How does string interpolation work in Kotlin?

Question

Does the Kotlin compiler translate \"Hello, $name!\" using something like

java.lang.String.format(\"Hello, %s!\", name)

or is

Answer 1

The Kotlin compiler translates this code to:

new StringBuilder().append("Hello, ").append(name).append("!").toString()

There is no caching performed: every time you evaluate an expression containing a string template, the resulting string will be built again.

Answer 2

As you know, in string interpolation, string literals containing placeholders are evaluated, yielding a result in which placeholders are replaced with their corresponding values. so interpolation (in KOTLIN) goes this way:

var age = 21

println("My Age Is: $age")

Remember: "$" sign is used for interpolation.

Answer 3

You could do this:

String.format("%s %s", client.firstName, client.lastName)

Answer 4

Regarding your 2nd question: If you need caching for fullName, you may and should do it explicitly:

class Client {
    val firstName: String
    val lastName: String
    val fullName = "$firstName $lastName"
}

This code is equivalent to your snipped except that the underlying getter getFullName() now uses a final private field with the result of concatenation.