true 调用 polyfil（没有扩展/剩余运算符）；奖金：申请&绑定polyfill

Question

我一直在练习呼叫，申请，绑定polyfills，并突然意识到，如果我使用传播/休息操作员，它并不是真正的多填充，无法弄清楚没有它的情况下如何通过呼叫中的参数，有什么想法吗？

//my attempt to write call polyfil without spread operator
//but it doesn't work
Function.prototype.call = function () {
  const obj = arguments[0];
  obj.fn = this;
  const args = Array(arguments).slice(1).reverse();
  while (args.length) {
    obj.fn.arguments.push(args.pop());
  }

  //obj.fn.arguments = Array(arguments).slice(1);
  obj.fn();
};

// that one works, but it uses spread operator
Function.prototype.call = function (obj, ...args) {
  obj.fn = this;
  obj.fn(...args);
};

Function.prototype.apply = function (obj, args) {
  obj.fn = this;
  obj.fn(...args);
};

Function.prototype.myBind = function (obj, ...args) {
  const fn = this;

  return function (...newArgs) {
    const allArgs = [...args, ...newArgs];
    fn.call(obj, ...allArgs);
    //fn.apply(obj, allArgs);
  };
};

let obj = {
  name: "Jack",
};

let myFunc = function (id, city) {
  console.log(`${this.name}, ${id}, ${city}`); 
};

let newFunc = myFunc.myBind(obj, "a_random_id");
newFunc("New York"); // Jack, a_random_id, New York