I am trying to understand how this custom Navigation bar / Paging View works, found here. What is tripping me up when I went through the README was setting up the tinder-like custom behavior:
// Tinder Like
controller?.pagingViewMoving = ({ subviews in
for v in subviews {
var lbl = v as UIImageView
var c = gray
if(lbl.frame.origin.x > 45 && lbl.frame.origin.x < 145) {
c = self.gradient(Double(lbl.frame.origin.x), topX: Double(46), bottomX: Double(144), initC: orange, goal: gray)
}
else if (lbl.frame.origin.x > 145 && lbl.frame.origin.x < 245) {
c = self.gradient(Double(lbl.frame.origin.x), topX: Double(146), bottomX: Double(244), initC: gray, goal: orange)
}
else if(lbl.frame.origin.x == 145){
c = orange
}
lbl.tintColor = c
}
})
I don't understand why there are parentheses around the closure that is being set to the controller?.pagingViewMoving property.
When I look in the SLPagingViewSwift.swift file, the .pagingViewMoving property is set to this alias:
public typealias SLPagingViewMoving = ((subviews: [UIView])-> ())
What are the extra set of parentheses doing outside the function type?
In this case the parentheses are purely for clarity and are unnecessary. Its just like saying let a = (2 + 2). Although this should not be confused with closures provided as arguments in the following case.
If a function itself takes an argument which is a closure, the closure is simply within the parentheses thats contain the functions parameters.
So thanks to the syntax higher order functions that take a closure as their last (or only) argument that be represented in two ways - inside the argument parentheses or as whats known as a trailing closure. Consider the following function:
func foo(bar: (Int -> Int)) {...}
it can be called in two ways - first with the closure in the parentheses like so:
foo({(i) in i + 2})
or as a trailing closure:
foo {(i) in i + 2}
Related
It seems logical to me that escaping closures would capture structs by copying. But if that was the case, the following code makes no sense and should not compile:
struct Wtf {
var x = 1
}
func foo(){
var wtf = Wtf()
DispatchQueue.global().async {
wtf.x = 5
}
Thread.sleep(forTimeInterval: 2)
print("x = \(wtf.x)")
}
Yet it compiles successfully and even prints 5 when foo is called. How is this possible?
While it might make sense for a struct to be copied, as your code demonstrates, it is not. That's a powerful tool. For example:
func makeCounter() -> () -> Int {
var n = 0
return {
n += 1 // This `n` is the same `n` from the outer scope
return n
}
// At this point, the scope is gone, but the `n` lives on in the closure.
}
let counter1 = makeCounter()
let counter2 = makeCounter()
print("Counter1: ", counter1(), counter1()) // Counter1: 1 2
print("Counter2: ", counter2(), counter2()) // Counter2: 1 2
print("Counter1: ", counter1(), counter1()) // Counter1: 3 4
If n were copied into the closure, this couldn't work. The whole point is the closure captures and can modify state outside itself. This is what separates a closure (which "closes over" the scope where it was created) and an anonymous function (which does not).
(The history of the term "close over" is kind of obscure. It refers to the idea that the lambda expression's free variables have been "closed," but IMO "bound" would be a much more obvious term, and is how we describe this everywhere else. But the term "closure" has been used for decades, so here we are.)
Note that it is possible to get copy semantics. You just have to ask for it:
func foo(){
var wtf = Wtf()
DispatchQueue.global().async { [wtf] in // Make a local `let` copy
var wtf = wtf // To modify it, we need to make a `var` copy
wtf.x = 5
}
Thread.sleep(forTimeInterval: 2)
// Prints 1 as you expected
print("x = \(wtf.x)")
}
In C++, lambdas have to be explicit about how to capture values, by binding or by copying. But in Swift, they chose to make binding the default.
As to why you're allowed to access wtf after it's been captured by the closure, that's just a lack of move semantics in Swift. There's no way in Swift today to express "this variable has been passed to something else and may no longer be accessed in this scope." That's a known limitation of the language, and a lot of work is going into fix it. See The Ownership Manifesto for more.
Here is a block of code that is equivalent (to my knowledge) to the other.
let f:()->() = brick
where brick is
func brick()->Void{ print("Throw Brick");}
but I can also write it as
let f = {return brick()}
What is this ^ code doing.
The first code I know makes sense to me. This is where I am defining a variable who's type is of the signature ()->() or ()->Void. And then passing the reference of brick function to the variable.
