此答案的版本图像: [ 解决]

只是忘记每件事的范围,并记住:当一个变量需要在某个地方,JavaScript不会破坏它。

例子1:

此分類上一篇

例子2:

此分類上一篇

例子3:

函数 foo(x) { var tmp = 3;函数栏(y) { console.log(x + y + (++tmp)); // 会记录 16 } 栏(10); } foo(2);

函数栏,以及其与函数 foo 的语法环境的联系,是关闭。

函数 foo(x) { var tmp = 3; 返回函数 (y) { console.log(x + y + (++tmp)); // 还会记录 16 } } var bar = foo(2); bar(10); // 16 bar(10); // 17

一个关闭的最简单的例子是:

var a = 10; 函数测试() { console.log(a); // 将输出 10 console.log(b); // 将输出 6 } var b = 6; 测试();

好吧,和一个6岁的孩子谈话,我可能会使用随后的协会。

想象一下 - 你正在玩你的小兄弟姐妹在整个房子,你正在移动周围与你的玩具,并将其中的一些带到你的哥哥的房间。 过了一会儿,你的哥哥从学校回来,去了他的房间,他锁在里面,所以现在你不能访问玩具留在那里再直接的方式。

比较一个情况,当一个门被草案锁定,没有人在里面(通用功能执行),然后一些当地的火灾发生并燃烧房间(垃圾收集器:D),然后一个新的房间被建造,现在你可以留下其他玩具在那里(新功能例子),但从来没有得到相同的玩具留在第一间房间例子。

对于一个先进的孩子,我会把这样的东西放在下面,这不是完美的,但它让你感觉到它是什么:

function playingInBrothersRoom (withToys) {
  // We closure toys which we played in the brother's room. When he come back and lock the door
  // your brother is supposed to be into the outer [[scope]] object now. Thanks god you could communicate with him.
  var closureToys = withToys || [],
      returnToy, countIt, toy; // Just another closure helpers, for brother's inner use.

  var brotherGivesToyBack = function (toy) {
    // New request. There is not yet closureToys on brother's hand yet. Give him a time.
    returnToy = null;
    if (toy && closureToys.length > 0) { // If we ask for a specific toy, the brother is going to search for it.

      for ( countIt = closureToys.length; countIt; countIt--) {
        if (closureToys[countIt - 1] == toy) {
          returnToy = 'Take your ' + closureToys.splice(countIt - 1, 1) + ', little boy!';
          break;
        }
      }
      returnToy = returnToy || 'Hey, I could not find any ' + toy + ' here. Look for it in another room.';
    }
    else if (closureToys.length > 0) { // Otherwise, just give back everything he has in the room.
      returnToy = 'Behold! ' + closureToys.join(', ') + '.';
      closureToys = [];
    }
    else {
      returnToy = 'Hey, lil shrimp, I gave you everything!';
    }
    console.log(returnToy);
  }
  return brotherGivesToyBack;
}
// You are playing in the house, including the brother's room.
var toys = ['teddybear', 'car', 'jumpingrope'],
    askBrotherForClosuredToy = playingInBrothersRoom(toys);

// The door is locked, and the brother came from the school. You could not cheat and take it out directly.
console.log(askBrotherForClosuredToy.closureToys); // Undefined

// But you could ask your brother politely, to give it back.
askBrotherForClosuredToy('teddybear'); // Hooray, here it is, teddybear
askBrotherForClosuredToy('ball'); // The brother would not be able to find it.
askBrotherForClosuredToy(); // The brother gives you all the rest
askBrotherForClosuredToy(); // Nothing left in there

正如你可以看到的那样,房间里留下的玩具仍然可以通过兄弟访问,不管房间是否被锁定。

有些人非常接近得到简单的解释,然后变得复杂或保持抽象,击败目的,并无法显示一个非常简单的现实世界使用。

关闭有访问包含函数的变量 关闭在自己的内存空间中持久(因此对所有类型的 oop-y 实例物品都是有用的)

function sayHello(name) {
  var text = 'Hello ' + name; // Local variable
  console.log(text);
  var sayAlert = function () {
      alert(text);
  }
  return sayAlert;
}

sayHello(); 
/* This will write 'Hello undefined' to the console (in Chrome anyway), 
but will not alert though since it returns a function handle to nothing). 
Since no handle or reference is created, I imagine a good js engine would 
destroy/dispose of the internal sayAlert function once it completes. */

// Create a handle/reference/instance of sayHello() using the name 'Bob'
sayHelloBob = sayHello('Bob');
sayHelloBob();

// Create another handle or reference to sayHello with a different name
sayHelloGerry = sayHello('Gerry');
sayHelloGerry();

/* Now calling them again demonstrates that each handle or reference contains its own 
unique local variable memory space. They remain in memory 'forever' 
(or until your computer/browser explode) */
sayHelloBob();
sayHelloGerry();

这表明了你应该关于关闭的两个基本概念。

在简单的说法中,我有一个基本函数,我可以创建参考或操作,其中包含独特的数据,留在记忆参考中。我不需要每次我想说别人的名字,我都需要重写函数。

默认情况下,JavaScript知道两种类型的目标:全球和本地。

var a = 1;

function b(x) {
    var c = 2;
    return x * c;
}

在上面的代码中,变量 a 和函数 b 可从代码中的任何地方(即全球)。变量 c 仅在 b 函数范围内可用(即本地)。

JavaScript 关闭有助于解决这个问题,通过将一个函数与一个背景:

function a(x) {
    return function b(y) {
        return x + y;
    }
}

在这里,函数 a 返回一个函数称为 b. 由于 b 在 a 中定义,它会自动访问在 a 中定义的任何内容,即 x 在此示例中。

var c = a(3);

var c = function b(y) {
    return 3 + y;
}

函数b 提醒 x = 3 在其背景下,因此:

var d = c(4);

关闭也可以用来限制全球宣布的变量和方法的范围:

(function () {
    var f = "Some message";
    alert(f);
})();

现在,有一个常见的JavaScript洞穴,在那里关闭可以帮助:

var a = new Array();

for (var i=0; i<2; i++) {
    a[i]= function(x) { return x + i ; }
}

a[0] = function (x) { return x + 0 ; }
a[1] = function (x) { return x + 1 ; }
a[2] = function (x) { return x + 2 ; }

a[0] = function (x) { return x + 2 ; }
a[1] = function (x) { return x + 2 ; }
a[2] = function (x) { return x + 2 ; }

var a = new Array();

for (var i=0; i<2; i++) {
    a[i]= function(tmp) {
        return function (x) { return x + tmp ; }
    } (i);
}

你睡了一会儿,你邀请丹,你告诉丹带来一个XBox控制器。

丹邀请保罗,丹要求保罗带来一个控制者,有多少控制者被带到派对?

function sleepOver(howManyControllersToBring) {

    var numberOfDansControllers = howManyControllersToBring;

    return function danInvitedPaul(numberOfPaulsControllers) {
        var totalControllers = numberOfDansControllers + numberOfPaulsControllers;
        return totalControllers;
    }
}

var howManyControllersToBring = 1;

var inviteDan = sleepOver(howManyControllersToBring);

// The only reason Paul was invited is because Dan was invited. 
// So we set Paul's invitation = Dan's invitation.

var danInvitedPaul = inviteDan(howManyControllersToBring);

alert("There were " + danInvitedPaul + " controllers brought to the party.");