shared_ptr：“在没有适当的operator（）或将函数转换为指针到函数的类型的情况下，调用类类型的对象”

Question

#include "boost\shared_ptr.hpp"

class A{

public:
A(){}
~A(){}

};

int main()
{
    boost::shared_ptr<A> ptrA;
    ptrA(new A); 

}

I would like to know why this code won't compile? I want to know the difference if I just use

boost::shared_ptr<A> ptrA(new A);?

Answer 1

boost::shared_ptr<A> ptrA(new A);

Calls conversion constructor which converts A* into the shared_ptr . This is a default way to construct the ptr.

ptrA(new A); 

Calls operator() . This is used for a lot of reasons, one being to make objects emulate functions, ie functors. But this is not used with shared_ptr .

The constructor exists, operator() doesnt.