I have a kotlin class with superclass: ```class MyClass(param: SomeType): Base(param.somecalculation...

# getting-started

[Pihentagy]

I have a kotlin class with superclass:

class MyClass(param: SomeType): Base(param.somecalculation) {
  private final val calculated = param.somecalculation()

How to do somecalculation just once?

[Sam]

fun MyClass(param: SomeType) = MyClass(param.someCalculation())

class MyClass(private val calculated: SomeCalculationResult): Base(calculated)

[Bogi Napoleon Wennerström]

You could try something like:

class MyClass(calculated: SomeCalculationResult): Base(calculated) {
    val calculated: SomeCalculationResult get() = super.calculated
}

[Pihentagy]

@Sam unfortunately MyClass is a @Component annotated spring class, so I'm afraid I cannot make it work that way.

[Pihentagy]

@Bogi Napoleon Wennerström maybe I can try with a by lazy property instead?

[Pihentagy]

Well, my original snippet was not accurate.

Instead of param.somecalculation there is someFunction("param") is more correct.

[Sam]

Can you make it a @Bean instead of a @Component?

@Configuration
class MyClassConfiguration {
  @Bean fun myClass(param: SomeType) = MyClass(param.someCalculation())
}

[Pihentagy]

Sure I can, but someCalculation should rather stick to MyClass, not the configuration (sorry for popping up with new requirements)

[Sam]

Can you modify the superclass?

[Sam]

If so, you could make this an abstract property rather than a constructor parameter

[Jacob]

The requirements don't make sense to me. You want the calculation to happen in MyClass but to construct myClass you need to have already done the calculation! (Because base requires the precalculated value)

[Matthew Feinberg]

What about something like this?

class MyClass private constructor(
    val param: SomeType,
    val calculated: CalculationResult
): Base(calculated) {
    constructor(param: SomeType) : this(param, param.calculate())
}