Pass a conditional as an argument in Swift? - swift

Question

I was thinking of abstracting some conditional logic in my application. Suppose I have two functions in this simplified example:
func1(val1: Int, val2: Int, threshold: Int) -> Bool {
return (val1 == threshold && val2 == threshold)
}
func2(val: Int, thresholdHi: Int, thresholdLo: Int) {
return (val < thresholdHi && val > thresholdLo)
}
My idea is to make a function that performs a conditional check on a set of values.
funcIdea(vals[Int], conditional: ???) -> Bool {
for val in vals {
if !conditional(val) { return false; }
}
return true
}
func1(...){
return funcIdea([val1, val2], ???)
}
I think this is possible with either a closure or a function.

Answer 1

It is possible with a closure.
func funcIdea(vals: [Int], conditional: (Int)->Bool) -> Bool {
for val in vals {
if !conditional(val) { return false; }
}
return true
}
func func1(val1: Int, val2: Int, threshold: Int) -> Bool {
//return (val1 == threshold && val2 == threshold)
return funcIdea(vals: [val1, val2], conditional: { (val) -> Bool in
return val > threshold
})
}

Answer 2

You need to pass in a closure to your function, then you can use contains(where:) to check if the closure is true for all elements of the array. The double negation is necessary to allow for an early exit in contains(where:).
extension Array {
func allElementsSatisfy(_ condition: (Element)->(Bool)) -> Bool {
return !self.contains(where: { !condition($0)})
}
}
Then you can simply call it on any Array like this:
[1,2,3].allElementsSatisfy({$0 < 1}) // false
[1,2,3].allElementsSatisfy({$0 < 4}) // true
["a","b","c"].allElementsSatisfy({$0.count == 1}) // true
["a","b","c"].allElementsSatisfy({$0.count > 1}) // false