简单的 JavaScript 继承插件错误

Question

我阅读了 John Resig 的博客文章，介绍了他基于 base2 和原型编写的简单 JavaScript 继承插件。

示例代码在这里：http://jsfiddle.net/rFfX9/

//'use strict';

/* Simple JavaScript Inheritance
 * By John Resig http://ejohn.org/
 * MIT Licensed.
 */
// Inspired by base2 and Prototype
(function () {
  var initializing = false, fnTest = /xyz/.test(function () { xyz; }) ? /\b_super\b/ : /.*/;

  // The base Class implementation (does nothing)
  this.Class = function () { };

  // Create a new Class that inherits from this class
  Class.extend = function (prop) {
    var _super = this.prototype;

    // Instantiate a base class (but only create the instance,
    // don't run the init constructor)
    initializing = true;
    var prototype = new this();
    initializing = false;

    // Copy the properties over onto the new prototype
    for (var name in prop) {
      // Check if we're overwriting an existing function
      prototype[name] = typeof prop[name] == "function" &&
        typeof _super[name] == "function" && fnTest.test(prop[name]) ?
        (function (name, fn) {
          return function () {
            var tmp = this._super;

            // Add a new ._super() method that is the same method
            // but on the super-class
            this._super = _super[name];

            // The method only need to be bound temporarily, so we
            // remove it when we're done executing
            var ret = fn.apply(this, arguments);
            this._super = tmp;

            return ret;
          };
        })(name, prop[name]) :
        prop[name];
    }

    // The dummy class constructor
    function Class() {
      // All construction is actually done in the init method
      if (!initializing && this.init)
        this.init.apply(this, arguments);
    }

    // Populate our constructed prototype object
    Class.prototype = prototype;

    // Enforce the constructor to be what we expect
    Class.prototype.constructor = Class;

    // And make this class extendable
    Class.extend = arguments.callee;

    return Class;
  };
})();

这行得通。但是，当我在顶部取消注释 'use strict'; 时，它会引发异常，但无法找出解决方案。有什么想法吗？

Answer 1

在这种情况下，当您调用匿名函数 this 将是 undefined 时，您不会将该函数作为对象的方法调用。在非严格模式下，this 将是 window 对象。

另请参阅：Why is "this" in an anonymous function undefined when using strict?