Question

shared_ptr<void> is special in that it, by definiton, will invoke undefined behavior by calling delete on a void*.

So, why is there not a shared_ptr<void> specialization which throws a compile error?

Answer 1

shared_ptr<T> is special in that it is by design allowed to hold a pointer to any pointer type which is convertible to T* and will use the proper deleter without UB! This comes into play with shared_ptr<Base> p(new Derived); scenarios, but also includes shared_ptr<void>.

For example:

#include <boost/shared_ptr.hpp>

struct T {
    T() { std::cout << "T()\n"; }
    ~T() { std::cout << "~T()\n"; }
};


int main() {
    boost::shared_ptr<void> sp(new T);
}

produces the output:

$ ./test
T()
~T()

If you visit http://www.boost.org/doc/libs/1_47_0/libs/smart_ptr/shared_ptr.htm, scroll down to the assignment section to see the very thing being demonstrated. See http://www.boost.org/doc/libs/1_47_0/libs/smart_ptr/sp_techniques.html#pvoid for more details.

EDIT as noted by trinithis, it is UB if the pointer type passed into the constructor is a void * pointer. Thanks for pointing that out!