How does the closure capture values?

Question

I post a question about closure but it was kept as on hold When to use closures in swift?

Now i want to go in detail and being more specific. How does this line work?

    let names = ["Chris", "Alex", "Ewa", "Barry", "Daniella"]
    func backwards(s1: String, s2: String) -> Bool {
        return s1 > s2
    }
   var reversed = sorted(names, backwards)

When I pass a function as an argument.How does this takes value from the names array?

There is documentation on Swift closures provided by Apple, you need to study it. — zaph, Apr 24 '15 at 05:06
It looks like you're getting this from the Swift closure documentation. I would look into the `sorted(_:_:)` function from the Swift standard library documented [here](https://developer.apple.com/library/ios/documentation/General/Reference/SwiftStandardLibraryReference/Algorithms.html#//apple_ref/doc/uid/TP40014608-CH15-DontLinkElementID_1) to see how it sorts sequences in combination with a closure. — Aaron, Apr 24 '15 at 05:20
@newacct perhaps if i pass the function in the sorted method then...?? — , Apr 25 '15 at 01:35
@copeME: I don't understand what you are saying. You have a function, `backwards`. It doesn't capture any variables (the only variables used in the function, `s1` and `s2`, are parameters; "captured" variables mean variables that are used in the function but declared outside the function). — newacct, Apr 25 '15 at 21:04

score 0 · Accepted Answer · answered Apr 24 '15 at 08:56

You should look at the documentation, but here is an example sorting function (selection sort because that is the simplest):

func selectionSort<T>(inout array: [T], comparator: (T,T) -> Bool) {
    var min: Int

    for n in 0..<array.count {
        min = n

        for x in (n + 1)..<array.count {
            if comparator(array[x], array[min]) {
                min = x
            }
        }

        if min != n {
            let temp = array[min]
            array[min] = array[n]
            array[n] = temp
        }
    }
}

selectionSort(&names, backwards)

As you can see the function uses the closure to check for order, the exact number of times your closure gets called depends on the implementation of the sorted function's implementation.

@copeME if I answered your question, don't forget to accept it. If you are happy with my answer, you can also upvote it. — vrwim, Apr 29 '15 at 13:16

score -8 · Answer 2 · edited May 04 '15 at 03:59

Most of the Java script Developer stuck in closure even they use closure in multiple places. There are many blogs to explain Closure. I came across most of the blogs and decided to make it more simple as I can. In a very simple way I can say Closure is a function which has the memory of Variables in which it was defined.

Ex.1:

function increment(i){  
    return function(){
        return i++;
    }
}

var increase_1 = increment(5); //passing a reference or scope variable 5

//increase_1 store value of i as 5 in memory
increase_1()                            //5
increase_1()                            //6
increase_1()  

                      //7

Let's have another simple example Ex.2:

add(3,5);
add(3)(5);

Now I need a function that can execute both expression....

function add(a,b){
  if(arguments.length===2){
    return a+b;
  } else {
   return function(b){
      return a+b;
    }    
  }
}

Ex:3:

// A typical example
function addLinks () {
    for (var i=0, link; i<5; i++) {        
        link = document.createElement("a");
        link.innerHTML = "Link " + i;     
        link.onclick = function(){
         console.log(i);
        }
        document.body.appendChild(link);
    };   
}
addLinks();

//What will be the output of this. When you execute this it create a link, and on click it will print some value. So what would be the value it print on click of link1. Think about it for a while..... I know, initially you get confused. yes it prints 5. Not only click on this particular link but on each link, it gives you 5. So, now I want to print 0,1,2,3,4. Yes that's right when I click on link0, it prints 0, on link1, it prints 1 and so on.. Now, I want to you check the example 1 closely and try to implement in the same way and try to memorise some thing self executing annonmys function.

function addLinks () {
    for (var i=0, link; i<5; i++) {        
        link = document.createElement("a");
        link.innerHTML = "Link " + i;     
        link.onclick = function(c){
          return function(){
            console.log(c);
          }
        }(i);//passing a reference 
        document.body.appendChild(link);
    };   
}

addLinks();

So, if you see this example closely, when an event handler is being bind with the link, a reference or scope is passed. This is the same way, in which jQuery was implemented.

This is a swift question, not a javascript one? – Bergi Apr 24 '15 at 13:28 — Bergi, Apr 24 '15 at 13:28

How does the closure capture values?

2 Answers2