unique_ptr to a derived class as an argument to a function that takes a unique_ptr to a base class

Question

I'm trying to use a unique_ptr to derived class in a function that takes a unique_ptr to a base class. Something like:

class Base {};

class Derived : public Base {};

void f(unique_ptr<Base> const &base) {}

…

unique_ptr<Derived> derived = unique_ptr<Derived>(new Derived);
f(derived);

If I understand this answer correctly, this code should work, but it causes the following compile errors:

error C2664: 'f' : cannot convert parameter 1 from 'std::unique_ptr<_Ty>' to 'const std::unique_ptr<_Ty> &'

IntelliSense: no suitable user-defined conversion from "std::unique_ptr<Derived, std::default_delete<Derived>>" to "const std::unique_ptr<Base, std::default_delete<Base>>" exists

If I change f to take unique_ptr<Derived> const &derived, it works fine, but that's not what I want.

Am I doing something wrong? What can I do to work around this?

I'm using Visual Studio 2012.

Solution 1

You have three options:

1. Give up ownership. This will leave your local variable without access to the dynamic object after the function call; the object has been transferred to the callee:

   f(std::move(derived));
   

2. Change the signature of f:

   void f(std::unique_ptr<Derived> const &);
   

3. Change the type of your variable:

   std::unique_ptr<base> derived = std::unique_ptr<Derived>(new Derived);
   

   Or of course just:

   std::unique_ptr<base> derived(new Derived);
   

   Or even:

   std::unique_ptr<base> derived = std::make_unique<Derived>();
   

4. Update: Or, as recommended in the comments, don't transfer ownership at all:

   void f(Base & b);
   
   f(*derived);
   

Solution 2

I had option #1 of the accepted answer and I still had the same compile error. I banged my head on the wall for over an hour and I finally realized that I had

class Derived : Base {};

instead of

class Derived : public Base {};

Solution 3

A possibile solution is to change the type of the argument to be a Base const*, and pass derived.get() instead. There is no transfer of ownership with unique_ptr const<Base>& (and the unique_ptr is not being modified), so changing to a Base const* does not change the meaning.

---

Herb Sutter discusses passing smart pointer arguments at length in Smart Pointer Parameters. An extract from the linked article refers to this exact situation:

Passing a const unique_ptr<widget>& is strange because it can accept only either null or a widget whose lifetime happens to be managed in the calling code via a unique_ptr, and the callee generally shouldn’t care about the caller’s lifetime management choice. Passing widget* covers a strict superset of these cases and can accept “null or a widget” regardless of the lifetime policy the caller happens to be using.

Comments

[svick]

In that case, I'm considering going with 4. Use shared_ptr.

[ltjax]

How about just using a reference instead of a unique_ptr for the function call?

[Jonathan Wakely]

@svick, why pass in a smart pointer to the function if it doesn't take ownership of the pointer? That's not what smart pointers are for.

[svick]

Doesn't Base const* mean that the function could store the pointer somewhere? I though smart pointers are trying to avoid that.

[svick]

@ltjax How exactly would I do that?

[hmjd]

@svick, the function could store base.get() just as easily (get() is const member function).

[svick]

@JonathanWakely So, what else am I supposed to do? Because I don't like any of the alternatives in this answer.

[etam1024]

@svick No. Smart pointers are to avoid memory leaks by making clear who is the owner of an object. If you pass raw pointer as argument to a function you just give access to read/modify it. Smart pointers are to avoid using raw new and delete, not pointers at all.

[metal]

@KerrekSB, Say one has a factory function: std::unique_ptr<Base> MakeDerived() { auto p = std::unique_ptr<Derived>( new Derived() ); /*...*/ return p; } My intuition says I shouldn't need std::move(p) for the return, but my compilers tell me I do. Is this intended behavior or a defect in the standard?

[metal]

Done here.

[Kerrek SB]

@metal: Great, thanks - yeah, important detail, the return types have to match. I missed that. Good stuff. You could have said return std::unique_ptr<Base>(std::move(p)), I suppose.

[Zitrax]

"This will destroy the object at the end of the function call": But we don't know what f() is doing (assuming that it's actually not empty) - it might pass the ownership elsewhere. Isn't all we can say that we give up or transfer the ownership ?

[Kerrek SB]

@Zitrax: You're right, that isn't phrased very well. Let me edit it.