How do smart pointers choose between delete and delete[]?

Question

Consider:

delete new std :: string [2];
delete [] new std :: string;

Everyone knows the first is an error. If the second wasn't an error, we wouldn't need two distinct operators.

Now consider:

std :: unique_ptr <int> x (new int [2]);
std :: unique_ptr <int> y (new int);

Does x know to use delete[] as opposed to delete?

---

Background: this question floated through my head when I thought array type qualification of pointers would be a handy language feature.

int *[] foo = new int [2]; // OK
int *   bar = new int;     // OK
delete [] foo;             // OK
delete bar;                // OK
foo = new int;             // Compile error
bar = new int[2];          // Compile error
delete foo;                // Compile error
delete [] bar;             // Compile error

Answer 1

Unfortunately, they don't know what delete to use therefore they use delete. That's why for each smart pointer we have a smart array counterpart.

std::shared_ptr uses delete
std::shared_array uses delete[]

So, your line

std :: unique_ptr <int> x (new int [2]);

actually causes undefined behavior.

Incidentally, if you write

std :: unique_ptr<int[]> p(new int[2]);
                     ^^

then delete[] will be used since you've explicitly requested that. However, the following line will still be UB.

std :: unique_ptr<int[]> p(new int);

The reason that they can't choose between delete and delete[] is that new int and new int[2] are exactly of the same type - int*.

Here's a related question of using correct deleters in case of smart_ptr<void> and smart_ptr<Base> when Base has no virtual destructor.