(新增官方回复)对【无类语言的OOP（JavaScript描述） 】一贴中第一个代码段的不同意见

function Darkness() {
}   

this.turnOn=function(){alert("bingo")}

1. Evaluate EqualityExpression.
2. Call GetValue(Result(1)).
3. Evaluate RelationalExpression.
4. Call GetValue(Result(3)).
5. Perform the comparison Result(4) === Result(2). (See below.)
6. Return Result(5).

1. If Type(V) is not Reference, return V.
2. Call GetBase(V).
3. If Result(2) is null, throw a ReferenceError exception.
4. Call the [[Get]] method of Result(2), passing GetPropertyName( V) for the property name.
5. Return Result(4).

	this.turnOn = test;

• GetBase(V). Returns the base object component of the reference V.

1. If O doesn't have a property with name P, go to step 4.
2. Get the value of the property.
3. Return Result(2).
4. ...

• Any time a non-internal property of an object O is created or set, the corresponding property is immediately also created or set with the same value and attributes in all objects joined with O.
• Any time a non-internal property of an object O is deleted, the corresponding property is immediately also deleted in all objects joined with O.
• ...（这两句跟主题无关）

function Light() {
	this.turnOn = function () {
		alert("bingo");
	}
}

var lightA = new Light();
var lightB = new Light();

alert(lightA.turnOn === lightB.turnOn); 

	function () {
		alert("bingo");
	}

1. If there already exists an object E that was created by an earlier call to this section's algorithm, and if that call to this section's algorithm was given a FunctionBody that is equated to the FunctionBody given now, then go to step 13. (If there is more than one object E satisfying these criteria, choose one at the implementation's discretion.)

• 这可以作为一种不同于 prototype 的“新”的给对象绑定方法的形式，更加安全。对于这种观点和使用方式效率没有任何问题。
• 当作普通 prototype 的另一种写法来使用，相对于在代码中多消耗了一个 O(n) 级的算法，而且此算法来自 JavaScript 底层实现且可被加速，效率下降忽略不计，去掉 JS 代码中所有为空的行、无效空格，把所有换行换成 ; 损失就回来了；仅当处于“大规模的”循环中时才可视为一个 O(n) 算法，这种情况存在的可能性微乎其微，不过，还是那句话，尽量避免。

13.1.1 Equated Grammar Productions
Two uses of the FunctionBody grammar production are defined to be equated when one of the following
is true:
• Both uses obtained their FunctionBody from the same location in the source text of the same
ECMAScript program. This source text consists of global code and any contained function codes
according to the definitions in 10.1.2.
• Both uses obtained their FunctionBody from the same location in the source text of the same call to
eval (15.1.2.1). This source text consists of eval code and any contained function codes according
to the definitions in 10.1.2.
NOTE
Two uses of FunctionBody obtained from a call to the Function constructor 15.3.1 and 15.3.2) are
never equated. Also, two uses of FunctionBody obtained from two different calls to eval are never
equated, even if those two calls to eval were passed the same argument.

我很不希望，你指的是这部分。

Step1 允许一个具体实现在在function A有一个不依赖于A的function B时进行优化。注意是允许

在给出的第一个例子中，规范也仅仅是说，允许但不需要把b1和b2 join。

再看最后一段，只有在具体实现能够证明两个function Object的scope属性的区别是看不出来的（真饶）情况下，两个function Object才可以join。根据这个规则，恐怕实现只能找到对象它自己是可已join的。

在回头看我们的相等性测试

两个值是同一对象，或者是可以join的，而关于这个join，规则本身也仅仅是一个非常宽松的规定。

再来看我引用的这对规范里面的这个例子，在客户端b1和b2能通过相等性测试么？很遗憾，至少在ie和ff，它都不能，也就是说

这两个客户端貌似没有实现这个规范。

至于说相等性测试对于函数是无意义的，一定返回false的，我希望谁能够告诉我具体在规范的哪里？

13.1.1 Equated Grammar Productions
Two uses of the FunctionBody grammar production are defined to be equated when one of the following
is true:
• Both uses obtained their FunctionBody from the same location in the source text of the same
ECMAScript program. This source text consists of global code and any contained function codes
according to the definitions in 10.1.2.
• Both uses obtained their FunctionBody from the same location in the source text of the same call to
eval (15.1.2.1). This source text consists of eval code and any contained function codes according
to the definitions in 10.1.2.
NOTE
Two uses of FunctionBody obtained from a call to the Function constructor 15.3.1 and 15.3.2) are
never equated. Also, two uses of FunctionBody obtained from two different calls to eval are never
equated, even if those two calls to eval were passed the same argument.

我很不希望，你指的是这部分。