Access default implementation of CustomStringConvertible in Swift

Question

I want to extend the default behaviour of the CustomStringConvertible protocol in my class, by adding some extra output.

For example, say you have this class in a project called Bar:

class Baz {
   let x = 42
}
let b = Baz()

When doing a po(b), the output will be Bar.Baz

I want to display some extra stuff, such as the value of x, without having to type again Bar.Baz

Is this possible?

Please show us an example of what you want to happen when you print a custom class. — Alex Popov
The protocol doesn't have a default implementation, each class/struct must implement it's own version — Cristik
Good question, but there are still no direct answer. Can it be reopened somehow? — kelin

Alex Popov Alex Popov · Accepted Answer · 2016-01-07T23:57:01

The reason why most of the classes you use have a default implementation of CustomStringConvertible is because NSObject (the superclass of all Obj-C classes) implements it.

If you have defined your own Swift base class, you must first declare that it conform to CustomStringConvertible and then implement it, e.g.

class Dog: CustomStringConvertible {
  // some code and stuff here
  var name = "Bobby"

  // conform to CustomStringConvertible
  var description: String {
    return "\(NSStringFromClass(self.dynamicType)) \(name)"
  }
}

print(Dog()) // outputs: "__lldb_expr_70.Dog Bobby" in Playground

A class that does not conform to CustomStringConvertible merely prints the class name.

Access default implementation of CustomStringConvertible in Swift

1 Answers