[swift-users] Subtract a set of a subclass?

[Zhao Xin]

>
> `A hash value, provided by a type’s hashValue property, is an integer
> that is the same for any two instances that compare equally. That is, for
> two instances a
> ​​
> and b of the same type`


So do you believe A and B are the same type if B inherits A. That's the
differences between you and me.

Zhaoxin

On Fri, Sep 2, 2016 at 6:55 AM, Jordan Rose <jordan_rose at apple.com> wrote:

>
> On Sep 1, 2016, at 15:44, Zhao Xin via swift-users <swift-users at swift.org>
> wrote:
>
> Hi Nick,
>
> Glad to help.
>
> but when using third party classes I don’t know if the hash values are
>> comparable
>>
>
> You can create an extension with a convenient init(:), which creates a new
> instance of  the super class basing on the instance of the sub class. That
> will guarantee the subtraction. Below code works in Xcode 7.3.1 with Swift
> 2.2.
>
> import Foundation
>
> func ==(lhs: Foo, rhs: Foo) -> Bool {
>     return lhs.id == rhs.id
> }
>
> class Foo:Hashable {
>     let id:Int
>     var hashValue: Int {
>         return id
>     }
>
>
>     required init(_ id:Int) {
>         self.id = id
>     }
> }
>
> class Bar:Foo {
>     override var hashValue: Int {
>         return id * 5
>     }
> }
>
> var fooSet:Set<Foo> = [Foo(10), Foo(9), Foo(8), Foo(7)]
> var barSet:Set<Bar> = [Bar(8), Bar(7), Bar(6), Bar(5)]
>
> //fooSet.subtract(barSet) // error: cannot invoke 'subtract' with an
> argument list of type '(Set<Bar>)'
> fooSet = fooSet.subtract(barSet as Set<Foo>) // works, but not what we
> want
> fooSet.forEach { print("\($0.dynamicType), id:\($0.id)") }
> /*
>  Foo, id:7
>  Foo, id:10
>  Foo, id:9
> */
>
>
> This isn't really a sensible thing to do. The rules for Hashable require
> that `a == b` implies `a.hashValue == b.hashValue`, and `a.hashValue !=
> b.hashValue` implies `a != b`. If you break these rules you're going to
> have problems no matter what static types you're using.
>
> Upcasting from Set<Bar> to Set<Foo> is the most concise way to solve this
> problem.
>
> Jordan
>
>
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