Question

In a couple of my older code projects when I had never heard of smart pointers, whenever I needed to check whether the pointer still pointed to a valid object, I would always do something like this…

object * meh = new object;
if(meh) 
    meh->member;

Or when I needed to delete the object safely, something like this

if(meh)
{
    delete meh;
    meh = 0;
}

Well, now I have learned about the problems that can arise from using objects and pointers in boolean expressions both with literal numbers, the hard way :. And now I’ve also learned of the not so new but pretty cool feature of C++, the nullptr keyword. But now I’m curious.

I’ve already gone through and revised most of my code so that, for example, when deleting objects I now write

if(meh)
{
    delete meh;
    meh = nullptr;
}

Now I’m wondering about the boolean. When you pass just say an int into an if statement like this,

int meh;
if(meh)

Then it implicitly checks for zero without you needing to write it.

if(meh == 0) // does the exact same check

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

char * meh;
if(meh)

Then will it implicitly compare it with nullptr? 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 its members. If this is not the functionality why not? Too difficult to implement? Would solve some problems by removing yet another tiny way you could mess up your code.

Answer 1

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.}