数组原型方法：`this` 是否可以为 `null`？

Question

我正在查看 Array.prototype.includes Polyfill, as shown on MDN。
顶部有几行对我来说意义不大：

// https://tc39.github.io/ecma262/#sec-array.prototype.includes
if (!Array.prototype.includes) {
  Object.defineProperty(Array.prototype, 'includes', {
    value: function(searchElement, fromIndex) {

      // 1. Let O be ? ToObject(this value).
      if (this == null) {
        throw new TypeError('"this" is null or not defined');
      }

具体来说，this == null 在那里检查。

this怎么会是null（甚至是假的）？这种情况怎么可能成立？

据我所知，没有一个可以调用 .includes 的虚假值，因为即使是空数组也是真实的。
我确定 null 支票存在的充分理由，但它可能是什么？

spec linked in the polyfill 没有明确说明任何 null 检查要求。

---

为了完整起见，这里是完整的 polyfill：

// https://tc39.github.io/ecma262/#sec-array.prototype.includes
if (!Array.prototype.includes) {
  Object.defineProperty(Array.prototype, 'includes', {
    value: function(searchElement, fromIndex) {

      // 1. Let O be ? ToObject(this value).
      if (this == null) {
        throw new TypeError('"this" is null or not defined');
      }

      var o = Object(this);

      // 2. Let len be ? ToLength(? Get(O, "length")).
      var len = o.length >>> 0;

      // 3. If len is 0, return false.
      if (len === 0) {
        return false;
      }

      // 4. Let n be ? ToInteger(fromIndex).
      //    (If fromIndex is undefined, this step produces the value 0.)
      var n = fromIndex | 0;

      // 5. If n ≥ 0, then
      //  a. Let k be n.
      // 6. Else n < 0,
      //  a. Let k be len + n.
      //  b. If k < 0, let k be 0.
      var k = Math.max(n >= 0 ? n : len - Math.abs(n), 0);

      function sameValueZero(x, y) {
        return x === y || (typeof x === 'number' && typeof y === 'number' && isNaN(x) && isNaN(y));
      }

      // 7. Repeat, while k < len
      while (k < len) {
        // a. Let elementK be the result of ? Get(O, ! ToString(k)).
        // b. If SameValueZero(searchElement, elementK) is true, return true.
        // c. Increase k by 1. 
        if (sameValueZero(o[k], searchElement)) {
          return true;
        }
        k++;
      }

      // 8. Return false
      return false;
    }
  });
}

Answer 1

我想我明白了

(function() {
  "use strict";

  if (!Array.prototype.includes2) {
  Object.defineProperty(Array.prototype, 'includes2', {
    value: function(searchElement, fromIndex) {

      // 1. Let O be ? ToObject(this value).
      if (this == null) {
        throw new TypeError('"this" is null or not defined');
      }

      var o = Object(this);

      // 2. Let len be ? ToLength(? Get(O, "length")).
      var len = o.length >>> 0;

      // 3. If len is 0, return false.
      if (len === 0) {
        return false;
      }

      // 4. Let n be ? ToInteger(fromIndex).
      //    (If fromIndex is undefined, this step produces the value 0.)
      var n = fromIndex | 0;

      // 5. If n ≥ 0, then
      //  a. Let k be n.
      // 6. Else n < 0,
      //  a. Let k be len + n.
      //  b. If k < 0, let k be 0.
      var k = Math.max(n >= 0 ? n : len - Math.abs(n), 0);

      function sameValueZero(x, y) {
        return x === y || (typeof x === 'number' && typeof y === 'number' && isNaN(x) && isNaN(y));
      }

      // 7. Repeat, while k < len
      while (k < len) {
        // a. Let elementK be the result of ? Get(O, ! ToString(k)).
        // b. If SameValueZero(searchElement, elementK) is true, return true.
        // c. Increase k by 1. 
        if (sameValueZero(o[k], searchElement)) {
          return true;
        }
        k++;
      }

      // 8. Return false
      return false;
    }
  });
}
})();

然后

console.log(Array.prototype.includes2.call(null));
// Uncaught TypeError: "this" is null or not defined

---

MDN。严格模式。 "Securing" JavaScript: (link)

For a normal function, this is always an object: either the provided object if called with an object-valued this; the value, boxed, if called with a Boolean, string, or number this; or the global object if called with an undefined or null this... Thus for a strict mode function, the specified this is not boxed into an object, and if unspecified, this will be undefined.