swift - Comparing optional arrays -

swift - Comparing optional arrays -

running next code snippet in playground gives error:

let a: [int]? = [1,2] allow b: [int]? = [1,2] == b // value of optional type '[int]?' not unwrapped; did mean utilize '!' or '?'?

while doing similar 'simpler' optional type works:

var x: int? = 10 var y: int? x == y // false

what reasoning behind first case, of optional arrays, not beingness allowed? why can't swift first see if either side if nil (.none) , if not, actual array comparison.

the reason works simpler types because there version of == defined optionals contain types equatable:

func ==<t : equatable>(lhs: t?, rhs: t?) -> bool

but while int equatable, array not (because might contain not equatable - in case how be). equatable things have == operator, not things == operator equatable.

you write special-case version of == optional arrays containing equatable types:

func ==<t: equatable>(lhs: [t]?, rhs: [t]?) -> bool { switch (lhs,rhs) { case (.some(let lhs), .some(let rhs)): homecoming lhs == rhs case (.none, .none): homecoming true default: homecoming false } }

you generalize cover collection containing equatable elements:

func ==<c: collectiontype c.generator.element: equatable> (lhs: c?, rhs: c?) -> bool { switch (lhs,rhs) { case (.some(let lhs), .some(let rhs)): homecoming lhs == rhs case (.none, .none): homecoming true default: homecoming false } }

swift