Why can't unique_ptr infer the type of the deleter?

Question

Let's say I want to use a custom deleter with an unique_ptr:

void custom_deleter(int* obj)
{
    delete obj; 
}

Why do I have to write this:

std::unique_ptr<int, void(*)(int*)> x(new int, custom_deleter);

instead of this:

std::unique_ptr<int> x(new int, custom_deleter); //does not compile

?

Can't the type of the deleter be inferred?

Answer 1

For unique_ptr, the deleter is part of the type:

template <
    class T,
    class Deleter = std::default_delete<T>
> class unique_ptr;

As such, when you're constructing an object, you need to specify its type. The line you're writing:

std::unique_ptr<int> x(new int, custom_deleter);

is equivalent to:

std::unique_ptr<int, std::default_delete<int> > x(new int, custom_deleter);

And you cannot construct a std::default_delete<int> from custom_deleter.

The only way to infer the deleter type is to use template deduction on that part too:

template <typename T, typename Deleter>
std::unique_ptr<T, Deleter> make_unique_ptr(T* ptr, Deleter deleter) {
    return std::unique_ptr<T, Deleter>(ptr, deleter);
}