Why can't Swift initializers call convenience initializers on their superclass?

Question

Consider the two classes:

class A {     var x: Int      init(x: Int) {         self.x = x     }      convenience init() {         self.init(x: 0)     } }  class B: A {     init() {         super.init() // Error: Must call a designated initializer of the superclass 'A'     } } 

I don't see why this isn't allowed. Ultimately, each class's designated initializer is called with any values they need, so why do I need to repeat myself in B's init by specifying a default value for x again, when the convenience init in A will do just fine?

Answer 1

This is Rule 1 of the "Initializer Chaining" rules as specified in the Swift Programming Guide, which reads:

Rule 1: Designated initializers must call a designated initializer from their immediate superclass.

https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/Initialization.html

Emphasis mine. Designated initializers cannot call convenience initializers.

There is a diagram that goes along with the rules to demonstrate what initializer "directions" are allowed:

Answer 2

Consider

class A {     var a: Int     var b: Int      init (a: Int, b: Int) {         print("Entering A.init(a,b)")         self.a = a; self.b = b     }      convenience init(a: Int) {         print("Entering A.init(a)")         self.init(a: a, b: 0)     }      convenience init() {         print("Entering A.init()")         self.init(a:0)     } }   class B : A {     var c: Int      override init(a: Int, b: Int)     {         print("Entering B.init(a,b)")         self.c = 0; super.init(a: a, b: b)     } }  var b = B() 

Because all designated initializers of class A are overridden, class B will inherit all convenience initializers of A. So executing this will output

Entering A.init() Entering A.init(a:) Entering B.init(a:,b:) Entering A.init(a:,b:) 

Now, if the designated initializer B.init(a:b:) would be allowed to call the base class convenience initializer A.init(a:), this would result in a recursive call to B.init(a:,b:).