class Base: Equatable {
static func == (lhs: Base, rhs: Base) -> Bool {
lhs.id == rhs.id
}
let id: String
init(id: String) {
self.id = id
}
}
class SubClass: Base {
public var id2: String?
public init(id1: String, id2: String? = nil) {
self.id2 = id2
super.init(id: id1)
}
static func == (lhs: SubClass, rhs: SubClass) -> Bool {
lhs.id2 == rhs.id2 && lhs.id == rhs.id
}
}
print(a != b) // result: false
// Calls `Base` class's static func ==
print(a == b) // result: false
// Calls `SubClass` class's static func ==
I have a simple super class and subclass, subclass inherits Base
and also implements
static func ==
When calling a != b
, it calls Base
class's ==
implementation instead of SubClass
's ==
implementation, why?
But when calling a == b
, it actually call's SubClass
's ==
implementation, why?
I expect both !=
and ==
calls SubClass
's ==
implementation
CodePudding user response:
This problem existed long time ago since the improvement of Swift language (https://github.com/apple/swift-evolution/blob/master/proposals/0091-improving-operators-in-protocols.md)
And the author also mentioned about this problem. You can check that. https://github.com/apple/swift-evolution/blob/master/proposals/0091-improving-operators-in-protocols.md#class-types-and-inheritance
One of the solution I found is define an isEqual
function.
class Superclass: Equatable {
var foo: Int
init(foo: Int) {
self.foo = foo
}
func isEqual(to instance: Superclass) -> Bool {
return self.foo == instance.foo
}
static func ==(lhs: Superclass, rhs: Superclass) -> Bool {
return type(of: lhs) == type(of: rhs) && lhs.isEqual(to: rhs)
}
}
class Subclass: Superclass {
var bar: String
init(foo: Int, bar: String) {
self.bar = bar
super.init(foo: foo)
}
override func isEqual(to instance: Superclass) -> Bool {
guard let instance = instance as? Subclass else {
return false
}
return self.foo == instance.foo && self.bar == instance.bar
}
}
let a = Subclass(foo: 1, bar: "a")
let b = Subclass(foo: 1, bar: "b")
print(a == b) // False
print(a != b) // True
Reference: