Duck Typing：利用 Duck Typing 将 Ruby 代码翻译成 Javascript

Question

这是《实用 Ruby 中的面向对象设计》一书中的一个示例。我有兴趣将这段 ruby​​ 代码翻译成 javascript，以便更好地理解 JS 中的鸭子类型。任何人都可以帮助翻译此代码以阐明如何最好地用 Javascript 编写此代码，或者至少帮助我入门吗？

这里的行程类充当其他类（Mechanic、Trip_Coordinator、Driver）的接口。其中 Trip 类中的 prepare 方法使用的是 Duck 类型的 preparer。

class Trip
    attr_reader :bicycles, :customers, :vehicle

    def prepare(preparers)
      preparers.each {|preparer|
         preparer.prepare_trip(self)}
    end
 end

class Mechanic
  def prepare_trip(trip)
    # Does something with trip.bicycles
  end
end


class Trip_Coordinator
  def prepare_trip(trip)
    # Does something with trip.customers
  end
end

class Driver
  def prepare_trip(trip)
    # Does something with trip.vehicle
  end
end

更新：

我添加了我认为可能是上述 Ruby 代码翻译的代码。但是，当我运行代码时，我得到以下输出和错误：

mountain
2
jeep

/home/ubuntu/user/tests/trip.js:3
       preparer.prepare_trip(this)

TypeError: Object [object Object] has no method 'prepare_trip'

建议或建议将不胜感激，以进一步提高对 JS 中鸭子类型的理解。

Javascript 代码：

var Trip = function(preparers){
    return preparers.forEach(function(preparer){
       preparer.prepare_trip(this)
    }
)};

var Mechanic = function(trip){
    // does something with trip.bicycles                                                                                                                                                           
    prepare_trip: console.log(trip.bicycles);
};

var TripCoordinator = function(trip){
    //does something with trip.customers                                                                                                                                                           
    prepare_trip: console.log(trip.customers);
};

var Driver = function(trip){
    //does something with trip.vehicle                                                                                                                                                             
    prepare_trip: console.log(trip.vehicle);
};

// customer wants to go on a trip for two and needs a car                                                                                                                                          
var planA = {
    bicycles: "mountain",
    customers: 2,
    vehicle: "jeep"
};

//customer wants to go a trip for one and only ride a bike                                                                                                                                         
var planB = {
    bicycles: "road",
    customers: 1
};

Trip([new Mechanic(planA), new TripCoordinator(planA), new Driver(planA)]);
Trip([new Mechanic(planB), new TripCoordinator(planB)]);

更新 2

根据下面 fgb 的解决方案和建议，我已将我的问题的最终解决方案放在下面。我添加了一个代理来消除调用者在创建旅行计划时必须知道他们需要什么准备的依赖性。

var Agent = function(plan){
    if("bicycles" && "customers" && "vehicle" in plan){
        Trip([new Mechanic(plan), new TripCoordinator(plan), new Driver(plan)]);
    }
    else if(!("vehicle" in plan) && "bicycles" && "customers" in plan){ //A driver is not needed                                                                                                                                        
        Trip([new Mechanic(plan), new TripCoordinator(plan)]);
    }
};

var Trip = function(preparers){
    return preparers.forEach(function(preparer){
       preparer.prepare_trip(this)
    }
)};
var Mechanic = function(trip){
    // does something with trip.bicycles                                                                                                                                                           
    this.prepare_trip = function() {
        console.log(trip.bicycles);
    }
};

var TripCoordinator = function(trip){
    //does something with trip.customers                                                                                                                                                           
    this.prepare_trip = function() {
        console.log(trip.customers);
    }
};

var Driver = function(trip){
    //does something with trip.vehicle                                                                                                                                                             
    this.prepare_trip = function() {
        console.log(trip.vehicle);
    }
};

// customer wants to go on a trip for two and needs a car                                                                                                                                          
var planA = {
    bicycles: "mountain",
    customers: 2,
    vehicle: "jeep"
};

//customer wants to go a trip for one and only ride a bike                                                                                                                                         
var planB = {
    bicycles: "road",
    customers: 1
};

Agent(planB);

相关SO讨论Example of Javascript Duck Typing?

Answer 1

代码几乎是正确的，但是您混淆了构造函数和对象字面量的语法。当你这样做时：

var Driver = function(trip){
    //does something with trip.vehicle                                                                                                                                                             
    prepare_trip: console.log(trip.vehicle);
};

“prepare_trip:”实际上是一个label，并没有定义对象的属性。

修复它的一种方法是：

var Driver = function(trip){
    //does something with trip.vehicle                                                                                                                                                             
    this.prepare_trip = function() {
        console.log(trip.vehicle);
    };
};

还要注意代码与 Ruby 代码不太一样：

preparer.prepare_trip(this);

这里this是全局对象而不是trip对象，方法中没有使用参数。在您的代码中，trip 参数改为在准备者的构造器中传递。