c++ - Return Optional value with ?: operator

Question

I often need to use optional type for functions:

std::optional<int32_t> get(const std::string& field)
{
    auto it = map.find(field);
    if (it != map.end()) return it->second;
    return {};
}

Is there a way to return optional value in one line? e.g. this:

std::optional<int32_t> get(const std::string& field)
{
    auto it = map.find(field);
    return it != map.end() ? it->second : {};
}

results in the error

error: expected primary-expression before '{' token
return it != map.end() ? it->second : {};
                                      ^

Answer 1

You can explicitly wrap the some-value return into an std::optional, and fall back on the constexpr std::nullopt for the no-value return.

std::nullopt:

std::nullopt is a constant of type std::nullopt_t that is used to indicate optional type with uninitialized state.

...

std::nullopt_t:

std::nullopt_t is an empty class type used to indicate optional type with uninitialized state. In particular, std::optional has a constructor with nullopt_t as a single argument, which creates an optional that does not contain a value.

With this approach, the true clause of the ternary operator call explicitly returns an std::optional with a some-value, so the compiler can deduce the template parameter/wrapped type (in this example: int32_t) from the type of the supplied wrapped value, meaning you needn't specify it explicitly.

Applied to your example:

return it != map.end() ? std::optional(it->second) : std::nullopt;

// alternatively
return it != map.end() ? std::make_optional(it->second) : std::nullopt;