I am not sure if the title is even relative as I do not have enough knowledge about move(). For this very reason it was hard to search and successfully find an answer to my question.
consider the following code:
using namespace std;
struct A
{
A(){cout << "Created!\n";}
~A(){cout << "Destroyed!\n";};
};
void f()
{
unique_ptr<A> q(new A);
unique_ptr<A> t (q.release()); // ----> (1)
// unique_ptr<A> t = move(q); // ----> (2)
}
int main() { f(); }
My understanding is: in both cases (1) and (2) the t will take the A ownership from q without destroying the object A.
(1) and (2)?I can see that move() is not part of the unique_ptr, instead it seems to me part of std so it has more general uses.
I am new to C++ and so in the simplest possible way:
Let me elaborate on question 3: what can move do?
std::move, as you noted, is not part of unique_ptr but of std. It is able to move any moveable object, not just pointers.
In the particular case of unique_ptr there is that handy release() that allows you to move the pointer manually, without using the move operators. But think for example of std::thread or std::fstream, there are no release/reacquire operators there.
Note that std::move by itself does nothing. Some people think that it returns a temporary while destroying the original, so a single:
std::unique_ptr<A> q(new A);
std::move(q); //this does nothing!!!
would move q to nothingness and destroy the object...
This is not true. All that std::move() does is a cast to an rvalue-reference. It is actually the move assignment operator (operator=(T&&)) or the move constructor (T(T&&)) that does the real moving.
You could also say:
std::unique_ptr<A> q(new A);
std::unique_ptr<A> r(static_cast<std::unique_ptr<A>&&>(q));
but that is cumbersome. std::move() is nicer to look at:
std::unique_ptr<A> q(new A);
std::unique_ptr<A> r(std::move(q));
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