Is use of empty std::optional<string> UB or not?

user_6483824 0
Viewed 1196

The following code:

    std::optional<std::string> so;
    std::cout << so->size() << std::endl;
    std::cout << so.has_value();

outputs:

   0
   0

My question is whether its safe to call : so->size() on an empty optional. I used clang sanitizer, but it didnt report any UB in above code.

c++
3 Answers

Using operator-> on an empty std::optional is Undefined Behavior, regardless of what type T is in std::optional<T>.

According to cppreference on std::optional<T>::operator-> :

The behavior is undefined if *this does not contain a value.

user_73590940

Quoting the current C++ working draft

20.6.3.6 Observers [optional.observe]

constexpr const T* operator->() const;
constexpr T* operator->();

Preconditions: *this contains a value.

Then:

16.3.2.4 Detailed specifications [structure.specifications]

Preconditions: the conditions that the function assumes to hold whenever it is called; violation of any preconditions results in undefined behavior.

Thus, it's undefined behavior.

user_65306950

You're calling the default optional constructor (1 in that link), which...

  1. Constructs an object that does not contain a value.

When you go to dereference...

The behavior is undefined if *this does not contain a value.

Which in your case it doesn't. So yes, you have UB.

user_26027180
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