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📝 documentation to avoid future issues like #1108

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Niels Lohmann 2018-05-28 18:14:44 +02:00
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@ -602,6 +602,7 @@ Likewise, when calling `get<your_type>()`, the `from_json` method will be called
Some important things: Some important things:
* Those methods **MUST** be in your type's namespace (which can be the global namespace), or the library will not be able to locate them (in this example, they are in namespace `ns`, where `person` is defined). * Those methods **MUST** be in your type's namespace (which can be the global namespace), or the library will not be able to locate them (in this example, they are in namespace `ns`, where `person` is defined).
* Those methods **MUST** be available (e.g., properly headers must be included) everywhere you use the implicit conversions. Look at [issue 1108](https://github.com/nlohmann/json/issues/1108) for errors that may occur otherwise.
* When using `get<your_type>()`, `your_type` **MUST** be [DefaultConstructible](http://en.cppreference.com/w/cpp/concept/DefaultConstructible). (There is a way to bypass this requirement described later.) * When using `get<your_type>()`, `your_type` **MUST** be [DefaultConstructible](http://en.cppreference.com/w/cpp/concept/DefaultConstructible). (There is a way to bypass this requirement described later.)
* In function `from_json`, use function [`at()`](https://nlohmann.github.io/json/classnlohmann_1_1basic__json_a93403e803947b86f4da2d1fb3345cf2c.html#a93403e803947b86f4da2d1fb3345cf2c) to access the object values rather than `operator[]`. In case a key does not exist, `at` throws an exception that you can handle, whereas `operator[]` exhibits undefined behavior. * In function `from_json`, use function [`at()`](https://nlohmann.github.io/json/classnlohmann_1_1basic__json_a93403e803947b86f4da2d1fb3345cf2c.html#a93403e803947b86f4da2d1fb3345cf2c) to access the object values rather than `operator[]`. In case a key does not exist, `at` throws an exception that you can handle, whereas `operator[]` exhibits undefined behavior.
* In case your type contains several `operator=` definitions, code like `your_variable = your_json;` [may not compile](https://github.com/nlohmann/json/issues/667). You need to write `your_variable = your_json.get<decltype your_variable>();` instead. * In case your type contains several `operator=` definitions, code like `your_variable = your_json;` [may not compile](https://github.com/nlohmann/json/issues/667). You need to write `your_variable = your_json.get<decltype your_variable>();` instead.