javascript - JS 绑定(bind)函数和函数调用运算符的性质

Question

var obj = {};

var r1 = (obj['toString'])();
var m1 = obj['toString'];
var r2 = m1();

var r3 = (obj.toString)();
var m2 = obj.toString;
var r4 = m2();

r1 和 r3 应该包含正确的结果:"[object Object]"，而 r2 和 r4 包含 "[object Undefined]" 表明 m1 和 m2 没有绑定(bind)到对象。

我无法完全理解 obj['toString']() 是如何执行的。我总是这样看，(obj['toString'])() -> (function obj)()。事实证明，函数调用运算符会回顾上下文。我希望运算符不知道操作数的来源。

谁能正确解释这种行为？

Answer 1

Turns out that function invocation operator looks back on what is the context. I would expect operator to not know where operands come from.

事实上，它确实知道。

在 EcmaScript 规范中，这由 property accessor operators 描述(以及一些类似的操作)返回一个“Reference object”，它包含了这个信息:属性被访问的上下文。任何“正常”操作通常只会获取引用的值 - 包括 assignment operators ，这确实会解除您案例中的引用。

call operator使用它来使方法调用变得特殊:

1. Let ref be the result of evaluating MemberExpression. which may return a Reference
2. Let func be GetValue(ref). which fetches the actual function object - you see this operation a lot in the spec
3. … fetch arguments, do some type assertions
4. If Type(ref) is Reference, then
   • If IsPropertyReference(ref) is true, then
     Let thisValue be GetBase(ref). <- here the method context is fetched
   • Else, the base of ref is an Environment Record
     … which basically describes variables inside a with statement
5. Else, Type(ref) is not Reference.
   Let thisValue be undefined.