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Flatten nested optionals resulting from 'try?'


Swift's try? statement currently makes it easy to introduce a nested optional. Nested optionals are difficult for users to reason about, and Swift tries to avoid producing them in other common cases.

This document proposes giving try? the same optional-flattening behavior found in other common Swift features, to avoid the common occurrence of a nested optional.

Swift-evolution thread: Make try? + optional chain flattening work together


It's currently quite easy to end up with a nested Optional type when using try?. Although it is valid to construct a nested optional, it is usually not what the developer intended.

Swift has various mechanisms to avoid accidentally creating nested optionals. For example:

// Note how 'as?' produces the same type regardless of whether the value
// being cast is optional or not.
let x = nonOptionalValue() as? MyType    // x is of type 'MyType?'
let y = optionalValue() as? MyType       // y is of type 'MyType?'

// Note how optional chaining produces the same type whether or not the
// call produces an optional value.
let a = optThing?.pizza()             // a is of type 'Pizza?'
let b = optThing?.optionalPizza()     // b is of type 'Pizza?'

However, try? behaves differently:

let q = try? harbor.boat()           // q is of type 'Boat?'
let r = try? harbor.optionalBoat()   // r is of type 'Boat??'

The above examples are contrived, but it's actually quite common to end up with a nested optional in production code. For example:

// The result of 'foo?.makeBar()' is 'Bar?' because of the optional
// chaining on 'foo'. The 'try?' adds an additional layer of 
// optionality. So the type of 'x' is 'Bar??'
let x = try? foo?.makeBar()

// JSONSerialization.jsonObject(with:) returns an 'Any'. We use 'as?' to 
// verify that the result is of the expected type, but the result is that 'dict' 
// is now of type '[String: Any]??' because 'try?' added an additional layer.
let dict = try? JSONSerialization.jsonObject(with: data) as? [String: Any]

Although it is fairly easy to produce a nested optional using try?, a survey of existing code suggests that it is almost never the desired outcome. Code that uses try? with nested optionals is almost always accompanied by one of the following patterns:

// Pattern 1: Double if-let or guard-let
if  let optionalX = try? self.optionalThing(),
    let x = optionalX {
    // Use 'x' here

// Pattern 2: Introducing parentheses to let 'as?' flatten for us
if let x = (try? somethingAsAny()) as? JournalEntry {
    // use 'x' here

// Pattern 3: Pattern matching
if case let x?? = try? optionalThing() {
    // use 'x' here

The need for these workarounds makes the language more difficult to learn and use, and they don't really give us any benefit in return.

Code using try? generally does not care to distinguish between the error case and the nil-result case, which is why all these patterns focus on extracting the value and ignore the error. If the developer does care to specifically detect errors, they should probably be using do/try/catch instead.

Proposed solution

In Swift 5, try? someExpr() will mirror the behavior of foo?.someExpr():

  • If someExpr() produces a non-optional value, it will be wrapped in an Optional.
  • If someExpr() produces an Optional, then no additional optional-ness is added.

This results in the following changes to the type of a try? expression:

// Swift 4: 'Int??'
// Swift 5: 'Int?'
let result = try? database?.countOfRows(matching: predicate)

// Swift 4: 'String??'
// Swift 5: 'String?'
let myString = try? String(data: someData, encoding: .utf8)
// Swift 4: '[String: Any]??'
// Swift 5: '[String: Any]?'
let dict = try? JSONSerialization.jsonObject(with: data) as? [String: Any]

There are no changes to the overall type when the sub-expression produces a non-optional.

// Swift 4: 'String?'
// Swift 5: 'String?'
let fileContents = try? String(contentsOf: someURL)

If the sub-expression already produces a nested optional, the result is equally nested:

func doubleOptionalInt() throws -> Int?? {
    return 3

// Swift 4: 'Int???'
// Swift 5: 'Int??'
let x = try? doubleOptionalInt()

A side note about try? and as?

Although as? often has the effect of flattening Optionals (as shown in the example in the Motivation section) it does not exhibit exactly the same behavior as proposed here for try?. Because as? takes an explicit type, it can actually flatten multiple levels of nested Optionals. foo as? T will always produce an Optional<T>, regardless of how many optionals were on foo. This can potentially add or subtract levels of optionals, depending on the type specified. (It can also cast between subtypes and supertypes, which is unrelated to the behavior under consideration.)

In practice, the most common use of as? with nested optionals is to reduce from T?? to T?, which makes it superficially similar to the use optional-chaining use case and the proposed behavior of try?. But as? is a more powerful and versatile construct than what is proposed for try? here.

Detailed design

The type of a try? expression in Swift 4 is defined as Optional<T>, where T is the type of the expression following the try? keyword.

In Swift 5, the type of a try? expression will be some type U, where U is an Optional<_> type and where the sub-expression type T is coercible to U. The type constraint system automatically chooses the smallest level of optional-nesting needed to satisfy this constraint, which results in the behavior described in this proposal.


