How to unwrap an optional value from Any type? - swift

Question

Given an array of [Any] that has a mix of optional and non optional values, e.g:
let int:Int? = 1
let str:String? = "foo"
let values:[Any] = [int,2,str,"bar"]
How can we extract the value of the Optional in the Any type (if there is one) so we can create a generic print function that only prints out the values.
E.g. this printArray function goes through and prints each element:
func printArray(values:[Any]) {
for i in 0..<values.count {
println("value[\(i)] = \(values[i])")
}
}
printArray(values)
Which will output:
value[0] = Optional(1)
value[1] = 2
value[2] = Optional("foo")
value[3] = bar
How can we change it so it only prints the underlying value so that it unwraps the value if it's Optional? e.g:
value[0] = 1
value[1] = 2
value[2] = foo
value[3] = bar
Update Progress...
It can work when changing the argument to [Any?], e.g:
let values:[Any?] = [int,2,str,"bar"]
func printArray(values:[Any?]) {
for i in 0..<values.count {
println("value[\(i)] = \(values[i]!)")
}
}
printArray(values)
Which will print the desired:
value[0] = 1
value[1] = 2
value[2] = foo
value[3] = bar
But would still like to see how we can unwrap an Optional from Any as this is what MirrorType.value returns making it difficult to extract the Optional value, e.g:
class Person {
var id:Int = 1
var name:String?
}
var person = Person()
person.name = "foo"
var mt:MirrorType = reflect(person)
for i in 0 ..< mt.count {
let (name, pt) = mt[i]
println("\(name) = \(pt.value)")
}
Prints out:
id = 1
name = Optional("foo")
When I need:
id = 1
name = foo

Answer 1

For Xcode 7 and Swift 2:
func unwrap(any:Any) -> Any {
let mi = Mirror(reflecting: any)
if mi.displayStyle != .Optional {
return any
}
if mi.children.count == 0 { return NSNull() }
let (_, some) = mi.children.first!
return some
}
let int:Int? = 1
let str:String? = "foo"
let null:Any? = nil
let values:[Any] = [unwrap(int),2,unwrap(str),"bar", unwrap(null)]
This will give you [1, 2, "foo", "bar", {NSObject}]
Change NSNull() to nil and the return value of unwrap func to Any? will always unwrap any type.

Answer 2

To maybe save somebody from cobbling it all together from the answers and comments, here is an answer including both "sane" ways and some what I consider to be improvements for Swift 3 coming with Xcode 8.2.1.
Using Reflection
func unwrap<T>(_ any: T) -> Any
{
let mirror = Mirror(reflecting: any)
guard mirror.displayStyle == .optional, let first = mirror.children.first else {
return any
}
return first.value
}
Discussion
The accepted answer from bubuxu fails to compile with Swift 3.
As walkline suggests in his comment, changing .Optional to .optional fixes this (see SE-0005 and Swift API Design Guidelines).
Reasons I thought this solution can be improved:
I find returning NSNull() weird.
I think the alternative of returning nil with return type Any? is also problematic because it turns everything (including non-optional values) into optional values
(e.g. unwrap(any: 42) returns Optional(42)).
When calling unwrap(any:) with anything but an Any value (any more any anybody?) the Swift 3 compiler warns about implicitly
coercing to Any.
Similiar thoughts apply to Sajjon's answer.
The solution I suggest addresses all those points. Be aware however that unwrap(_:) returns nil as type Any so using the nil
coalescing operator does not work anymore. This means that this just shifts around what I think is problematic about the second point. But I found this to be just the right thing to do for the (to me) more interesting use case regarding reflection.
Using an Extension on Optional
protocol OptionalProtocol {
func isSome() -> Bool
func unwrap() -> Any
}
extension Optional : OptionalProtocol {
func isSome() -> Bool {
switch self {
case .none: return false
case .some: return true
}
}
func unwrap() -> Any {
switch self {
case .none: preconditionFailure("trying to unwrap nil")
case .some(let unwrapped): return unwrapped
}
}
}
func unwrapUsingProtocol<T>(_ any: T) -> Any
{
guard let optional = any as? OptionalProtocol, optional.isSome() else {
return any
}
return optional.unwrap()
}
Discussion
This is bascially LopSae's solution updated to Swift 3. I also changed the precondition failure message and added unwrapUsingProtocol(_:).
Usage
class Person {
var id:Int = 1
var name:String?
}
var person = Person()
person.name = "foo"
let mirror = Mirror(reflecting: person)
for child in mirror.children.filter({ $0.label != nil }) {
print("\(child.label!) = \(unwrap(child.value))")
}
No matter if you're using unwrap() or unwrapUsingProtocol(), this will print
id = 1
name = foo
If you're looking for a way to neatly align the output, see Is there a way to use tabs to evenly space out description strings in Swift?

Answer 3

To check if a Any variable is an optional a protocol can be used as a means of a typeless Optional.
Just as its currently imposible (as of Swift 2) to check against a typeless Optional it is also not posible to cast an into a typeless optional:
let anyType: Any.Type = Optional<String>.self
let anyThing: Any = Optional.Some("string")
anyType is Optional.Type // Causes error
let maybeString = anything as? Optional // Also causes error
// Argument for generic parameter 'Wrapped' could not be inferred
However, the proposed OptionalProtocol can also be used to provide a generic-less interface to access the Optional values and even unwrap them:
protocol OptionalProtocol {
func isSome() -> Bool
func unwrap() -> Any
}
extension Optional : OptionalProtocol {
func isSome() -> Bool {
switch self {
case .None: return false
case .Some: return true
}
}
func unwrap() -> Any {
switch self {
// If a nil is unwrapped it will crash!
case .None: preconditionFailure("nill unwrap")
case .Some(let unwrapped): return unwrapped
}
}
}
// With this we can check if we have an optional
let maybeString: String? = "maybe"
let justString: String = "just"
maybeString is OptionalProtocol // true
justString is OptionalProtocol // false
With the methods provided the optionals can be checked and accessed in quite a natural way, without needing the impossible cast to Optional:
let values:[Any] = [
Optional.Some(12),
2,
Optional<String>.None, // a "wrapped" nil for completeness
Optional.Some("maybe"),
"something"
]
for any in values {
if let optional = any as? OptionalProtocol {
if optional.isSome() {
print(optional.unwrap())
} else {
// nil should not be unwrapped!
print(optional)
}
continue
}
print(any)
}
Which will print:
12
2
nil
maybe
something

