When programming in Swift, sometimes I see error messages containing $T0, $T1, $T2, etc. It seems like they have something to do with closures.
This code works fine:
let test = { (argument: Int) in
println("Argument is \(argument)")
}
test(42)
Xcode's autocomplete tells me that the type of test is (Int) -> (), so the return type appears to be inferred.
But if I add a return statement to the closure:
let test = { (argument: Int) in
println("Argument is \(argument)")
return
}
test(42)
I get an error saying:
error: cannot convert the expression's type '(Int) -> (Int) -> $T0' to type '(Int) -> (Int) -> $T0'
What does $T0 mean?
$T0 refers to an anonymous, or unknown type, generally speaking, if you see it, it means that the type inference failed and you can eliminate the problem by being more specific about typing.
In the specific case you refer to, the issue is that the compiler currently has problems resolving types for closures, particularly when they have return statements and no explicit type. To get your example to work (maybe you want a return in the middle of a block) you can be explicit about the return type:
let test = { (argument: Int) -> Void in
println("Argument is \(argument)")
return
}
(Note: apparently Swift 1.2 fixes several major classes of similar bugs)
Related
I have a method:
func allRegions() -> [MappedRegion] {
return self.items.lazy.compactMap { item in item.valveAny }.flatMap { valve in valve.regions }
}
I was frankly surprised this worked. I'm doing lazy stuff here, but it's apparently having a lazy sequence that turns into a sequence of MappedRegion be the same.
Then I was doing some poor mans timing and modified the function to read:
func allRegions() -> [MappedRegion] {
let startTime = Date()
let result = self.items.lazy.compactMap { item in item.valveAny }.flatMap { valve in valve.regions }
self.sumRender += (Date() - startTime)
return result
}
But that created an error:
Cannot convert return expression of type 'LazySequence<FlattenSequence<LazyMapSequence<LazyMapSequence<LazyFilterSequence<LazyMapSequence<LazySequence<[StatusRowItem]>.Elements, ValveAbstract?>>, ValveAbstract>.Elements, [MappedRegion]>>>' (aka 'LazySequence<FlattenSequence<LazyMapSequence<LazyMapSequence<LazyFilterSequence<LazyMapSequence<Array<StatusRowItem>, Optional<ValveAbstract>>>, ValveAbstract>, Array<MappedRegion>>>>') to return type '[MappedRegion]'
That was initially a surprise. I found that if I specified the return type of result as [MappedRegion] all was happy (e.g. let result:[MappedRegion] = ...).
What is going on here? I get that the original one line function is inferring the result type as [MappedRegion], so I'm probably not getting much benefits from the lazy use. But what confuses me, is that this coercion from a lazy sequence to a fixed array automagically is reminiscent of casting in C, and I thought that Swift didn't do casting?
No, there is no casting going on. There are simply two different flatMap functions being called. LazyMapSequence has two flatMap(_:) functions (well, technically four, but two are deprecated).
In your first code block, this function is inferred (because this version of flatMap has a return type that matches your allRegions function's return type):
func flatMap<SegmentOfResult>(_ transform: (Element) throws -> SegmentOfResult) rethrows -> [SegmentOfResult.Element] where SegmentOfResult : Sequence
And in your second code block, this function is inferred (because there is no type annotation on your local variable that's forcing it to choose the above version of flatMap):
func flatMap<SegmentOfResult>(_ transform: #escaping (Element) -> SegmentOfResult) -> LazySequence<FlattenSequence<LazyMapSequence<LazyMapSequence<Base, Element>, SegmentOfResult>>> where SegmentOfResult : Sequence
// Xcode 11.6 / Swift 5
import Foundation
func f<T>(_: (T) -> Void) { }
#discardableResult
func g(_: Int) -> Int { 0 }
f { g($0) } // compiles fine
f { let _ = g($0) } // Generic parameter 'T' could not be inferred
In the above code, the generic function f expects as its argument a function that takes an argument of type T.
The function g takes an argument of type Int.
When I write f { g($0) }, the code compiles. I believe (please correct me if I'm wrong) this compiles because the compiler can infer that T is an Int based on g's argument type.