Some questions have been raised regarding the interoperability with generic code, as in the following example:

func test<T>(fn: () throws -> T) -> T? {

    // Will this line change behavior if T is an Optional?
    if let result = try? fn() {
        print("We got a result!")
        return result
    else {
        print("There was an error")
        return nil

// T is inferred as 'Int' here
let value  = test({ return 15 })

// T is inferred as 'Int?' here
let value2 = test({ return 15 as Int? })

The answer is that it does not matter if T is optional at runtime. At compile time, result has a clearly-defined type: T. This is true in both Swift 4 and Swift 5 modes; because T is not known to be an Optional type at compile-time, a single layer of Optional is added via the try? expression and then unwrapped via if let.

Generic code that uses try? can continue to use it as always without concern for whether the generic type might be optional at runtime. No behavior is changed in this case.

Source compatibility

This is a source-breaking change for try? expressions that operate on an Optional sub-expression if they do not explicitly flatten the optional themselves. It appears that those cases are rare, though; see the analysis below for details. We can provide backward-compatible behavior when the compiler is running in Swift 4 mode, and a migration should be possible for most common cases.

The bar for including source-breaking changes in Swift 5 is high, but I believe it passes the bar. These are the criteria listed in the swift-evolution README:

1. The current syntax/API must be shown to actively cause problems for users.

Nested optionals are a complex concept. They have value in the language, but their use should be intentional. Currently, it's far too easy for beginners to create a nested optional without understanding why, due to the current interaction of try? and optional chaining or as? casting.

Code that uses try? to actually detect errors is more difficult to understand than just using try. Compare:

// Using 'try?'
if let result = try? foo?.bar() {
    // Do something with 'result', which may be nil
    // even though we are in an 'if let'.
else {
    // There was an error, but we don't know what it is
// Using 'try/catch'
do {
    let result = try foo?.bar()
    // Do something with 'result', which may be nil due to optional chaining
catch {
    // Handle the error

The variant using try? is significantly less clear (what is the type of result?), and has no obvious advantages. Using try? is better if you don't care about the else clause and only want to handle a value if one exists, but that use case is better served by the proposed change.

The current syntax is also harmful because of its interaction with as? casting, as seen here:

if let x = try? foo() as? String {
    // We specifically called out `String` as the desired type here,
    // so it unexpected that `x` is of type `Optional<String>`

2. The new syntax/API must be clearly better and must not conflict with existing Swift syntax.

The proposed change resolves all of the above problems, which is better. This change also clarifies the role of try? as a means for accessing values when possible, rather than as an alternative error-handling mechanism to try/catch.

3. There must be a reasonably automated migration path for existing code.

As shown in the analysis below, most source code will require no migration; developers who have encountered code that produces nested Optionals are likely already using patterns that are source-compatible with this change. This proposal simply provides a way to simplify that code.

Automated migration is implemented for the double if/guard let and case let value??: patterns mentioned above.

Swift Source Compatibility Suite analysis

The Swift Source Compatibility Suite suggests that this is unlikely to be a breaking change for most users. I manually inspected the use cases of try? in the compatibility suite. Here are the results:

  • There are 613 total instances of try? in the compatibility suite. The vast majority of those appear to use non-optional sub-expressions, and would be unaffected by this proposal.

  • There are 4 instances of try? ... as?. All four of them wrap the try? in parentheses to get the flattening behavior of as?, and are source-compatible with this change. They all look something like this:

    (try? JSONSerialization.jsonObject(with: $0)) as? NSDictionary
  • There are 12 cases of try? foo?.bar() across 3 projects. 10 of those assign it to _ = try? foo?.bar() , so the resulting type does not matter. 2 of those cases have a Void sub-expression type, and do not assign the result to any variable.

  • There are 6 instances of try? somethingReturningOptional() . They all flatten it manually using flatMap { $0 }, and are thus source-compatible with this change, as the return type of that expression is the same under either behavior.

    (try? get(key)).flatMap { $0 }
  • As far as I can tell, there are zero cases in the entire suite where a double-optional is actually used to distinguish between the error case and the nil-as-a-value case.

  • As far as I can tell, there are zero cases of source incompatibility found in the compatibility suite.

Effect on ABI stability

No impact on ABI.

Effect on API resilience

A try? expression is never exposed at function boundaries, so API resilience should be unaffected.

Alternatives considered

Alter the binding precedence of try?

For expressions like let x = try? getObject() as? Foo, the nested optional can be eliminated by parsing the expression as (try? getObject) as? Foo. Adding explicit parentheses like this is already a common workaround for the double-optional problem.

However, this change would not resolve cases of try? with optional chaining (e.g. try? foo?.bar?.baz()), nor cases where an optional result is returned directly from a function (e.g. try? cachedValue(for: key)).

Altering the binding precedence of try? would also be far more source-breaking than this proposal.

Do nothing

It is possible to write correct code under the current model. We are not proposing to eliminate nested Optionals from the language entirely, so we could just expect users to figure them out.

This is a workable solution, but it is odd to have such a complex structure produced as part of a syntactic sugar designed to simplify a common case.