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.
Parameters
p | - | a shared pointer whose deleter needs to be accessed |
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.
Exceptions
noexcept
specification: noexcept
Notes
The returned pointer may outlive the last shared_ptr
if, for example, std::weak_ptr
s remain and the implementation doesn't destroy the deleter until the entire control block is destroyed.
Example
demonstrates that shared_ptr deleter is independent of the shared_ptr's type.
#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!
See also
std::shared_ptr constructors (public member function) |
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