c++ - 这个 shared_ptr 是如何自动转换为原始指针的？

Question

我现在正在学习 C++11 的 enable_shared_from_this；一个例子让我感到困惑:shared_from_this() 返回的 shared_ptr 类型如何转换为这个原始指针？

#include <iostream>
#include <memory>
#include <functional>

struct Bar {
    Bar(int a) : a(a) {}
    int a;
};

struct Foo : public std::enable_shared_from_this<Foo> {
    Foo() { std::cout << "Foo::Foo\n"; }
    ~Foo() { std::cout << "Foo::~Foo\n"; }

    std::shared_ptr<Bar> getBar(int a)
    {
        std::shared_ptr<Bar> pb(
            new Bar{a}, std::bind(&Foo::showInfo, shared_from_this(), std::placeholders::_1)
        );
        return pb;
    }

    void showInfo(Bar *pb)
    {
        std::cout << "Foo::showInfo()\n";
        delete pb;
    }

};

int main()
{
    std::shared_ptr<Foo> pf(new Foo);
    std::shared_ptr<Bar> pb = pf->getBar(10);
    std::cout << "pf use_count: " << pf.use_count() << std::endl;
}

Answer 1

这是 std::bind 变得聪明，而不是指针。

As described in Callable, when invoking a pointer to non-static member function or pointer to non-static data member, the first argument has to be a reference or pointer (including, possibly, smart pointer such as std::shared_ptr and std::unique_ptr) to an object whose member will be accessed.

bind 已实现，因此它可以接受智能指针来代替原始指针。

在glibc++ implementation than bind internally calls invoke 中可以看到:

  // Call unqualified
  template<typename _Result, typename... _Args, std::size_t... _Indexes>
_Result
__call(tuple<_Args...>&& __args, _Index_tuple<_Indexes...>)
{
  return std::__invoke(_M_f,
      _Mu<_Bound_args>()(std::get<_Indexes>(_M_bound_args), __args)...
      );
}

std::invoke 可以开箱即用地处理智能事物(指针、reference wrappers 等):

INVOKE(f, t1, t2, ..., tN) is defined as follows:

If f is a pointer to member function of class T:

• If std::is_base_of<T, std::decay_t<decltype(t1)>>::value is true, then INVOKE(f, t1, t2, ..., tN) is equivalent to (t1.*f)(t2, ..., tN)
• If std::decay_t<decltype(t1)> is a specialization of std::reference_wrapper, then INVOKE(f, t1, t2, ..., tN) is equivalent to (t1.get().*f)(t2, ..., tN)
• If t1 does not satisfy the previous items, then INVOKE(f, t1, t2, ..., tN) is equivalent to ((*t1).*f)(t2, ..., tN).