Tuples composed of Equatable types should automatically be equatable

Originator:GriotSpeak
Number:rdar://17823187 Date Originated:2014-07-27
Status:Open Resolved:
Product:Swift Product Version:
Classification:Enhancement Reproducible:
 
Summary:
If I try to compare a double of Ints, the compiler gives an error.

Steps to Reproduce:
// this method makes it compile
//private func ==(first:(quotient:Int, modulus:Int), second:(quotient:Int, modulus:Int)) -> Bool {
//    return (first.0 == second.0) && (first.1 == second.1)
//}

internal extension Int {
    
    func divMod(other:Int) -> (quotient:Int, modulus:Int) {
        let remainder = self % other
        let quotient = self / (other - remainder)
        return (quotient, remainder)
    }
    
    public func testDivMod() -> Bool {
        var success = true
        success =  success || (13.divMod(4) == (quotient:3, modulus:1))
        
        return success
    }
}

Expected Results:
I shouldn't need the implementation for ==(a:(Int, Int), b:(Int, Int))->Bool

Actual Results:
I do need the implementation for ==(a:(Int, Int), b:(Int, Int))->Bool

Version:
Version 6.0 (6A267n)

Notes: I think that Comparability could also be inferred.


Configuration:


Attachments:

Comments

Possible implementation

You can implement this yourself, and you may want to add the following to the radar as a proposed implementation. It's tedious and cannot be made generic up to an arbitrary length tuple (which is the same problem Scala has. Scala can handle up to a 22-element tuple because that's how many times they copied the code: http://www.scala-lang.org/api/2.11.2/#scala.Product22). But still, it can be made useful for any real-world tuples.

https://gist.github.com/rnapier/411a710ebb725677b25d 

func ==(lhs:(A,B), rhs:(A,B)) -> Bool {
  return lhs.0 == rhs.0
    && lhs.1 == rhs.1
}
 
func ==(lhs:(A,B,C), rhs:(A,B,C)) -> Bool {
  return lhs.0 == rhs.0
    && lhs.1 == rhs.1
    && lhs.2 == rhs.2
}
 
func ==(lhs:(A,B,C,D), rhs:(A,B,C,D)) -> Bool {
  return lhs.0 == rhs.0
    && lhs.1 == rhs.1
    && lhs.2 == rhs.2
    && lhs.3 == rhs.3
}
 
...
By robnapier at July 27, 2014, 5:26 p.m. (reply...)

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