Passing shared_ptr as shared_ptr

Question

What is the best method to go about passing a shared_ptr of a derived type to a function that takes a shared_ptr of a base type?

I generally pass shared_ptrs by reference to avoid a needless copy:

int foo(const shared_ptr<bar>& ptr);

but this doesn't work if I try to do something like

int foo(const shared_ptr<Base>& ptr);

...

shared_ptr<Derived> bar = make_shared<Derived>();
foo(bar);

I could use

foo(dynamic_pointer_cast<Base, Derived>(bar));

but this seems sub-optimal for two reasons:

• A dynamic_cast seems a bit excessive for a simple derived-to-base cast.
• As I understand it, dynamic_pointer_cast creates a copy (albeit a temporary one) of the pointer to pass to the function.

Is there a better solution?

Update for posterity:

It turned out to be an issue of a missing header file. Also, what I was trying to do here is considered an antipattern. Generally,

• Functions that don't impact an object's lifetime (i.e. the object remains valid for the duration of the function) should take a plain reference or pointer, e.g. int foo(bar& b).

• Functions that consume an object (i.e. are the final users of a given object) should take a unique_ptr by value, e.g. int foo(unique_ptr<bar> b). Callers should std::move the value into the function.

• Functions that extend the lifetime of an object should take a shared_ptr by value, e.g. int foo(shared_ptr<bar> b). The usual advice to avoid circular references applies.

See Herb Sutter's Back to Basics talk for details.

Answer 1

This will also happen if you've forgotten to specify public inheritance on the derived class, i.e. if like me you write this:

class Derived : Base
{
};

Instead of:

class Derived : public Base
{
};

Answer 2

Although Base and Derived are covariant and raw pointers to them will act accordingly, shared_ptr<Base> and shared_ptr<Derived> are not covariant. The dynamic_pointer_cast is the correct and simplest way to handle this problem.

(Edit: static_pointer_cast would be more appropriate because you're casting from derived to base, which is safe and doesn't require runtime checks. See comments below.)

However, if your foo() function doesn't wish to take part in extending the lifetime (or, rather, take part in the shared ownership of the object), then its best to accept a const Base& and dereference the shared_ptr when passing it to foo().

void foo(const Base& base);
[...]
shared_ptr<Derived> spDerived = getDerived();
foo(*spDerived);

As an aside, because shared_ptr types cannot be covariant, the rules of implicit conversions across covariant return types does not apply when returning types of shared_ptr<T>.

Answer 3

Also check that the #include of the header file containing the full declaration of the derived class is in your source file.

I had this problem. The std::shared<derived> would not cast to std::shared<base>. I had forward declared both classes so that I could hold pointers to them, but because I didn't have the #include the compiler could not see that one class was derived from the other.

Answer 4

Sounds like you're trying too hard. shared_ptr is cheap to copy; that's one of its goals. Passing them around by reference doesn't really accomplish much. If you don't want sharing, pass the raw pointer.

That said, there are two ways to do this that I can think of off the top of my head:

foo(shared_ptr<Base>(bar));
foo(static_pointer_cast<Base>(bar));