How does one downcast a std::shared_ptr?

C++C++11Shared PtrDowncast

C++ Problem Overview


Consider:

struct SomethingThatsABase
{
    virtual bool IsChildOne() const { return false; }
    virtual bool IsChildTwo() const { return false; }
};

struct ChildOne : public SomethingThatsABase
{
    virtual bool IsChildOne() const { return true; }
};

struct ChildTwo : public SomethingThatsABase
{
    virtual bool IsChildTwo() const { return true; }
};

void SomeClientExpectingAChildOne(std::shared_ptr<ChildOne> const& ptrOne)
{
    //Does stuff
}

void SomeClient(std::shared_ptr<SomethingThatsABase> const& ptr)
{
    if (ptr->IsChildOne())
    {
        SomeClientExpectingAChildOne(ptr); //Oops.
        //Hmm.. can't static_cast here, because we need a `shared_ptr` out of it.
    }
}

(Note that I can't simply do a std::shared_ptr<ChildOne>(static_cast<ChildOne*>(ptr.get())), because then the reference counts don't get shared between the two shared_ptrs)

C++ Solutions


Solution 1 - C++

This ought to work:

if (ptr->IsChildOne())
{
    SomeClientExpectingAChildOne(std::static_pointer_cast<ChildOne>(ptr));
}

Solution 2 - C++

The shared_ptr equivalent of static_cast is static_pointer_cast, and the shared_ptr equivalent of dynamic_cast is dynamic_pointer_cast.

Solution 3 - C++

Starting from C++11, §20.10.2.2.9 ([util.smartptr.shared.cast]) of the C++ standard specifies the equivalents of static_cast, const_cast and dynamic_cast for std::shared_ptr to be as follows:

std::static_pointer_cast:

template <class T, class U>
shared_ptr<T> static_pointer_cast(shared_ptr<U> const & r) noexcept;

static_pointer_cast requires static_cast<T *>(r.get()) to be well formed. If r is empty, an empty shared_ptr<T> is returned, otherwise returns a pointer w sharing ownership with r where w.get() == static_cast<T *>(r.get()) and w.use_count() == r.use_count().

std::const_pointer_cast:

template <class T, class U>
shared_ptr<T> const_pointer_cast(shared_ptr<U> const & r) noexcept;

const_pointer_cast has similar requirements and semantics to static_pointer_cast, except that const_cast is used instead of static_cast.

std::dynamic_pointer_cast:

template <class T, class U>
shared_ptr<T> dynamic_pointer_cast(shared_ptr<U> const & r) noexcept;

dynamic_pointer_cast is a bit different as it requires dynamic_cast<T *>(r.get()) to be well formed and have well defined semantics. If dynamic_cast<T *>(r.get()) is a non-zero value, returns a pointer w sharing ownership with r where w.get() == dynamic_cast<T *>(r.get()) and w.use_count() == r.use_count(), otherwise an empty shared_ptr<T> is returned.

std::reinterpret_pointer_cast:

For C++17, N3920 (adopted into Library Fundamentals TS in February 2014) also proposed a std::reinterpret_pointer_cast similar to the above, which would only require reinterpret_cast<T *>((U *) 0) to be well formed and returns shared_ptr<T>(r, reinterpret_cast<typename shared_ptr<T>::element_type *>(r.get())). Note N3920 also changed the wording for the other shared_ptr casts and extended shared_ptr to support arrays.

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionBilly ONealView Question on Stackoverflow
Solution 1 - C++mwigdahlView Answer on Stackoverflow
Solution 2 - C++JoelView Answer on Stackoverflow
Solution 3 - C++jotikView Answer on Stackoverflow