std::get_deleter

Defined in header <memory>.

Declarations

C++11

template< class Deleter, class T >
Deleter* get_deleter( const std::shared_ptr<T>& p ) noexcept;

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.

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.

Example

demonstrates that std::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

Result

shared_ptr<int> owns a deleter
...and it equals &foo_deleter
foo_deleter called!