Extending the SequenceType in Swift - swift

Question

I wondered why map() and filter() in SequenceType return both an Array.
Actually, I don't think that's necessary. Returning a sequence again feels much more sensible to me.
However, I got stuck when trying to add sequential versions. Here's my attempt with map:
extension SequenceType {
func seqMap<T, S: SequenceType where S.Generator.Element == T>(
transform: Self.Generator.Element -> T) -> S
{
var sourceGen = generate()
let tGen: AnyGenerator<T> = anyGenerator {
if let el = sourceGen.next() {
return transform(el)
} else {
return nil
}
}
return AnySequence { tGen }
}
}
XCode tells me at the last return statement the following error:
cannot invoke initializer for type 'AnySequence<T>' with an argument list of type '(() -> AnyGenerator<T>)'
note: overloads for 'AnySequence<T>' exist with these partially matching parameter lists: (S), (() -> G)
Actually, my tGen is of type () -> G, so why does XCode think it is ambiguous?

Answer 1

The problem becomes more apparent if you split the return statement:
let tSeq = AnySequence { tGen }
return tSeq // error: cannot convert return expression of type 'AnySequence<T>' to return type 'S'
The compiler would infer the placeholder type S from the context
of a method call, and that could be any sequence
type with element type T, and not necessarily an AnySequence.
Here is a simple example demonstrating the same problem:
protocol MyProtocol { }
struct MyType { }
extension MyType : MyProtocol { }
func foo<P : Protocol>() -> P {
return MyType() // error: cannot convert return expression of type 'MyType' to return type 'P'
}
To solve the problem, change the return type to AnySequence<T>
and drop the generic type S:
extension SequenceType {
func seqMap<T>(transform: Self.Generator.Element -> T) -> AnySequence<T>
{
var sourceGen = generate()
let tGen: AnyGenerator<T> = anyGenerator {
if let el = sourceGen.next() {
return transform(el)
} else {
return nil
}
}
return AnySequence { tGen }
}
}
which can be written more compactly as
extension SequenceType {
func seqMap<T>(transform: Self.Generator.Element -> T) -> AnySequence<T>
{
var sourceGen = generate()
return AnySequence(anyGenerator {
sourceGen.next().map(transform)
})
}
}
using the map() method of the Optional type.
But note that SequenceType already has a lazy method which returns
a LazySequenceType:
/// A sequence containing the same elements as a `Base` sequence, but
/// on which some operations such as `map` and `filter` are
/// implemented lazily.
///
/// - See also: `LazySequenceType`
public struct LazySequence<Base : SequenceType>
and you can use
someSequence.lazy.map { ... }
to get a (lazily evaluated) sequence of the mapped values.