在封閉範圍

Question

我想我明白爲什麼變量，他們被宣佈在函數之外存在設定的變量，因爲你返回另一個功能：

myFunction = function() { 
    var closure = 'closure scope' 
    return function() { 
     return closure; 
    } 
} 
A = myFunction(); // myFunction returns a function, not a value 
B = A(); // A is a function, which when run, returns: 
console.log(B); // 'closure scope' 

，它的現在寫的，調用的方式（）就像一個getter。

問：如何編寫myFunction以便調用A（123）是一個setter？

Answer 1

嘗試以下操作：

myFunction = function() { 
    var closure = 'closure scope' 

    // value is optional 
    return function(value) { 
     // if it will be omitted 
     if(arguments.length == 0) { 
      // the method is a getter 
      return closure; 
     } else { 
      // otherwise a setter 
      closure = value; 
      // with fluid interface ;) 
      return this; 
     } 
    } 
} 
A = myFunction(); // myFunction returns a function, not a value 
A(123); // set value 
B = A(); // A is a function, which when run, returns: 
console.log(B); // '123' 

Answer 2

應儘可能簡單：

myFunction = function() { 
    var closure = 'closure scope' 
    return function(setTo) { 
     if (typeof setTo !== "undefined") { 
      closure = setTo; 
      return this; //support call chaining, good idea hek2mgl 
     } else { 
      return closure; 
     } 
    } 
} 

由於closure變量函數的範圍的封閉內，你應該能夠分配給它，你可以從中讀出同樣的方式。

見的jsfiddle：http://jsfiddle.net/WF4VT/1/

Answer 3

如果你想例如getter和setter方法你可以做這樣的事情：

var func = function() { 
    var closure = 'foo'; 

    return { 
    get: function() { return closure; }, 
    set: function(value) { closure = value; } 
    } 
}; 

var A = func(); 

A.set('foobar'); 
console.log(A.get()); //=> "foobar" 

Answer 4

另一種方法是使用一個類並定義getter和setter：

function MyClass(p){ 
    this._prop = p; 
} 
MyClass.prototype = { 
    constructor: MyClass, 
    get prop(){ 
     return this._prop; 
    }, 
    set prop(p){ 
     this._prop = p; 
    } 
} 

var myObject = new MyClass("TEST"); 
console.log(myObject.prop); 
myObject.prop = "test"; 
console.log(myObject.prop); 

演示：http://jsfiddle.net/louisbros/bMkbE/

Answer 5

jsFiddle Demo

讓你的返回函數接受參數。使用它作爲一個二傳手：

myFunction = function() { 
var closure = 'closure scope'; 
return function(val) { 
    closure = val; 
    return closure; 
} 
} 
A = myFunction(); // myFunction returns a function, not a value 
B = A(123); // A is a function, which when run, returns: 
console.log(B); // 'closure scope' 

Answer 6

重新審視這個問題，我知道我能做到這一點是這樣的：

function outside() { 
 
    \t var result = 'initialized' 
 
    \t return inside 
 
    \t function inside(argVariable) { 
 
    \t \t if(arguments.length) { 
 
    \t \t \t result = argVariable 
 
    \t \t \t return this 
 
    \t \t } else { 
 
    \t \t \t return result 
 
    \t \t } 
 
    \t } 
 
    } 
 
    myFunction = outside() // outside returns a function 
 
    X = myFunction() // returns: 'initialized' 
 
    $('body').append(X + '<br>') 
 
    myFunction(123) // setter 
 
    X = myFunction() // returns: 123 
 
    $('body').append(X)

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>