However, when I try to do something with the return value of g, for example in the let _ = g($0) line, the compiler complains that it can no longer infer the type of T.
It seems to me the return type of g should have no bearing on how the compiler infers T's type, but clearly it does.
Can anyone shed some light on why this happens, and what (if anything) can be done to correct it?
This may or may not be a compiler bug.
It is known that Swift does not try to infer the types of some closures, namely, multi-statement ones, as said in SR-1570:
This is correct behavior: Swift does not infer parameter or return types from the bodies of multi-statement closures.
However, your closure consists of only one statement, one declaration to be specific. It is possible, albeit weird, that they designed it so that Swift doesn't try to infer types if the closure contains one declaration as well. For example, this does not compile either
f { let x: Int = $0 } // nothing to do with "g"! The issue seems to be with declarations
If this were by-design, the rationale behind it might be because a single declaration in a closure doesn't make much sense anyway. Whatever is declared, won't be used.
But again, this is just speculation, and this could be a bug as well.
To fix it, simply make it a not-a-declaration:
f { _ = g($0) } // this, without the "let", is IMO the idiomatic way to ignore the function result
Or
f { g($0) } // g has #discardableResult anyway, so you don't even need the wildcard
The function f takes in a function as a parameter which in turn takes a parameter of type T and returns nothing (Void). For a closure to infer types automatically, it has to consist of single (and sometimes simple) expression. Anything complex makes it difficult for the compiler to infer (which makes sense from the compiler's standpoint). Apparently, let _ = g($0) is a complex statement as far as the compiler is concerned. For further information, see this mailing list discussion
it looks like #discardableResult gives you an ability to have 2 types of functions:
g(_: Int) -> Int and g(_: Int) -> Void (it is when you don't want to use a result of function)
I think that
f { g($0) } - here your f can infer a type because it has the same Type
(_: (T) -> Void) and (_: (Int) -> Void)
f { let _ = g($0) } - in this case the type of g function is different from f function
(_: (T) -> Void) and (_: (Int) -> Int)
If you will remove "let" it will compile again:
f { _ = g($0) }
I think that
f { let _ = g($0) } - return only Int value
f { _ = g($0) } - return function (_: (Int) -> Int)
Maybe it is a key here
As you may know, Swift can infer types from usage. For example, you can have overloaded methods that differ only in return type and freely use them as long as compiler is able to infer type. For example, with help of additional explicitly typed variable that will hold return value of such method.
I've found some funny moments. Imagine this class:
class MyClass {
enum MyError: Error {
case notImplemented
case someException
}
func fun1() throws -> Any {
throw MyError.notImplemented
}
func fun1() -> Int {
return 1
}
func fun2() throws -> Any {
throw MyError.notImplemented
}
func fun2() throws -> Int {
if false {
throw MyError.someException
} else {
return 2
}
}
}
Of course, it will work like:
let myClass = MyClass()
// let resul1 = myClass.fun1() // error: ambiguous use of 'fun1()'
let result1: Int = myClass.fun1() // OK
But next you can write something like:
// print(myClass.fun1()) // error: call can throw but is not marked with 'try'
// BUT
print(try? myClass.fun1()) // warning: no calls to throwing functions occur within 'try' expression
so it looks like mutual exclusive statements. Compiler tries to choose right function; with first call it tries to coerce cast from Int to Any, but what it's trying to do with second one?
Moreover, code like
if let result2 = try? myClass.fun2() { // No warnings
print(result2)
}
will have no warning, so one may assume that compiler is able to choose right overload here (maybe based on fact, that one of the overloads actually returns nothing and only throws).
Am I right with my last assumption? Are warnings for fun1() logical? Do we have some tricks to fool compiler or to help it with type inference?
Obviously you should never, ever write code like this. It's has way too many ways it can bite you, and as you see, it is. But let's see why.
First, try is just a decoration in Swift. It's not for the compiler. It's for you. The compiler works out all the types, and then determines whether a try was necessary. It doesn't use try to figure out the types. You can see this in practice here:
class X {
func x() throws -> X {
return self
}
}
let y = try X().x().x()
You only need try one time, even though there are multiple throwing calls in the chain. Imagine how this would work if you'd created overloads on x() based on throws vs non-throws. The answer is "it doesn't matter" because the compiler doesn't care about the try.
