Question: **EDIT:** I understand unique\_ptr is non-copyable and can be only moved. i do not understand what happens with the initialization list.

Why unique\_ptr in member initialization list can work as in the code snipt?

```
#include <memory>

class MyObject
{
public:
    MyObject() : ptr(new int) // this works.
    MyObject() : ptr(std::unique_ptr<int>(new int)) 
    // i found this in many examples. but why this also work? 
    // i think this is using copy constructor as the bottom.        
    {
    }

    MyObject(MyObject&& other) : ptr(std::move(other.ptr))
    {
    }

    MyObject& operator=(MyObject&& other)
    {
        ptr = std::move(other.ptr);
        return *this;
    }

private:
    std::unique_ptr<int> ptr;
};

int main() {
    MyObject o;
    std::unique_ptr<int> ptr (new int);
    // compile error, of course, since copy constructor is not allowed. 
    // but what is happening with member initialization list in above?
    std::unique_ptr<int> ptr2(ptr); 
}

```

Answer: 
In your example, `std::unique_ptr<int>(new int)` is an rvalue, so the move-constructor of `ptr` is used.

The second time (in `main`), `std::unique_ptr<int> ptr2(ptr)` doesn't work because `ptr` is an lvalue, and cannot be moved directly (you can use `std::move`).
```
MyObject() : ptr(std::unique_ptr<int>(new int))

```

that uses the move constructor of `std::unique_ptr` (from a temporary).

You might change your `main` into

```
std::unique_ptr<int> ptr2(std::move(ptr));

```

to compile