Hi all! please tell me why? Spring Boot Applcation...

# getting-started

[Shumilin Alexandr]

Hi all! please tell me why? Spring Boot Applcation Main class (for start)

can not start, application

@SpringBootApplication
class ApiApplication {

    fun main(args: Array<String>) {
        runApplication<ApiApplication>(*args) {
            setBannerMode(Banner.Mode.OFF)
        }
    }
}

and this work! why?

@SpringBootApplication
class ApiApplication {

    companion object {
        @JvmStatic
        fun main(args: Array<String>) {
            SpringApplication.run(ApiApplication::class.java, *args)
        }
    }

}

[Klitos Kyriacou]

In your first example, main is not one of the ways to specify an entry point in Kotlin. As you have it, it's just an ordinary instance function of the class. You can fix it by moving it out of the class:

@SpringBootApplication
class ApiApplication

fun main(args: Array<String>) {
    runApplication<ApiApplication>(*args) {
        setBannerMode(Banner.Mode.OFF)
    }
}

[Shumilin Alexandr]

thank you!

[Shumilin Alexandr]

I didn’t notice

[Shumilin Alexandr]

magic)

[Vishnu Ravi]

To add a bit more context, If you compare this with the entry point in a Java class, you could see a public and static keyword as part of the main method signature. Your second example example get translated to this because of that companion object and static annotation, and that's why it worked.

In case of Kotlin, the entry point is a top level function (without enclosing that inside a class). And like @Klitos Kyriacou said, when you wrap it inside a class, it becomes an ordinary instance method.

Also, runApplication<ApiApplication>(*args) is just a Kotlin idiomatic alternative to SpringApplication.run(ApiApplication::class.java, *args)