How to write any(s : SequenceType) in Swift
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?
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.
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