C ++智能指针

Question

consider the following code:

class Base{

};

class Derived : public Base{

};

int main(int argc, char **argv)
{
        std::unique_ptr<Base> b(new Derived());//1 // b is Base object but holds Derived pointer
        Base *b1 = new Derived();//Base obj points to Derived //3
        //std::unique_ptr<Base*> b(new Derived());//2
        return 0;
}

In the second statement(//2), I am getting compilation error, but if we consider syntax(//3), why I am getting this error, unique_ptr should be supplied pointer type instead of class type.

I am new to smart pointers in c++.

Thanks in Advance.

Answer 1

std::unique_ptr<T> is a smart pointer to T ; that is, it resembles T* but smarter.

Therefore, std::unique_ptr<Base*> is a smart pointer to Base* ; that is, it resembles Base** but smarter.

Answer 2

std::unique_ptr (and other pointers) take the type of the object it should point to, not of the pointer.

Think of it like this:

• Regular pointer: T becomes T*
• Unique pointer: T becomes unique_ptr<T>

Answer 3

The class template unique_ptr<T> manages a pointer to an object of type T.

This is how unique_ptr is defined:

template<
    class T,
    class Deleter = std::default_delete<T>
> class unique_ptr;

It should be given the type T.

So if you need a pointer to Base , you should give it type Base . If you give it Base* , it will become pointer to pointer to Base .

See more details here: http://en.cppreference.com/w/cpp/memory/unique_ptr or here http://www.cplusplus.com/reference/memory/unique_ptr/unique_ptr/