std::get_deleter
From cppreference.com
< cpp | memory | shared ptr
| template< class Deleter, class T >
Deleter* get_deleter( const std::shared_ptr<T>& p ); |
(since C++11) | |
Access to the p's deleter. If the shared pointer p owns a deleter of type cv-unqualified Deleter (e.g. if it was created with one of the constructors that take a deleter as a parameter), then returns a pointer to the deleter. Otherwise, returns a null pointer.
Contents |
[edit] Parameters
| p | - | a shared pointer whose deleter needs to be accessed |
[edit] Return value
A pointer to the owned deleter or nullptr. The returned pointer is valid at least as long as there remains at least one shared_ptr instance that owns it.
[edit] Exceptions
noexcept specification:
noexcept
[edit] Notes
The returned pointer may outlive the last shared_ptr if, for example, std::weak_ptrs remain and the implementation doesn't destroy the deleter until the entire control block is destroyed.
[edit] Example
demonstrates that shared_ptr deleter is independent of the shared_ptr's type
Run this code
#include <iostream> #include <memory> struct Foo { int i; }; void foo_deleter(Foo * p) { std::cout << "foo_deleter called!\n"; delete p; } int main() { std::shared_ptr<int> aptr; { // create a shared_ptr that owns a Foo and a deleter auto foo_p = new Foo; std::shared_ptr<Foo> r(foo_p, foo_deleter); aptr = std::shared_ptr<int>(r, &r->i); // aliasing ctor // aptr is now pointing to an int, but managing the whole Foo } // r gets destroyed (deleter not called) // obtain pointer to the deleter: if(auto del_p = std::get_deleter<void(*)(Foo*)>(aptr)) { std::cout << "shared_ptr<int> owns a deleter\n"; if(*del_p == foo_deleter) std::cout << "...and it equals &foo_deleter\n"; } else std::cout << "The deleter of shared_ptr<int> is null!\n"; } // deleter called here
Output:
shared_ptr<int> owns a deleter ...and it equals &foo_deleter foo_deleter called!
[edit] See also
| std::shared_ptr constructors (public member function) |