Spring Boot 如何在这里配置我的 Jackson ObjectMapper？

我试图了解 Spring/Spring Boot 如何为我负责维护的应用程序管理 DI。在该应用程序中，我看到一个MyAppInjectorGroovy 文件，如下所示：

@Configuration
class MyAppInjector {
    @Autowired
    void configureJackson(ObjectMapper objectMapper) {
        SimpleModule jacksonModule = new SimpleModule()
                .addDeserializer(AccountDeserializer, new AccountDeserializer())
                .addDeserializer(PhoneNumberDeserializer, new PhoneNumberDeserializer())
                .addDeserializer(AddressDeserializer, new AddressDeserializer())
                .addDeserializer(ContactDeserializer, new ContactDeserializer())
                .addDeserializer(CustomerDeserializer, new CustomerDeserializer())
                .addDeserializer(DeploymentInfoDeserializer, new DeploymentInfoDeserializer())
                .addDeserializer(ServiceAgreementDeserializer, new ServiceAgreementDeserializer())

        objectMapper.registerModule(jacksonModule)
    }
}

但是我没有看到configureJackson(...)从我的应用程序源代码中的任何地方调用。我假设此方法正在配置 Jackson ObjectMapper，Spring 使用它来将 JSON 序列化为 POJO。

但是，Spring 怎么知道在ObjectMapper这里配置呢？它是否寻找任何注释的东西@Configuration，然后看到它在ObjectMapper某处注册了一个 Jackson ，并将它传递给这个configureJackson(...)方法？对我来说看起来像魔术......

对于用 注释的类中的任何方法都是如此@Configuration吗？意思是我可以创建一个方法，如：

@Configuration
class MyAppInjector {
    @Autowired
    Fizz fizz(Buzz buzz) {
      new Fizz(buzz)
    }

    @Autowired
    void configureJackson(ObjectMapper objectMapper) {
        SimpleModule jacksonModule = new SimpleModule()
                .addDeserializer(AccountDeserializer, new AccountDeserializer())
                .addDeserializer(PhoneNumberDeserializer, new PhoneNumberDeserializer())
                .addDeserializer(AddressDeserializer, new AddressDeserializer())
                .addDeserializer(ContactDeserializer, new ContactDeserializer())
                .addDeserializer(CustomerDeserializer, new CustomerDeserializer())
                .addDeserializer(DeploymentInfoDeserializer, new DeploymentInfoDeserializer())
                .addDeserializer(ServiceAgreementDeserializer, new ServiceAgreementDeserializer())

        objectMapper.registerModule(jacksonModule)
    }
}

...并期望 Spring 智能地寻找一个Buzz实例以Fizz在构造时传入？