Thanks
Though in both code snippets, f will behave in the same way when you call it, the code snippets are semantically different.
let f:()->() = brick
This assigns the function brick to the let constant f. Note that the type annotation is not required since the compiler knows that brick is a function that takes no parameters and returns Void, so it can infer that f must also be such a function too. Therefore, you can write it as:
let f = brick
Another way to write functions is to use a closure expression. For example, the following closure expression represents a function that calls brick:
{ return brick() }
Since you omitted the in keyword and didn't use any shorthand argument names ($0, $1 etc), the compiler infers that the closure expression takes no arguments. And since brick() returns Void, the closure expression returns Void too - return brick() means "return what brick() returns". The compiler is able to infer the type of the closure expression, so this is valid:
let f = { return brick() }
This assigns a "closure expression that calls simply calls brick() and returns what it returns" to f.
Depending on how you look at it, this is a bit different from let f = brick, where you are directly assigning brick to f. But in the end, in both cases calling f will do the same thing - you will end up calling brick.
The difference is sort of similar to the difference between let x: Double = 1 and let x = cos(0) - x = 1 in both cases, but one of those ways is more direct.
First of all your two versions are not equivalent. This:
func brick()->Void{ print("Throw Brick");}
let f = {return brick()}
is equated by this:
func brick()->Void{ print("Throw Brick");}
func brickCaller() { return brick() }
let f = brickCaller
Whereas, the equivalent of your
let f:()->() = brick
func brick()->Void{ print("Throw Brick");}
would be simply
let f:()->() = { print("Throw Brick") }
with no intermediary call to brick at all.
So, anyway, I'd put it differently from Sweeper's answer. I'd reply that a function can be declared-with-a-name, using func, or be nameless (anonymous), using a mere function body.
So the point is that in each pair of equivalents, one function has a name-and-declaration of its own, with func, and the other function doesn't — its body is simply assigned directly to the variable f.
I'm having an issue with evaluation of one line of code
if i break it down to two lines, it's working , but in one line of code, it's just evaluate in a 'new' to a 'wrong' way.
my main reason for asking this question, is not to solve it, I know I can use parenthesis to solve it, and break it to Two line, but don't want to solve it, I just want to know why its evaluated like this , and if there's a solution for this : some setting to patch , in Order THAT it will work in ONE LINE OF CODE :
Heres the code that working in Two lines
Heres the code that trying to do the same thing, but rise an error as you can see:
full code of both working and not working :
class ClosuresStack {
var dic = Dictionary<String,(()->String)->String >()
subscript(_ str:String)-> (()->String)->String {
get {
return dic[str]!
}
set {
dic[str] = newValue
}
}
}
func createClosuresStak() -> ClosuresStack {
let cs = ClosuresStack()
func takesAClosureReturnA_string(_ closure:()->String) ->String {
return closure() + " Two"
}
cs["C"] = takesAClosureReturnA_string
return cs
}
let c = createClosuresStak()["C"]
let str = c{"One"}
print(str) // print: One Two
let c = createClosuresStak()["C"]{"One"} // error -->
now, I want to somehow understand how to change it that it will work in ONE LINE OF CODE : meaning that the evaluation of 'createClosuresStak()["C"]{"One"}' will create a closure after ["C"] , and then from that point writing the {"One"}
will make it a full evaluate of the line :
let c = createClosuresStak()["C"]{"One"}
making 'c' a String
if that's not possible, I need to know it Too , tnx :)
UPDATE
tnx for the comments , its help me understand the problem more clearly :
1) im understanding that the createClosuresStak()["C"]{"One"}
acutely trying to add the string 'One' as another parameter to the sub script , and there for the error from the compiler was that is cannot subscript (String,()->String} , 'C' as the string inside the [] , and the other parameter {"One"} -> BUT , isn't that some kind of a bug?, been that i'm using [] ,Cleary the compiler need to 'understand' that I want to subscript a String, also by power of inferring that swift has,
2) now I'm still trying to get that syntax to work as it is so I try to change some things, in order to get it to work :
so I created a function that take a string, and return a dictionary of type : Dictionary<String,()->String>, and then trying so subscript it
and the compiler don't rise an error that way :
func closuresDictionary(_ s:String) -> Dictionary<String,()->String> {
var dic = Dictionary<String,()->String>()
func foo()->String {
return s + " Two"
}
dic["C"] = foo
return dic
}
let c = closuresDictionary("One")["C"]{ "SomeString" }
c is now a closure of type ()->String which does noting with string that I put inside, so the syntax works, but the outcome is not doing anything.
when im changing the return type of the dictionary to a different closure : (String)->String instead of ()->String , im getting the same old error, that I'm trying to subscript a (String,(String)->String)
and I need a function that will take the string inside the {} , and create something from it meaning that I need to subscript to return a closure of (String)->String
its seems like there's no way to do that
im adding two more pictures of my last trying in order to get this line of code in current syntax to work
the wanted syntax working but the outcome is not an outcome not doing any thing with the string inside the {}:
same error, by changing the function to (String)->String
Your example:
let c = createClosuresStak()["C"]{"One"}
is using trailing closure syntax.
