I want to use them as a parameter to a method of my Region struct:
private func combineWith(region: RegionProtocol, combine: (Bool, Bool) -> Bool) -> Region {
return Region() {point in
combine(self.contains(point), region.contains(point))
}
}
But apparently, (Bool, Bool) -> Bool) is not what && or || are. If you know, please let me know how you found out.
If you "cmd-click" on the word "Swift" in the statement
import Swift
in Xcode and search for || then you'll find that it is declared as
func ||<T : BooleanType>(lhs: T, rhs: @autoclosure () -> Bool) -> Bool
The reason is the "short-circuiting" behaviour of the || operator: If the first
operand is true, then the second operand must not be evaluated at all.
So you have to declare the parameter as
combine: (Bool, @autoclosure () -> Bool) -> Bool
Example:
func combineWith(a : Bool, b : Bool, combine: (Bool, @autoclosure () -> Bool) -> Bool) -> Bool {
return combine(a, b)
}
let result = combineWith(false, true, ||)
println(result)
Note: I tested this with Xcode 6.1.1. The syntax for autoclosure parameters changed in Swift 1.2 (Xcode 6.3) and I haven't been able yet to translate the above code for Swift 1.2 (see also Rob's comments below).
The only thing that I can offer at the moment are extremely ugly
workarounds. You could wrap || into a closure that does not have
autoclosure parameters:
func combineWith(a : Bool, b : Bool, combine: (Bool, () -> Bool) -> Bool) -> Bool {
return combine(a, { b })
}
let result = combineWith(false, true, { $0 || $1() } )
Or you go without the short-circuiting behaviour:
func combineWith(a : Bool, b : Bool, combine: (Bool, Bool) -> Bool) -> Bool {
return combine(a, b)
}
let result = combineWith(false, true, { $0 || $1 } )
