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kotlinlang

```
interface Some {
  fun calculate(str: String? = null)
}
class SomeImpl : Some {
  fun calculate(str: String? = null)
}
```
An overriding function is not allowed to specify default values for its parameters
Why?

That does not stop you from calling it without parameters, no?

The overloading is “handled” by the interface

but why I can not specify same default parameter in implementation?

I guess it’s to not break the interface’s contract

<@U33S3CR1A> That would violate one of the SOLID principles, in this case the L(iskov substitution principal).

How it can break if it is the same default value?

If it is the same, you don’t have to specify it in its implementation.

I open the implementation class and I want to see that it can have a default parameter, if I will not open the interface than I will not know this

But the code that calls the `calculate(...)` functions, calls it on a variable that is declared to be of type `Some`, correct, not of type `SomeImpl`?

```
val mySome : Some
...  
...
mySome = ... SomeImpl() ...
...
mySome.calculate()
```

it could be also a type of `SomeImpl`, why not?

Yup, but usually it won't be for the *users* of your `Some` instances (they should not be concerned with the specific implementation of an interface).