I am new in swift . let' try to understand closures
declare a function like closures:
let sayHello = {
return "hello"
}
unable to infer closures return type of current context ()->()
My question is here what is the context ?
let sayHello : String = {
return "hello"
}
Function produce expected type of "string" , did you mean to call it with ()
honestly i did not understand this error ? anyone help me to understand and why need to specify this one ()
however . this is working fine and without error it is called closures
let sayHello : String = {
return "hello"
}()
thank you
The context the error message refers to means the initialization statement for the closure, along with its surrounding.
Swift can often infer (figure out from things that it already knows) the type of closure, without requiring you to specify it explicitly.
For example, if you are calling a function that takes a closure taking an Int as its argument, Swift figures out that the type of your closure's argument must be an Int. Similarly, if your closure returns a captured local variable of type String, Swift does not need you to specify the return type, because it figures it must be String as well.
None of this works when you create closures for future use, and assign them to a variable. In situations like that you need to tell Swift what kind of closure you are creating:
let sayHello : ()->String = {
return "hello"
}
Now you have a variable sayHello of type "a closure that takes no arguments and returns a String" (i.e. ()->String closure). You can call it later to say hello:
let hello = sayHello()
Note: Your example that "worked" also defined a closure. However, the closure was used immediately after its definition, and then discarded.
Related
I'm currently learning type as a value in functions and wrote this sample code to play around:
import Foundation
class Animal {
func sound() {
print("Generic animal noises")
}
}
func foo(_ t:Animal) {
print("Hi")
}
foo(Animal) //Cannot convert value of type 'Animal.Type' to expected argument type 'Animal'
I'm not surprised by this result. Obviously you cant pass the type itself as an argument where an instance of that type is expected. But notice that the compiler says that the argument I passed was of type Animal.Type. So if I did this, it should compile right?
func foo(_ t:Animal.Type) {
print("Hi")
}
foo(Animal) //Expected member name or constructor call after type name
This is what really confuses me a heck ton, the compiler told me it was of type Animal.Type *but after making this change it once again shows an error.
Of course I listened to the fix Swift suggests and do:
foo(Animal.self) //Works correctly
But my biggest question is: WHY? Isn't Animal itself the type? Why does the compiler require me to use Animal.self to get the type? This really confuses me, I would like for some guidance.
Self-answering, with help of comments, I was able to find out the reason:
Using .self after the type name is called Postfix Self Expression:
A postfix self expression consists of an expression or the name of a
type, immediately followed by .self. It has the following forms:
expression.self
type.self
The first form evaluates to the value of the expression. For example, x.self evaluates to x.
The second form evaluates to the value of the type. Use this form to access a type as a value. For example, because SomeClass.self evaluates to the SomeClass type itself, you can pass it to a function or method that accepts a type-level argument.
Thus, the .self keyword is required to consider the type as a value capable of being passed as an argument to functions.
How can I access the str label inside the function body? If it's impossible then why the compiler doesn't throw an error ?
class Test {
func methodA(str _: String) {
}
func methodB() {
methodA(str: "")
}
}
Using _ for the parameter name means you have no need to use the parameter within the method. The str argument label has no use inside the implementation of the method itself. It's only useful in the calling syntax.
If you actually have a need to use a parameter in a method, don't give it the anonymous name of _. Give it a useful name so it can be referenced in the implementation of the method. Or simply remove the _ so both the parameter name and label are the same.
Either do:
func methodA(str: String) {
// do something with str
}
or something like:
func methodA(str val: String) {
// do something with val
}
In both cases the caller is the same as you currently have:
methodA(str: "")
The only time you would want an anonymous parameter name is if your implementation of the method doesn't actually make use of the parameter. Maybe the implementation isn't finished or over time it's changed and a parameter isn't needed any more and you don't want to update lots of existing code calling the method.
There is a compiler setting to get warnings about unused parameters. If you have an unused parameter, changing the name to _ eliminates the warning.
See Function Argument Labels and Parameter Names in the Swift book (though it doesn't cover this case of anonymous parameter names).
If I have the following code, why am I not required to use the argument names in the function call and instead an error is thrown when I attempt to add them?
func foo(bar: Any) {}
var faz = foo
foo(1) // Missing argument label 'bar:' in call
foo(bar: 1)
faz(1)
faz(bar: 1) // Extraneous argument label 'bar:' in call
When you assign your method as closure, you just assign method's types of paramters and return type, so type of your closure is actually (Types of arguments) -> Return type
In your specific case: (Any) -> Void
As you can see, this closure doesn't have names for arguments and you can't add them. So adding argument label when you call your closure won't work.
But, you can name parameter which closure takes when you declare it and then you can work with it inside closure scope ... similar to declaring function
faz = { bar in
... // you can work with bar of type `Any`
}
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.