Swift protocol for string interpolation
You need to implement the Printable
protocol:
This protocol should be adopted by types that wish to customize their textual representation. This textual representation is used when objects are written to an
OutputStreamType
.
protocol Printable {
var description: String { get }
}
There's also the DebugPrintable
protocol when it's only for debugging purposes:
This protocol should be adopted by types that wish to customize their textual representation used for debugging purposes. This textual representation is used when objects are written to an
OutputStreamType
.
protocol DebugPrintable {
var debugDescription: String { get }
}
Documentation (Thanks @MartinR)
Note: As @Antonio and @MartinR mentioned in the comments, this doesn't work in the playground (as of Xcode6 GM anyway); that's a known bug. It does work in compiled apps.
From the Xcode6 GM Release Notes:
In Playgrounds, println() ignores the Printable conformance of user-defined types. (16562388)
As of Swift 2.0 Printable
has now become CustomStringConvertible
. Everything stays the same as before, you still need to implement
var description: String { get }
But now its called CustomStringConvertible. And debug is CustomDebugStringConvertible
In Swift 5 Apple introduced Custom String Interpolation.
Suppose you have person struct with two properties name and age.
struct Person {
var name: String
var age: Int
}
If you wanted to add a special string interpolation for that so that we can print persons in descriptive way, we can add an extension to String.StringInterpolation with a new appendInterpolation() method.
extension String.StringInterpolation {
mutating func appendInterpolation(_ person: Person) {
appendInterpolation("My name is \(person.name) and I'm \(person.age) years old.")
}
}
Now If we print the person details like:
let person = Person(name: "Yogendra", age: 28)
print("Person Details: \(person)")
Output will be:
Person Details: My name is Yogendra and I'm 28 years old.