Question

这是我的抽象类：

var Animal = function(){
    this.name = ''
    this.legs = 0;
    throw new Error("cannot instantiate abstract class");
}
Animal.prototype.walk = function(){ console.log(this.name+ " walked")};

创建一个新的具体类：

var Dog = function(){} ;

现在我想创建一个具体的类Dog来继承抽象类Animal。我试过以下两种方式。哪一个是标准？：

Dog.prototype = Object.create(Animal.prototype)

OR

Dog.prototype = Animal.prototype

我也试过了var Dog = Object.create(Animal)这给了我一个错误。

Answer 1

在Animal内部，您可以if检查当前constructor是否为Animal，如果是Dog则抛出错误。当您在Dog内调用父构造函数并从原型继承时，您还需要将其构造函数指针设置为var Animal = function() { this.name = '' this.legs = 0; if (this.constructor == Animal) { throw new Error("cannot instantiate abstract class"); } } Animal.prototype.walk = function() { console.log(this.name + " walked") }; function Dog() { Animal.call(this); } Dog.prototype = Object.create(Animal.prototype); Dog.prototype.constructor = Dog; var inst = new Dog; inst.name = 'lorem'; inst.walk()。

class CompetenceGroupeType extends AbstractType
{
    /**
     * @param FormBuilderInterface $builder
     * @param array $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('titre')
            ->add('competence_items', CollectionType::class, array(
                'entry_type'   => CompetenceItemType::class,
                'allow_add'    => true,
                'allow_delete' => true,
                'label' => false,
                'prototype' => true,

            ))
            //->add('save',      SubmitType::class)
            ;
    }

    /**
     * @param OptionsResolver $resolver
     */
    public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults(array(
            'data_class' => 'AppBundle\Entity\CompetenceGroupe',
        ));
    }
}

Javascript抽象类与对象创建

1 个答案: