Does this use of std::make_unique lead to non-unique pointers?

Question

Suppose I have the following code in C++:

#include <memory>
#include <iostream>

struct Some {
        Some(int _a) : a(_a) {}
        int a;
};

int main() {
        Some some(5);

        std::unique_ptr<Some> p1 = std::make_unique<Some>(some);
        std::unique_ptr<Some> p2 = std::make_unique<Some>(some);

        std::cout << p1->a << " " << p2->a << std::endl;
        return 0;
}

As I understand, unique pointers are used to guarantee that resources are not shared. But in this case both p1 and p2 point to the same instance some.

Please unveil the situation.

Answer 1

They don't point to the same resource, they each point to a different copy of it. You can illustrate it by deleting the copy constructor to see the error:

#include <memory>
#include <iostream>

struct Some {
        Some(int _a) : a(_a) {}
        Some(Some const&) = delete;
        int a;
};

int main() {
        Some some(5);

        std::unique_ptr<Some> p1 = std::make_unique<Some>(some); //error here
        std::unique_ptr<Some> p2 = std::make_unique<Some>(some);

        std::cout << p1->a << " " << p2->a << std::endl;
        return 0;
}

Answer 2

std::make_unique creates objects, calling constructor with specified arguments.

You passed Some& as parameter, and here copy constructor was invoked, and new object constructed.

So, p1 and p2 are 2 absolutely different pointers, but constructed from same object, using copy constructor

Answer 3

both p1 and p2 point to the same instance some

No, they don't.

#include <memory>
#include <iostream>

struct Some {
        Some(int _a) : a(_a) {}
        int a;
};

int main() {
        Some some(5);

        std::unique_ptr<Some> p1 = std::make_unique<Some>(some);
        std::unique_ptr<Some> p2 = std::make_unique<Some>(some);

        std::cout << p1->a << " " << p2->a << std::endl;
        p1->a = 42;
        std::cout << p1->a << " " << p2->a << std::endl;
        return 0;
}

output:

5 5
42 5