How does string interpolation work in Kotlin?

问题

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

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

or is there some other mechanism?

And if I have a class like this for example:

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

Will this getter return a cached string or will it try to build a new string? Should I use lazyOf delegate instead?

I know that there will be no performance issue unless there will be millions of calls to fullName, but I haven't found documentation about this feature except for how to use it.

回答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.

回答2:

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.

回答3:

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.

回答4:

You could do this:

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

来源：https://stackoverflow.com/questions/37442198/how-does-string-interpolation-work-in-kotlin

标签

kotlin

string-interpolation