Some problems with prototypal inheritance-JS Tutorial-php.cn

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Some problems with prototypal inheritance

一个新手

Sep 06, 2017 am 11:48 AM

When we use prototype chain inheritance, we need to be careful in defining methods and properties on the prototype, because this may bring unexpected results.

1. Define methods on the prototype carefully.
When we want to define a method on the prototype of a constructor, we must define it after changing the prototype, otherwise the new prototype object will not have this method defined, resulting in the results we expected. different. Example:

1 function superObj(){}
2 superObj.prototype.sayHi=function sayHi(){
3     console.log('hi');
4 };
5 superObj.prototype={
6 name:'Poly'
7 };
8 var obj=new superObj();
9 obj.sayHi();//报错!! superObj.sayHi is not a function

The correct operation is as follows

1 function superObj(){}
2 superObj.prototype={
3     name:'Poly'
4 };
5 superObj.prototype.sayHi=function sayHi(){6     console.log('hi');7 };8 var obj=new superObj();9 obj.sayHi();// 'hi'

2. Do not use object literals to create properties/methods for the prototype.
Using object literals will create a new object and assign the reference address of the new object to the prototype of the constructor. Example

1 function superObj(){}
2 superObj.prototype={
3     sayHi:function sayHi(){
4         console.log('hi');
5     }
6 }

The correct operation is as follows:

1 function superObj(){}
2 superObj.prototype.sayHi=function sayHi(){
3     console.log('hi');
4 }

3. There is a direct correspondence between object instances and prototypes.
This means that when a reference to __proto__ is made, the reference address of the prototype will be saved. Even if the prototype of the constructor changes, it will not affect the __proto__ in the previously created instance. Example

 1 function superObj(){} 
 2 superObj.prototype.say=function() { 
 3     console.log('hello'); 
 4 } 
 5 var obj=new superObj();
 6 superObj.prototype={ 
 7     say:function() { 
 8         console.log('world'); 
 9     }
 10 };
 11 var obj2=new superObj();
 12 obj.say();//'hello'
 13 obj2.say();//'world'

Fourth, it is best not to define attributes with reference types on the prototype.
If a property with a value of a reference type is defined on the prototype, then all instances will share the property value (reference type value, pointing to the same object), and when one of the instances modifies the value or property on the reference type , the changes will occur on all instances. Therefore, properties whose values are reference types are best defined in the constructor. Example

1 function superObj(){}
2 superObj.prototype.ary=[1,2,3];
3 var obj1=new superObj();
4 var obj2=new superObj();
5 obj1.ary[0]=0;//obj1.ary和obj2.ary指向的是同一个数组，当obj1修改此数组时,obj2.ary也会发生改变
6 console.log(obj2.ary[0]);//0

<br/>

If you don’t want instances to share the same reference object, you should define it in the constructor. Example

1 function superObj(){
2     this.ary=[1,2,3];
3 }
4 var obj1=new superObj();
5 var obj2=new superObj();
6 obj1.ary[0]=0;//obj1.ary和obj2.ary指向的不是同一个数组，所以修改obj1.ary不会影响obj2.ary
7 console.log(obj2.ary[0]);//1

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