Next there's the issue of type inference vs type coercion. This is type inference:
let resul1 = myClass.fun1() // error: ambiguous use of 'fun1()'
Swift will never infer an ambiguous type. This could be Any or it could beInt`, so it gives up.
This is not type inference (the type is known):
let result1: Int = myClass.fun1() // OK
This also has a known, unambiguous type (note no ?):
let x : Any = try myClass.fun1()
But this requires type coercion (much like your print example)
let x : Any = try? myClass.fun1() // Expression implicitly coerced from `Int?` to `Any`
// No calls to throwing function occur within 'try' expression
Why does this call the Int version? try? return an Optional (which is an Any). So Swift has the option here of an expression that returns Int? and coercing that to Any or Any? and coercing that to Any. Swift pretty much always prefers real types to Any (and it properly hates Any?). This is one of the many reasons to avoid Any in your code. It interacts with Optional in bizarre ways. It's arguable that this should be an error instead, but Any is such a squirrelly type that it's very hard to nail down all its corner cases.
So how does this apply to print? The parameter of print is Any, so this is like the let x: Any =... example rather than like the let x =... example.
A few automatic coercions to keep in mind when thinking about these things:
Every T can be trivially coerced to T?
Every T can be explicitly coerced to Any
Every T? can also be explicitly coerce to Any
Any can be trivially coerced to Any? (also Any??, Any???, and Any????, etc)
Any? (Any??, Any???, etc) can be explicitly coerced to Any
Every non-throwing function can be trivially coerced to a throwing version
So overloading purely on "throws" is dangerous
So mixing throws/non-throws conversions with Any/Any? conversions, and throwing try? into the mix (which promotes everything into an optional), you've created a perfect storm of confusion.
Obviously you should never, ever write code like this.
The Swift compiler always tries to call the most specific overloaded function is there are several overloaded implementations.
The behaviour shown in your question is expected, since any type in Swift can be represented as Any, so even if you type annotate the result value as Any, like let result2: Any = try? myClass.fun1(), the compiler will actually call the implementation of fun1 returning an Int and then cast the return value to Any, since that is the more specific overloaded implementation of fun1.
You can get the compiler to call the version returning Any by casting the return value to Any rather than type annotating it.
let result2 = try? myClass.fun1() as Any //nil, since the function throws an error
This behaviour can be even better observed if you add another overloaded version of fun1 to your class, such as
func fun1() throws -> String {
return ""
}
With fun1 having 3 overloaded versions, the outputs will be the following:
let result1: Int = myClass.fun1() // 1
print(try? myClass.fun1()) //error: ambiguous use of 'fun1()'
let result2: Any = try? myClass.fun1() //error: ambiguous use of 'fun1()'
let stringResult2: String? = try? myClass.fun1() // ""
As you can see, in this example, the compiler simply cannot decide which overloaded version of fun1 to use even if you add the Any type annotation, since the versions returning Int and String are both more specialized versions than the version returning Any, so the version returning Any won't be called, but since both specialized versions would be correct, the compiler cannot decide which one to call.
// this declaration / definition of variable is OK, as expected
var i = Optional<Int>.None
var j:Int?
// for the next line of code compiler produce a nice warning
// Variable 'v1' inferred to have type 'Optional<Void>' (aka 'Optional<()>'), which may be unexpected
var v1 = Optional<Void>.None
// but the next sentence doesn't produce any warning
var v2:Void?
// nonoptional version produce the warning 'the same way'
// Variable 'v3' inferred to have type '()', which may be unexpected
var v3 = Void()
// but the compiler feels fine with the next
var v4: Void = Void()
What is the difference? Why is Swift compiler always happy if the type is anything else than 'Void' ?
The key word in the warning is "inferred." Swift doesn't like inferring Void because it's usually not what you meant. But if you explicitly ask for it (: Void) then that's fine, and how you would quiet the warning if it's what you mean.