Trailing closure syntax works by including the trailing closure as an additional parameter to a function call. Subscripting an array is really a function call under the hood (to a function called subscript), and Swift is trying to pass that closure as a second parameter to the subscripting call, which is what the error is explaining:
Cannot subscript a value of type 'ClosuresStack' with an argument of type '(String, () -> String)'.
In other words, you can't pass both "C" and the closure {"One"} to the subscripting function.
There are at least 3 ways to fix this and still put it on one line:
Option 1: Use an explicit call to pass the closure instead of using trailing closure syntax
Wrap the closure in () to make the call explicit:
let c1 = createClosuresStak()["C"]({"One"})
print(c1)
Option 2: Wrap the createClosureStak()["C"] in parentheses
That lets Swift know the subscripting only gets "C" as a parameter and allows trailing closure syntax to work as expected:
let c2 = (createClosuresStak()["C"]){"One"}
print(c2)
Option 3: Add .self to the result before the trailing closure syntax:
That again finishes the subscripting call and avoids the confusion.
let c3 = createClosuresStak()["C"].self {"One"}
print(c3)
Personally, I would choose option one, because trailing closure syntax is unnecessary syntactic sugar that clearly is not working here.
Let's say I do the following in C++:
int i = 1;
int* ptr = &i;
*ptr = 2;
cout << i << '\n';
And I want to do something similar in swift. Could I do the following?
var i : Int = 1
var iptr : UnsafeMutablePointer<Int> = &i
iptr.memory = 2
print(i)
And achieve the same result?
Yes-ish.
You can't do it exactly as you've attempted in the question. It won't compile. Swift won't let you directly access the address of a value like this. At the end of the day, the reason is mostly because there's simply no good reason to do so.
We do see the & operator in Swift however.
First of all, there is the inout keyword when declaring function parameters:
func doubleIfPositive(inout value: Float) -> Bool {
if value > 0 {
value *= 2
return true
}
return false
}
And to call this method, we'd need the & operator:
let weMadeARadian = doubleIfPositive(&pi)
We can see it similarly used when we have a function which takes an argument of type UnsafeMutablePointer (and other variants of these pointer structs). In this specific case, it's primarily for interoperability with C & Objective-C, where we could declare a method as such:
bool doubleIfPositive(float * value) -> bool {
if (value > 0) {
value *= 2;
return true;
}
return false;
}
The Swift interface for that method ends up looking somethin like this:
func doubleIfPositive(value: UnsafeMutablePointer<Float>) -> Bool
And calling this method from Swift actually looks just like it did before when using the inout approach:
let weMadeARadian = doubleIfPositive(&pi)
But these are the only two uses of this & operator I can find in Swift.
With that said, we can write a function that makes use of the second form of passing an argument into a method with the & operator and returns that variable wrapped in an unsafe mutable pointer. It looks like this:
func addressOf<T>(value: UnsafeMutablePointer<T>) -> UnsafeMutablePointer<T> {
return value
}
And it behaves about as you'd expect from your original code snippet:
var i: Int = 1
var iPtr = addressOf(&i)
iPtr.memory = 2
print(i) // prints 2
As noted by Kevin in the comments, we can also directly allocate memory if we want.
var iPtr = UnsafeMutablePointer<Int>.alloc(1)
The argument 1 here is effectively the mount of space to allocate. This says we want to allocate enough memory for a single Int.
This is roughly equivalent to the following C code:
int * iPtr = malloc(1 * sizeof(int));
BUT...
If you're doing any of this for anything other than interoperability with C or Objective-C, you're most likely not Swifting correctly. So before you start running around town with pointers to value types in Swift, please, make sure it's what you absolutely need to be doing. I've been writing Swift since release, and I've never found the need for any of these shenanigans.
Like this (not the only way, but it's clear):
var i : Int = 1
withUnsafeMutablePointer(&i) {
iptr -> () in
iptr.memory = 2
}
print(i)
Not a very interesting example, but it is completely parallel to your pseudo-code, and we really did reach right into the already allocated memory and alter it, which is what you wanted to do.
This sort of thing gets a lot more interesting when what you want to do is something like cycle thru memory just as fast as doing pointer arithmetic in C.
