Kotlin: why do constructor parameters have "internal" visibility by default?

Question

If we have a class:

class Customer(val customerName: String) { }

Its contructor parameter customerName is accessible through getCustomerName() (because it's also a property). If we want to restrict access to this property we should declare it as private.

Since in many cases from Java world (and if a class is not intended to be data class) fields which are assigned from constructor parameters are for private / protected use it feels like an additional effort to explicitely declare them private in Kotlin.

Also, Kotlin classes are final by default, so why not follow this principle for properties? Am I missing something?

Andrey Breslav · Accepted Answer · 2015-03-10 08:42:30Z

9

In our experience, and from some empirical studies of existing codebases, internal/public visibility is best for properties.

Also, Kotlin classes are final by default, so why not follow this principle for properties? Am I missing something?

Properties are final by default, i.e. they can not be overridden unless you supply the open modifier explicitly.

answered Mar 10, 2015 at 8:42

Andrey Breslav

26k10 gold badges70 silver badges61 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Pouria Hemi · Accepted Answer · 2022-11-12 05:53:25Z

0

A Protected Modifier in Kotlin: CANNOT be set on top-level declarations. Declarations that are protected in a class, can be accessed only in their subclasses.

for more details you can use this link

answered Nov 12, 2022 at 5:53

Pouria Hemi

7355 gold badges15 silver badges30 bronze badges

Collectives™ on Stack Overflow

Kotlin: why do constructor parameters have "internal" visibility by default?

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related