms, member function pointers are crippled by the standard because they do not allow for calls after casts; this makes them quite pointless for the sort of behavior 90% of people want from member function pointers: delegates.

For this reason, std::function is used to represent an abstract "callable" type, with std::bind being used to bind a this to the member function pointer. You should absolutely not mess with member function pointers, and instead use std::bind and std::function.

回答2:

I believe the simplest soultion would be to replace 'structure dereference' (->) operator with a pair of derefence(*) and structure reference(.) operators:

template <typename Pointer, typename Function, typename... Args>
auto invoke1(Pointer p, Function f, Args... args) -> decltype(((*p).*f)(args...))
{
  return ((*p).*f)(args...);
}

回答3:

I believe shared_ptr does not have operator ->* because it's impossible to implement it for arbitrary number of arguments (which C++11 allows to do for other use cases). Also, you can easily add an overload of invoke function for smart pointers that calls get(), so complicating the interface is not desirable.

来源：https://stackoverflow.com/questions/17696664/about-shared-ptr-and-pointer-to-member-operator-and-stdbind

标签

c++

operator-overloading

shared-ptr

boost-bind

pointer-to-member