Does Kotlin have primitive types?

人走茶凉 提交于 2020-07-17 11:55:14

问题


Does Kotlin have primitive types?. When I declare the variable: val myAge: Int = 18 then the myAge variable stores the actual values is 18 or stores the addresses of the objects in the memory?. If Int is primitive type then why we can use its method like myAge.minus(10)?

Image source: stackoverflow.com


回答1:


No... and yes.

Kotlin doesn't have primitive type (I mean you cannot declare primitive directly). It uses classes like Int, Float as an object wrapper for primitives.

When kotlin code is converted to jvm code, whenever possible, "primitive object" is converted to java primitive. In some cases this cannot be done. Those cases are, for example, collection of "primitives". For example, List<Int> cannot contains primitive. So, compiler knows when it can convert object to primitive. And, again, it's very similar to java:

List<Integer> numbers = new ArrayList<>;

numbers.add(0); // <-- you use primitive, but in fact, JVM will convert this primitive to object.
numbers.add(new Integer(0)); // <-- We don't need do that.

Also, when you declare "nullable primitive" it is never converted to primitive (what is kind of obvious, as primitive cannot be null). In java it works very similar:

int k = null; // No way!
Integer kN = null; // That's OK.

One more thing - what docs are saying about it?

For Common, JVM, JS

Represents a 32-bit signed integer. On the JVM, non-nullable values of this type are represented as values of the primitive type int.

For Native

Represents a 32-bit signed integer.

@see: https://kotlinlang.org/api/latest/jvm/stdlib/kotlin/-int/index.html

So, the last conclusion. Kotlin doesn't have primitive types out of the box. You treat all objects like objects. Converting to primitive is done at some lower level than code. This design is caused to keep compatibility with JVM.

I did a little deep dive and published it on medium. For interested: https://medium.com/@przemek.materna/kotlin-is-not-primitive-primitives-in-kotlin-and-java-f35713fda5cd




回答2:


Short answer - yes and depends on the declaration.

val myAge: Int = 18           // this is primitive
val myAge2: Int? = 18         // this is not

There's a very informative video about that https://www.youtube.com/watch?v=Ta5wBJsC39s




回答3:


On the Java platform, numbers are physically stored as JVM primitive types, unless we need a nullable number reference (e.g. Int?) or generics are involved. In the latter cases numbers are boxed.

Note that boxing of numbers does not necessarily preserve identity:

val a: Int = 10000
println(a === a) // Prints 'true'
val boxedA: Int? = a
val anotherBoxedA: Int? = a
println(boxedA === anotherBoxedA) // !!!Prints 'false'!!!

Note "===" used to compare reference ....

On the other hand, it preserves equality:

val a: Int = 10000
println(a == a) // Prints 'true'
val boxedA: Int? = a
val anotherBoxedA: Int? = a
println(boxedA == anotherBoxedA) // Prints 'true'


来源:https://stackoverflow.com/questions/57408327/does-kotlin-have-primitive-types

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