为什么带有自定义删除器的unique_ptr对于nullptr不起作用，而shared_ptr呢？

Question

Simple code to use either unique_ptr or shared_ptr as a scope guard. All information about what to clear is captured in the deleter , so I though it is safe to use nullptr for constructor.

Apparently, with Visual C++ 2017 (14.1), it is not working as expected for unique_ptr , but works for shared_ptr . Is it a Microsoft quirk, or does the standard prevent calling the deleter of a unique_ptr when holding nullptr ?

In the code below, I'm forced to construct a unique_ptr with (void*)1 . If I construct it with nullptr , cleaner won't be called. For shared_ptr , there is no difference, cleaner is always called.

#include <memory>
#include <iostream>

int main()
{
    int ttt = 77;

    auto cleaner = [&ttt](void*) {
        std::cout << "cleaner: " << ttt << "\n"; // do something with capture here instead of print
    };

    std::unique_ptr<void, decltype(cleaner)> p((void*)1, cleaner);

    std::shared_ptr<void> q(nullptr, [&ttt](void*) {
        std::cout << "shared: " << ttt << "\n"; // do something with capture here instead of print
    });

    std::cout << "done\n";
    return 0;
}

Answer 1

unique_ptr 's destructor is required to do so:

23.11.1.2.2 unique_ptr destructor [unique.ptr.single.dtor]

2 Effects: If get() == nullptr there are no effects. Otherwise get_deleter()(get()) .

actually shared_ptr 's destructor is required to do the same:

23.11.2.2.2 shared_ptr destructor [util.smartptr.shared.dest]

— (1.1) If *this is empty or shares ownership with another shared_ptr instance ( use_count() > 1 ), there are no side effects.

— (1.2) Otherwise, if *this owns an object p and a deleter d, d(p) is called.

So relying on smart pointers to perform arbitrary actions at scope exit while passing null pointers is not reliable.