I know about the ampersand as a bit operation but sometimes I see it in front of variable names. What does putting an & in front of variables do?
It works as an inout to make the variable an in-out parameter. In-out means in fact passing value by reference, not by value. And it requires not only to accept value by reference, by also to pass it by reference, so pass it with & - foo(&myVar) instead of just foo(myVar)
As you see you can use that in error handing in Swift where you have to create an error reference and pass it to the function using & the function will populate the error value if an error occur or pass the variable back as it was before
Why do we use it? Sometimes a function already returns other values and just returning another one (like an error) would be confusing, so we pass it as an inout. Other times we want the values to be populated by the function so we don't have to iterate over lots of return values, since the function already did it for us - among other possible uses.
It means that it is an in-out variable. You can do something directly with that variable. It is passed by address, not as a copy.
For example:
var temp = 10
func add(inout a: Int){
a++
}
add(inout:&temp)
temp // 11
There's another function of the ampersand in the Swift language that hasn't been mentioned yet. Take the following example:
protocol Foo {}
protocol Bar {}
func myMethod(myVar: Foo & Bar) {
// Do something
}
Here the ampersand syntax is stating that myVar conforms to both the Foo and Bar protocol.
As another use case, consider the following:
func myMethod() -> UIViewController & UITableViewDataSource {
// Do something
}
Here we're saying that the method returns a class instance (of UIViewController) that conforms to a certain protocol (UITableViewDataSource). This is rendered somewhat obsolete with Swift 5.1's Opaque Types but you may see this syntax in pre-Swift 5.1 code from time to time.
If you put & before a variable in a function, that means this variable is inout variable.
#Icaro already described what it means, I will just give an example to illustrate the difference between inout variables and in variables:
func majec(inout xValue:Int, var yValue:Int) {
xValue = 100
yValue = 200
}
var xValue = 33
var yValue = 33
majec(&xValue, yValue: yValue)
xValue //100
yValue //33
As noted in other answers, you use prefix & to pass a value to an inout parameter of a method or function call, as documented under Functions > Function Argument Labels and Parameter Names > In-Out Parameters in The Swift Programming Language. But there's more to it than that.
You can, in practice, think about Swift inout parameters and passing values to them as being similar to C or C++ pass-by-address or pass-by-reference. In fact, the compiler will optimize many uses of inout parameters down to roughly the same mechanics (especially when you're calling imported C or ObjC APIs that deal in pointers). However, those are just optimizations — at a semantic level, inout really doesn't pass addresses around, which frees the compiler to make this language construct more flexible and powerful.
For example, here's a struct that uses a common strategy for validating access to one of its properties:
struct Point {
private var _x: Int
var x: Int {
get {
print("get x: \(_x)")
return _x
}
set {
print("set x: \(newValue)")
_x = newValue
}
}
// ... same for y ...
init(x: Int, y: Int) { self._x = x; self._y = y }
}
(In "real" code, the getter and setter for x could do things like enforcing minimum/maximum values. Or x could do other computed-property tricks, like talking to a SQL database under the hood. Here we just instrument the call and get/set the underlying private property.)
Now, what happens when we pass x to an inout parameter?
func plusOne(num: inout Int) {
num += 1
}
var pt = Point(x: 0, y: 1)
plusOne(num: &pt.x)
// prints:
// get x: 0
// set x: 1
So, even though x is a computed property, passing it "by reference" using an inout parameter works the same as you'd expect it to if x were a stored property or a local variable.
This means that you can pass all sorts of things "by reference" that you couldn't even consider in C/C++/ObjC. For example, consider the standard library swap function, that takes any two... "things" and switches their values:
var a = 1, b = 2
swap(&a, &b)
print(a, b) // -> 2 1
var dict = [ "Malcolm": "Captain", "Kaylee": "Mechanic" ]
swap(&dict["Malcolm"], &dict["Kaylee"])
print(dict) // -> ["Kaylee": "Captain", "Malcolm": "Mechanic"], fanfic ahoy
let window1 = NSWindow()
let window2 = NSWindow()
window1.title = "window 1"
window2.title = "window 2"
var windows = [window1, window2]
swap(&windows[0], &windows[1])
print(windows.map { $0.title }) // -> ["window 2", "window 1"]
The the way inout works also lets you do fun stuff like using the += operator on nested call chains:
window.frame.origin.x += 10
... which is a whole lot simpler than decomposing a CGRect just to construct a new one with a different x coordinate.
This more nuanced version of the inout behavior, called "call by value result", and the ways it can optimize down to C-style "pass by address" behavior, is covered under Declarations > Functions > In-Out Parameters in The Swift Programming Language.