How can I pass std::unique_ptr into a function

Question

How can I pass a std::unique_ptr into a function? Lets say I have the following class:

class A
{
public:
    A(int val)
    {
        _val = val         


        

Answer 1

As MyFunc doesn't take ownership, it would be better to have:

void MyFunc(const A* arg)
{
    assert(arg != nullptr); // or throw ?
    cout << arg->GetVal() << endl;
}

or better

void MyFunc(const A& arg)
{
    cout << arg.GetVal() << endl;
}

If you really want to take ownership, you have to move your resource:

std::unique_ptr<A> ptr = std::make_unique<A>(1234);
MyFunc(std::move(ptr));

or pass directly a r-value reference:

MyFunc(std::make_unique<A>(1234));

std::unique_ptr doesn't have copy on purpose to guaranty to have only one owner.

Answer 2

There's basically two options here:

Pass the smart pointer by reference

void MyFunc(unique_ptr<A> & arg)
{
    cout << arg->GetVal() << endl;
}

int main(int argc, char* argv[])
{
    unique_ptr<A> ptr = unique_ptr<A>(new A(1234));
    MyFunc(ptr);
}

Move the smart pointer into the function argument

Note that in this case, the assertion will hold!

void MyFunc(unique_ptr<A> arg)
{
    cout << arg->GetVal() << endl;
}

int main(int argc, char* argv[])
{
    unique_ptr<A> ptr = unique_ptr<A>(new A(1234));
    MyFunc(move(ptr));
    assert(ptr == nullptr)
}

Answer 3

Since unique_ptr is for unique ownership, if you want to pass it as argument try

MyFunc(move(ptr));

But after that the state of ptr in main will be nullptr.

Answer 4

Passing std::unique_ptr<T> as value to a function is not working because, as you guys mention, unique_ptr is not copyable.

What about this?

std::unique_ptr<T> getSomething()
{
   auto ptr = std::make_unique<T>();
   return ptr;
}

this code is working

Answer 5

Why can I not pass a unique_ptr into a function?

You cannot do that because unique_ptr has a move constructor but not a copy constructor. According to the standard, when a move constructor is defined but a copy constructor is not defined, the copy constructor is deleted.

12.8 Copying and moving class objects

...

7 If the class definition does not explicitly declare a copy constructor, one is declared implicitly. If the class definition declares a move constructor or move assignment operator, the implicitly declared copy constructor is defined as deleted;

You can pass the unique_ptr to the function by using:

void MyFunc(std::unique_ptr<A>& arg)
{
    cout << arg->GetVal() << endl;
}

and use it like you have:

or

void MyFunc(std::unique_ptr<A> arg)
{
    cout << arg->GetVal() << endl;
}

and use it like:

std::unique_ptr<A> ptr = std::unique_ptr<A>(new A(1234));
MyFunc(std::move(ptr));

Important Note

Please note that if you use the second method, ptr does not have ownership of the pointer after the call to std::move(ptr) returns.

void MyFunc(std::unique_ptr<A>&& arg) would have the same effect as void MyFunc(std::unique_ptr<A>& arg) since both are references.

In the first case, ptr still has ownership of the pointer after the call to MyFunc.

Answer 6

Why can I not pass a unique_ptr into a function?

You can, but not by copy - because std::unique_ptr<> is not copy-constructible.

Surely this is the primary purpose of the construct?

Among other things, std::unique_ptr<> is designed to unequivocally mark unique ownership (as opposed to std::shared_ptr<> ).

And most strangely of all, why is this an OK way of passing it?

Because in that case, there is no copy-construction.