CODEWITHSUNDEEP
c++stdoption-type
Andreas T
Andreas T

Reputation: 753

What is the rationale behind std::optional?

To me, std::optional<T> always was a "cleaner" version of std::unique_ptr<T>: both have an empty state (optional() and nullptr) and own a T object otherwise. This analogy, however, breaks down when considering std::optional<const T>. The first code block is valid, while the second one should not be, as the type (const T) needs to be MoveAssignable.

std::unique_ptr<const int> ptr1, ptr2;
ptr1 = std::move(ptr2);

std::optional<const int> opt1, opt2;
opt1 = std::move(opt2);

With a similar reasoning, I would expect std::optional<T&> to be equivalent to a non-owning pointer T*. While this analogy is a bit blurry for general T, I think it would make much sense for const T.

const int ZERO = 0;

void AssignPtrIfNull(const int*& ptr) {
  ptr = (ptr == nullptr ? &ZERO : ptr);
}

void AssignOptIfNull(std::optional<const int&>& ptr) {
  ptr = (ptr ? make_optional<const int&>(ZERO) : ptr);
}

So I am wondering, what is the thought process behind making optional the way it is? Because it seems really odd to me. Are there some pitfalls I am overseeing?

Upvotes: 1

Views: 412

Answers (1)

Richard Hodges
Richard Hodges

Reputation: 69864

Like everything in the c++ standard library post c++11, std::optional was lifted straight out of the boost library suite.

The motivation is laid out here

Upvotes: 1

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