返回构造函数为private的对象的shared_ptr实例

Question

让我困惑的是，为什么Foo *foo = new Foo(5);编译，而std::make_shared<Foo>(5)不编译?

下面是代码片段。请注意代码片段中的注释

#include <iostream>
#include <memory>

class Foo {
private:     //the user should not construct an instance through the constructor below.                    
Foo(int num):num_(num) { std::cout << "Foo::Foon"; }
public:
~Foo() { std::cout << "Foo::~Foon"; } 
static std::shared_ptr<Foo> Create() {
#if WORKS_WELL
Foo *foo = new Foo(5);  //Why this works?  The constructor is private indeed.
return std::shared_ptr<Foo>(foo);
#else
return std::make_shared<Foo>(5);
#endif
}
private:
int num_;
};

int main() {
auto pf = Foo::Create(); 
}

Answer 1

静态函数仍然是类的成员，因此可以访问私有部分。