Function.prototype.call 在严格模式之外改变 this 的类型；为什么？

Question

var example = function () {
  console.log(typeof this);
  return this;
};

在严格模式下：example.call('test') # prints 'string'

否则，example.call('test') # prints 'object'

但是，console.log(example.call('test')) 打印出 test（如您所料）

为什么Function.call将typeof 'test' === 'string'绑定到example内的this？

Answer 1

当使用 call() 并将 this 参数设置为原始值时，该原始值总是转换为对象，因此您得到的是字符串对象而不是原始字符串

String {0: "t", 1: "e", 2: "s", 3: "t", length: 4, ...

MDN 上 call() 的文档指出

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

所以在非严格模式下原始字符串值被转换为一个对象，这也在ECMA standard, Annex C

中指定

strict mode restriction and exceptions
If this is evaluated within strict mode code, then the this value is not coerced to an object.
A this value of null or undefined is not converted to the global object and primitive values are not converted to wrapper objects.
The this value passed via a function call (including calls made using Function.prototype.apply and Function.prototype.call) do not coerce the passed this value to an object

Function.prototype.call 在严格模式之外改变 this 的类型；为什么？

Function.prototype.call alters typeof this, outside strict mode; why?

javascript

function

call

strict

strict-mode