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->*
?