如何使用'Function.prototype.call'來調用未附加到任何對象原型的函數

Question

以下代碼的輸出是undefined has undefined wheels而不是Bike has 2 wheels。 Instantiated object不可能調用未附加到任何對象的prototype的函數嗎？

var vehicle = function(name, wheels){ 
 
\t name = "car", 
 
\t wheels = 4, 
 
\t fuel = "petrol" 
 
} 
 

 
function drive(name, wheels) { 
 
\t console.log(this.name + " has " + this.wheels + " wheels"); 
 
} 
 

 
var vehicle1 = new vehicle('Bike' , 2); 
 
drive.call(vehicle1); //undefined is driven with undefined wheels

Answer 1

不幸的是，幾乎所有關於你的代碼不正確。

首先，「車輛」的功能必須在this設置屬性，它應該使用傳入的參數，而不是常量：

var vehicle = function(name, wheels) { 
    this.name = name, 
    this.wheels = wheels, 
    this.fuel = "petrol" 
} 

然後，你的功能「驅動器」應該沒有任何參數：

function drive() { 
    console.log(this.name + " is driven with " + this.wheels + " wheels"); 
} 

要撥打drive()與您的汽車作爲this值：

var vehicle1 = new vehicle("Bike", 2); 
drive.call(vehicle1); 

.call()的第一個參數將用作被調用函數內部的this的值。

Answer 2

您需要使用this這樣的：

var vehicle = function(name, wheels){ 
 
\t this.name = "car", 
 
\t this.wheels = 4, 
 
\t this.fuel = "petrol" 
 
} 
 

 
function drive(name, wheels) { 
 
\t console.log(this.name + " is driven with " + this.wheels + " wheels"); 
 
} 
 

 
var vehicle1 = new vehicle('Bike' , 2); 
 
drive.call(vehicle1); //undefined is driven with undefined wheels

Answer 3

幾個問題：

1. 您使用vehicle作爲一個構造函數，而不是this指定的任何屬性。

2. 你不使用name或wheels參數vehicle

3. 你定義name和wheels參數drive但不使用他們，而不是調用它時向他們提供。

4. 不是一個真正的問題，只是約定（如我上your last question提到的）：如果你打算使用new與vehicle，應該首先上限：Vehicle通過壓倒性約定。

看評論：

var vehicle = function(name, wheels){ 
 
    // Actually use name and wheels and put them on `this` 
 
    this.name = name; 
 
    this.wheels = wheels; 
 
}; // <== Added missing ; 
 

 
function drive(/* No parameters here*/) { 
 
\t console.log(this.name + " is driven with " + this.wheels + " wheels"); 
 
} 
 

 
var vehicle1 = new vehicle('Bike' , 2); 
 
drive.call(vehicle1);