如何将返回类型作为参数传递给新函数(就好像函数是在返回类型内部)?js

Question

假设任意一个对象，例如返回字符串。我有以下内容:

function alterString(str){
return `${str} was altered!`
}
//I want call my method as if it were be inside the return type
let new_string = object.returnString().alterString();

我已经在c#中这样做了，但我甚至不知道要研究

的名称

Answer 1

String.prototype.alterString=function(){
return `${this} was altered!`;
}
var obj = {
returnString : function(){ return 'Hi'; }
};
let new_string = obj.returnString().alterString();

或者你可以直接将alterString()添加到returnString()中，如下所示。

var obj = {
returnString : function(){
this.text = "Hey";
}
};
obj.returnString.prototype.alterString = function(){
return `${this.text} was altered!`;
}
let new_string = new obj.returnString().alterString();