It's important to recognize which types are inferred and which are explicit. To infer is to "deduce or conclude (information) from evidence and reasoning rather than from explicit statements." It is not a synonym for "guess" or "choose." If the type is ambiguous, then the compiler will generate an error. The type must always be well-defined. The question is whether it is explicitly defined, or defined via inference based on explicit information.
This statement has a type inference:
let x = Foo()
The type of Foo() is explicitly known, but the type of x is inferred based on the type of the entire expression (Foo). It's well defined and completely unambiguous, but it's inferred.
This statement has no type inference:
let x: Foo = Foo()
But also, there are no type inferences here:
var x: Foo? = nil
x = Foo()
The type of x (Foo?) in the second line is explicit because it was explicitly defined in the line above.
That's why some of your examples generate warnings (when there is a Void inference) and others do not (when there is only explicit use of Void). Why do we care about inferred Void? Because it can happen by accident very easily, and is almost never useful. For example:
func foo() {}
let x = foo()
This is legal Swift, but it generates an "inferred to have type '()'" warning. This is a very easy error to make. You'd like a warning at least if you try to assign the result of something that doesn't return a result.
So how is it possible that we assign the result of something that doesn't return a result? It's because every function returns a result. We just are allowed to omit that information if the return is Void. It's important to remember that Void does not mean "no type" or "nothing." It is just a typealias for (), which is a tuple of zero elements. It is just as valid a type as Int.
The full form of the above code is:
func foo() -> () { return () }
let x = foo()
This returns the same warning, because it's the same thing. We're allowed to drop the -> () and the return (), but they exist, and so we could assign () to x if we wanted to. But it's incredibly unlikely that we'd want to. We almost certainly made a mistake and the compiler warns us about that. If for some reason we want this behavior, that's fine. It's legal Swift. We just have to be explicit about the type rather than rely on type inference, and the warning will go away:
let x: Void = foo()
Swift is being very consistent in generating warnings in your examples, and you really do want those warnings. It's not arbitrary at all.
EDIT: You added different example:
var v = Optional<Void>()
This generates the error: ambiguous use of 'init()'. That's because the compiler isn't certain whether you mean Optional.init() which would be .None, or Optional.init(_ some: ()), which would be .Some(()). Ambiguous types are forbidden, so you get a hard error.
In Swift any value will implicitly convert with it's equivalent 1-tuple. For example, 1 and (1) are different types. The first is an Int and the second is a tuple containing an Int. But Swift will silently convert between these for you (this is why you sometimes see parentheses pop up in surprising places in error messages). So foo() and foo(()) are the same thing. In almost every possible case, that doesn't matter. But in this one case, where the type really is (), that matters and makes things ambiguous.
var i = Optional<Int>()
This unambiguously refers to Optional.init() and returns nil.
The compiler is warning you about the "dummy" type Void, which is actually an alias for an empty tuple (), and which doesn't have too many usages.
If you don't clearly specify that you want your variable to be of type Void, and let the compiler infer the type, it will warn you about this, as it might be you didn't wanted to do this in the first place.
For example:
func doSomething() -> Void {
}
let a = doSomething()
will give you a warning as anyway there's only one possible value doSomething() can return - an empty tuple.
On the other hand,
let a: Void = doSomething()
will not generate a warning as you explicitly tell the compiler that you want a Void variable.
So I was trying to write some basic Swift, and I wrote:
func timesByHundred(d: Int) {
return d * 100
}
and the compiler said "cannot invoke '*' with an argument list of type '(Int, IntegerLiteralConvertible)'". So I changed it to:
func timesByHundred(d: Int) {
let e: Int = 100
return d * e
}
and the compiler said "cannot invoke '*' with an argument list of type '(Int, Int)'". What can I even multiply if not two ints?? There's some similar questions on here but they all have people trying to operate on different types.
The compiler error is misleading.
The real issue is that you missed the declaration of the function return type, so the compiler infers Void and it gets confused when tries (and fails) to find a suitable overloading for * that returns Void.
Change your function to
func timesByHundred(d: Int) -> Int {
return d * 100
}