function Base() {
this.name="name"
this.age=18
}
Base.prototype={
say() {
alert("hi")
},
ex:"lucy"
}
function Student() {
Base.call(this,arguments)//////
this.add="US"
}
Student.prototype=new Base()//call
let tom=new Student()
tom.say()remove the following code
Base.call(this,arguments)////// with or without the above code, the content of the variable tom is different. I don't know if you found out.
essentially, if you don't write this line of code, the base constructor won't execute, and if you don't need the logic of the parent constructor, that's fine.
but I bet you need it.
I hope my answer will be helpful to you. My personal homepage is https://azl397985856.github.io/, and the Nuggets homepage https://juejin.im/user/58af98.
Base.call(this,arguments) this sentence executes the constructor of the parent class, which is equivalent to super ()
class .
inheritance of
JS there are many kinds of prototype object inheritance and modified constructor inheritance,
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