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_ptr
s 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:
dynamic_cast
seems a bit excessive for a simple derived-to-base cast.dynamic_pointer_cast
creates a copy (albeit a temporary one) of the pointer to pass to the function.Is there a better solution?
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.
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>
.