如何将变量声明为函数并为其赋值[重复]

Question

我试过了：

 var fun = function(){ console.log("booyah"} , {
  hell: function(){
    console.log("hello");
  },
  boy: ()=>{
    alert("blah");
  }
};
fun.boy();

我希望首先调用 fun 函数，然后调用我存储为对象的 Access 函数。但我收到错误。我该如何解决？或者有没有其他方法可以做到这一点？ 请帮忙。

Answer 1

你可以通过编辑函数原型来达到这种效果。

例子：

function foo() {
        const something = 'the thing';
    
        return something;
    }
    
    const customPrototype = {
        greeting: (name) => {
            const greetingTemplate = `Hello ${name}`;
    
            return greetingTemplate;
        },
    
        randomNumber: () => {
            return Math.random();
        },
    };
    
    // Extending prototype to access our custom prototype functions
    Object.assign(foo, customPrototype);
    
    
    console.log(foo()); // the thing
    console.log(foo.greeting('People')); // Hello People
    console.log(foo.randomNumber()); // 0.26138311987993545
    // Default prototype functions are working as well
    console.log(foo.toString()); // [object Function]

编辑：感谢@Bergi 纠正我！

Answer 2

查看以下代码，参考取自Can you create functions with custom prototypes in JavaScript?

function MyFun() {
    if (!this || this==window) {
        return new MyFun();
    }

    var f = function() {
        return "thanks for calling!";
    }
    f.__proto__ = MyFun.prototype;
    f.constructor = MyFun;

    return f;
}

MyFun.prototype = {
    foo: function() {
        return "foo:" + this();
    },
    __proto__: Function.prototype
};

var f = new MyFun();
alert("proto method:"+f.foo()); // try our prototype methods
alert("function method:"+f.call()); // try standard function methods
alert("function call:"+f()); // try use as a function
alert('typeof:' + typeof f); // "function", not "object". No way around it in current js versions
alert('is MyFun:' + (f instanceof MyFun)); // true
alert('is Function:' + (f instanceof Function)); // true