C++17: std::optional

The C++17 standard library ships with a very interesting class template: std::optional<T>.

The idea behind it is to make explicit the fact that a variable can hold or not an actual value.

Before the existence of std::optional<T>, the only way to implement such semantics was through pointers or tagged unions (read about C++17 std::variant here).

For example, if I want to declare a struct person that stores a person’s first name, last name and nickname; and since not all people have or not a nickname, I would have to implement that (in older C++) in this way:

struct person
{
  std::string first_name;
  std::string last_name;
  std::string* nickname; //no nickname if null
};

To make explicit that the nickname will be optional, I need to write a comment stating that “null” represents “no nickname” in this scenario.

And it works, but:

  • It is error prone because the user can easily do something like: p.nickname->length(); and run into an unexpected behavior when the nickname is null.
  • Since the nickname will be stored as a pointer, the instance needs to be created in heap, adding one indirection level and one additional dynamic allocation/deallocation only to support the desired behavior (or the programmers need to have the nickname manually handled by them and set a pointer to that nickname into this struct).
  • Because of the last reason, it is not at all obvious if the instance pointed to by said pointer should be explicitly released by the programmer or it will be released automatically by the struct itself.
  • The “optionalness” here is not explicit at all at code level.

std::optional<T> provides safeties for all these things:

  • Its instances can be created at stack level, so there will not be extra allocation, deallocation or null-references: RAII will take care of them (though this depends on the actual Standard Library implementation).
  • The “optionalness” of the attribute is completely explicit when used: Nothing is more explicit than marking as “optional” to something… optional, isn’t it?
  • Instances of std::optional<T> hide the direct access to the object, so to access its actual value they force the programmer to do extra checks.
  • If we try to get the actual value of an instance that is not storing anything, a known exception is thrown instead of unexpected behavior.

Refactoring my code, it will look like this one:

#include <optional>
#include <string>

struct person
{
  std::string first_name;
  std::string last_name;
  std::optional<std::string> nickname;
};

The code is pretty explicit and no need to further explanation or documentation about the optional character of “nickname”.

So let’s create two people, one with nickname and the other one with no nickname:

int main()
{
  person p1 { "John", Doe", std::nullopt };
  person p2 { "Robert", "Balboa", "Rocky" };
}

In the first instance, I have used “std::nullopt” which represents an std::optional<T> instance with no value (i.e. : an “absence of value”).

In the second case, I am implicitly invoking to the std::optional<T> constructor that receives an actual value.

The verbose alternative would be:

int main()
{
    person p1 { "John", "Doe", std::optional<std::string> { } };
    person p2 { "Robert", "Balboa", std::optional<std::string> {"Rocky"} };
}

The parameterless constructor represents an absence of value (std::nullopt) and the other constructor represents an instance storing an actual value.

Next I will overload the operator<< to work with my struct person, keeping in mind that if the person has a nickname, I want to print it out.

This could be a valid implementation:

std::ostream& operator<<(std::ostream& os, const person& p)
{
    os << p.last_name << ", " << p.first_name;
    
    if (p.nickname.has_value())
    {
        os << " (" << p.nickname.value() << ")";
    }
    
    return os;
}

The has_value() method returns true if the optional<T> instance is storing an actual value. The value can be retrieved using the value() method.

There is an overload for the operator bool that does the same thing that the has_value() method does: Verifying if the instance stores an actual value or not.

Also there are overloads for operator* and operator-> to access the actual values.

So, a less verbose implementation of my operator<< shown above would be:

std::ostream& operator<<(std::ostream& os, const person& p)
{
    os << p.last_name << ", " << p.first_name;
    
    if (p.nickname)
    {
        os << " (" << *(p.nickname) << ")";
    }
    
    return os;
}

Other way to retrieve the stored value, OR return an alternative value would be using the method “value_or()” method.

void print_nickname(const person& p)
{
    std::cout << p.first_name << " " << p.last_name << "'s nickname: "
              << p.nickname.value_or("[no nickname]") << "\n";
}

For this example, if the nickname variable stores an actual value, it will return it, otherwise, the value returned will be, as I coded: “[no nickname]”.

What will happen if I try to access to the optional<T>::value() when no value is actually stored? A std::bad_optional_access exception will be thrown:

try
{
    std::optional<int> op {};
    std::cout << op.value() << "\n";
}
catch (const std::bad_optional_access& e)
{
    std::cerr << e.what() << "\n";
}

Notice I have used the value() method instead of operator*. When I use operator* instead of value(), the exception is not thrown and the user runs into an unexpected behavior.

So, use std::optional<T> in these scenarios:

  • You have some attributes or function arguments that may have no value and are, therefore, optional. std::optional<T> makes that decision explicit at the code level.
  • You have functions that may OR may not return something. For example, what will be the minimum integer found in an empty list of ints? So instead of returning an int with its minimum value (std::numeric_limits<int>::min()), it would be more accurate to return an std::optional<int>.

Note that std::optional<T> does not support reference types (i.e. std::optional<T&>) so if you want to store an optional reference, probably you want to use a std::reference_wrapper<T> instead of type T (i.e. std::optional<std::reference_wrapper<T>>).

2 thoughts on “C++17: std::optional

  1. While the explanation is great, the example is very unfortunate. Strings can be empty. There’s even a method to query just that: std::string::empty(). So std::string has the “optional” element already in-built and doesn’t need std::optional.

    1. Hi “That Guy”,

      Yes, I already received that feedback on Reddit about my example and yes, you all are right :)

      I will update the post to show a far better example on this topic. Thanks a lot.

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