How to write any(s : SequenceType) in Swift

Question

Swift doesn't seem to have functions for any() or all() so I am trying to make my own. The closest I have gotten is

func any<S: SequenceType
    where S.Generator.Element: BooleanType>
    (sequence: S) -> Bool{

    var __g = sequence.generate() 
    var first_element = __g.next()
    if first_element == nil{
        return false
    }
    return reduce(sequence, first_element!, {$0 || $1})
}

Edit: As per comments, this function should really look like this

func any<S: SequenceType
    where S.Generator.Element: BooleanType>
    (sequence: S) -> Bool{

    return reduce(sequence, false, {$0 || $1})
}

The compiler tells me

'S.Generator.Element' is not convertible to 'Bool'

It seems to me I am explicitly telling the compiler that it is on the second line. What am I missing?

Answer 1

Your problem is that next returns an optional. That means reduce is expecting to return an optional and $0 is an optional. You would have to unwrap the optional to be able to use the || operator. You also don't know the the type is a Bool, just that it is a BooleanType. You would have to also convert it to an actual Bool by calling boolValue on it:

func any<S: SequenceType
    where S.Generator.Element: BooleanType>
    (sequence: S) -> Bool{

    var __g = sequence.generate() 
    var first_element = __g.next()
    if first_element == nil{
        return false
    }
    return reduce(sequence, first_element!.boolValue, {$0 || $1})
}

However, in this case, I think simplicity is best:

func any<S: SequenceType
    where S.Generator.Element: BooleanType>
    (sequence: S) -> Bool{

    for element in sequence {
        if element.boolValue {
            return true
        }
    }
    return false
}

This is actually more efficient too because it returns as soon as it finds the first true instead of always going through the whole thing.