Answer 4

Slight alteration on #thm to completely unwrap:
func unwrap<T>(_ any: T) -> Any {
let mirror = Mirror(reflecting: any)
guard mirror.displayStyle == .optional, let first = mirror.children.first else {
return any
}
return unwrap(first.value)
}

Answer 5

I think this is a kind of bug.
In general, to discover and extract the specific type from Any, down casting with as is the only supported method. But :
let int:Int? = 1
let any:Any = int
switch any {
case let val as Optional<Int>: // < [!] cannot downcast from 'Any' to a more optional type 'Optional<Int>'
print(val)
default:
break
}
This means, there is no supported way to do that.
Anyway, apparently you can do that with reflect:
func printArray(values:[Any]) {
for i in 0..<values.count {
var val = values[i]
var ref = reflect(val)
// while `val` is Optional and has `Some` value
while ref.disposition == .Optional && ref.count > 0 && ref[0].0 == "Some" {
// replace `val` with unwrapped value
val = ref[0].1.value;
ref = reflect(val)
}
println("value[\(i)] = \(val)")
}
}
let int:Int? = 1
let str:String? = "foo"
let values:[Any] = [int,2,str,"bar"]
printArray(values)
outputs:
value[0] = 1
value[1] = 2
value[2] = foo
value[3] = bar
ADDED: minor tweaked version
func printArray(values:[Any]) {
for i in 0..<values.count {
var ref = reflect(values[i])
// while `val` is Optional and has `Some` value
while ref.disposition == .Optional && ref.count > 0 && ref[0].0 == "Some" {
// Drill down to the Mirror of unwrapped value
ref = ref[0].1
}
let val = ref.value
println("value[\(i)] = \(val)")
}
}
Factoring out into a function:
func unwrapAny(val:Any) -> Any {
var ref = reflect(val)
while ref.disposition == .Optional && ref.count > 0 && ref[0].0 == "Some" {
ref = ref[0].1
}
return ref.value
}
func printArray(values:[Any]) {
for i in 0..<values.count {
println("value[\(i)] = \(unwrapAny(values[i]))")
}
}

Answer 6

Not a complete answer. It boils down to this:
let int:Int? = 1
let str:String? = "foo"
let values:[Any] = [int,2,str,"bar"]
func printArray(values:[Any]) {
for i in 0..<values.count {
let v = values[i]
if _stdlib_demangleName(_stdlib_getTypeName(v)) == "Swift.Optional" {
println("value[\(i)] = "it's optional: \(v)") // here I'm stuck
}else {
println("value[\(i)] = \(values[i])")
}
}
}
printArray(values)

Answer 7

how about this solution, I made a generic version of previous answer.
fileprivate func unwrap<T>(value: Any)
-> (unwraped:T?, isOriginalType:Bool) {
let mirror = Mirror(reflecting: value)
let isOrgType = mirror.subjectType == Optional<T>.self
if mirror.displayStyle != .optional {
return (value as? T, isOrgType)
}
guard let firstChild = mirror.children.first else {
return (nil, isOrgType)
}
return (firstChild.value as? T, isOrgType)
}
let value: [Int]? = [0]
let value2: [Int]? = nil
let anyValue: Any = value
let anyValue2: Any = value2
let unwrappedResult:([Int]?, Bool)
= unwrap(value: anyValue) // ({[0]}, .1 true)
let unwrappedResult2:([Int]?, Bool)
= unwrap(value: anyValue2) // (nil, .1 true)
let unwrappedResult3:([UInt]?, Bool)
= unwrap(value: anyValue) // (nil, .1 false)
let unwrappedResult4:([NSNumber]?, Bool)
= unwrap(value: anyValue) ({[0]}, .1 false)
The following is code on Playground.

Answer 8

Based on the solution by #bubuxu, one can also:
func unwrap<T: Any>(any: T) -> T? {
let mirror = Mirror(reflecting: any)
guard mirror.displayStyle == .optional else { return any }
guard let child = mirror.children.first else { return nil }
return unwrap(any: child.value) as? T
}
But you need to check against nil using ?? nil when using unwrap, as done in foo
func foo<T>(_ maybeValue: T?) {
if let value: T = unwrap(any: maybeValue) ?? nil {
print(value)
}
}
Still neat though!
(Anyone got a solution for the ?? nil check?)

Answer 9

Without making it too complicated, why not:
let int:Int? = 1
let str:String? = "foo"
let values:[Any?] = [int,2,str,"bar"]
for var i:Int = 0; i < values.count; i++
{
println("\(values[i]!)")
}
This prints:
1
2
foo
bar

Answer 10

According to Using Enumeration case patterns in Swift 2.0
those might be look like this:
let pattern :[Int?] = [nil, 332, 232,nil,55]
for case let number? in pattern {
print(number)
}
Output:
332,
232,
55