Nullptr and checking if a pointer points to a valid object

Question

In C, anything that’s not 0 is true. So, you certainly can use:

if (ptrToObject) 
    ptrToObject->doSomething();

to safely dereference pointers.

C++11 changes the game a bit, nullptr_t is a type of which nullptr is an instance; the representation of nullptr_t is implementation specific. So a compiler may define nullptr_t however it wants. It need only make sure it can enforce proper restriction on the casting of a nullptr_t to different types–of which boolean is allowed–and make sure it can distinguish between a nullptr_t and 0.

So nullptr will be properly and implicitly cast to the boolean false so long as the compiler follows the C++11 language specification. And the above snippet still works.

If you delete a referenced object, nothing changes.

delete ptrToObject;
assert(ptrToObject);
ptrToObject = nullptr;
assert(!ptrToObject);

Because of how long I have been writing these ifs like this, it is second nature at this point to check if the pointers valid before using by typing if (object *) and then calling it’s members.

No. Please maintain a proper graph of objects (preferably using unique/smart pointers). As pointed out, there’s no way to determine if a pointer that is not nullptr points to a valid object or not. The onus is on you to maintain the lifecycle anyway.. this is why the pointer wrappers exist in the first place.

In fact, because the life-cycle of shared and weak pointers are well defined, they have syntactic sugar that lets you use them the way you want to use bare pointers, where valid pointers have a value and all others are nullptr:

Shared

#include <iostream>
#include <memory>

void report(std::shared_ptr<int> ptr) 
{
    if (ptr) {
        std::cout << "*ptr=" << *ptr << "\n";
    } else {
        std::cout << "ptr is not a valid pointer.\n";
    }
}

int main()
{
    std::shared_ptr<int> ptr;
    report(ptr);

    ptr = std::make_shared<int>(7);
    report(ptr);
}

Weak

#include <iostream>
#include <memory>

void observe(std::weak_ptr<int> weak) 
{
    if (auto observe = weak.lock()) {
        std::cout << "\tobserve() able to lock weak_ptr<>, value=" << *observe << "\n";
    } else {
        std::cout << "\tobserve() unable to lock weak_ptr<>\n";
    }
}

int main()
{
    std::weak_ptr<int> weak;
    std::cout << "weak_ptr<> not yet initialized\n";
    observe(weak);

    {
        auto shared = std::make_shared<int>(42);
        weak = shared;
        std::cout << "weak_ptr<> initialized with shared_ptr.\n";
        observe(weak);
    }

    std::cout << "shared_ptr<> has been destructed due to scope exit.\n";
    observe(weak);
}

Now, will C++ do the same for pointers? If pass in a char * like this to an if statement?

So to answer the question: with bare pointers, no. With wrapped pointers, yes.

_{Wrap your pointers, folks.}

Leave a Comment Cancel reply