Converting JSON data to a struct with optional fields

[jozefizso]

We are using the latest develop code of the json library so we can use structs with std::optional<> data types.

The incoming JSON we are parsing may not include the fields which are optional. Right now the code generated by NLOHMANN_DEFINE_TYPE_INTRUSIVE macros will throw exception out_of_range.403 if the field is not found in the JSON data.

Is it possible to skip such field and not throw?

Compiler Explorer sample: https://godbolt.org/z/3P58EWhqd

namespace ns {
    struct info {
        std::optional<int> start;
        std::optional<int> end;

        NLOHMANN_DEFINE_TYPE_INTRUSIVE(info, start, end)
    };
}

// this will throw `out_of_range.403` because the `end` field is not present in the JSON data
json j = R"( { "start": null } )"_json;
ns::info i = j.template get<ns::info>();

terminate called after throwing an instance of 'nlohmann::json_abi_v3_11_3::detail::out_of_range'
  what():  [json.exception.out_of_range.403] key 'end' not found
Program terminated with signal: SIGSEGV

I have found these similar question:

• Don't throw exception in serialize/deserialize macros when we have no some field #2395
• json can't convert to struct/class when not contains vars(use NLOHMANN_DEFINE_TYPE_INTRUSIVE) #2718

[nlohmann]

It is a difference between (1) the field is null and (2) the field is not provided in the JSON object at all. You probably want to try NLOHMANN_DEFINE_TYPE_INTRUSIVE_WITH_DEFAULT.

[jozefizso]

Thank you, this indeeds work. I'm sorry we did not try this variant of the macro first.