unique_ptr < 0 OR what does less than operator do?

Question

I am dealing with code that has been written not by me. I have this statement:

// p is type of std::unique_ptr<uint8_t[]>
if (p < 0) { /* throw an exception */ }

So what does p < 0 mean in this context?
On the documentation page, I believe my case is 16) y < nullptr, where 0 is nullptr.

But what does it do?

Answer 1

unique_ptr < 0 OR what does less than operator do?

It matches the overload (11) on cppreference operator<(const unique_ptr&, nullptr_t);. 0 implicitly converts to std::nullptr_t. As per the documentation, the result is std::less<unique_ptr<T,D>::pointer>()(x.get(), nullptr).

The result is implementation defined, but unconditionally false on probably most systems. Presumably on an exotic system where null doesn't have the binary representation of 0, the result might be true.

I believe my case is 16)

(16) is the same other way around: 0 > unique_ptr. The result is same.

Answer 2

Check that operator < is not overloaded somewhere in your code base. That seems to be the only way how (p < 0) could be true.

Example:

bool operator< (const std::unique_ptr<uint8_t[]>&, int) { return true; }

int main() {
    std::unique_ptr<uint8_t[]> p;
    std::cout << (p < 0) << std::endl;
}

Prints:

1

live demo

Otherwise as others have said, 0 implicitly converts to std::nullptr_t, which would select the bool operator<(const unique_ptr<T, D>& x, nullptr_t) overload which would call std::less(p, 0) which would return false (even on Windows with a -1 pointer value).