为什么在 `this == null` 时不触发 TypeError

Question

Array.prototype._find = function(callbackfn, thisArg) {
  if (this == null) {
    throw new TypeError('this is null or not defined')
  }
  if (typeof callbackfn !== 'function') {
    throw new TypeError('callbackfn is not a function')
  }

  for (let i in this) {
    if (callbackfn.call(thisArg, this[i], i, this)) return this[i]
  }

  return undefined
}

console.log(Array.prototype._find.call(null, x => x > 21))

这是一个 Array.prototype.find polyfill，我除了在运行 console.log(Array.prototype._find.call(null, x => x > 21)) 时触发 TypeError，我很困惑为什么不触发 TypeError

Answer 1

您可能运行您的函数处于 "non-strict" 模式。根据文档：

function.call(thisArg, arg1, arg2, ...)

thisArg

Optional. The value of this provided for the call to a function. Note that this may not be the actual value seen by the method: if the method is a function in non-strict mode, null and undefined will be replaced with the global object and primitive values will be converted to objects.

_{From the MDN page on Function.prototype.call，强调我的}

这是一个例子。第一个块以严格模式运行，并将记录 null。第二个将记录 window，因为这是您浏览器的全局范围。（请给堆栈片段一些时间来记录 window 对象，它很慢）

(function() {
  "use strict";
  
  function logThis() { console.log("strict:", this); }
  
  logThis.call(null);

}());

(function() {
  
  function logThis() { console.log("non-strict:", this); }
  
  logThis.call(null);

}());

为什么在 `this == null` 时不触发 TypeError

Why not trigger the TypeError when `this == null`

javascript

arrays

find

polyfills