JavaScript 继承与 _.extend()
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JavaScript inheritance with _.extend()
提问by Jiew Meng
Whats the difference between
有什么区别
Employee.prototype = Object.create(Person.prototype);
and
和
_.extend(Employee.prototype, Person.prototype);
Both give similar results (output), but the underscore method seems to add the Person.prototype to the Employee.constructor.prototype, and quite abit extra stuff here and there, why?
两者都给出了相似的结果（输出），但是下划线方法似乎将 Person.prototype 添加到 Employee.constructor.prototype 中，并且到处都有很多额外的东西，为什么？
pure JS
纯JS
underscoreJS
下划线JS
A nice side effect of _.extendis I can easily do multiple inheritance: seems like it doesnt make the prototype chain longer too ...
一个很好的副作用_.extend是我可以轻松地进行多重继承：似乎它不会使原型链变得更长......
_.extend(Employee.prototype, Person.prototype);
_.extend(Employee.prototype, {
doSomething: function() {
return "hi ...";
}
});
But ...
但 ...
Why is there 2 sayHi and doSomething functions? (actually its the same when I just do 1 extend).
为什么有 2 个 sayHi 和 doSomething 函数？（实际上，当我只做 1 个扩展时，它是一样的）。
回答by rdiazv
With Employee.prototype = Object.create(Person.prototype);you are completely replacing the Employee.prototype.
随着Employee.prototype = Object.create(Person.prototype);您完全取代Employee.prototype.
But with _.extend(Employee.prototype, Person.prototype);you are adding the Person.prototypeon top of the Employee.prototype.
但是随着_.extend(Employee.prototype, Person.prototype);您Person.prototype在Employee.prototype.
For example,
例如，
var a = {var1:1, var2:2};
var b = {var2:4, var3:3};
console.log(_.extend(a, b)); // {var1:1, var2:4, var3:3}
As you see, ait's not completely replaced by b, it's just extendedby the properties defined in b.
如您所见，a它并没有完全被 替换b，它只是被 中定义的属性扩展了b。
