为什么对Function.prototype函数的引用不起作用

Question

Why does the second case not working?

// 1. works
Object.prototype.hasOwnProperty.call({a:1}, 'a');

// 2. does not work
var hasProp = Object.prototype.hasOwnProperty.call;
hasProp({a:1}, 'a');

http://jsbin.com/ramenaxame/2/edit?js,console

Answer 1

Note that all functions share the same call method, inherited from Function.prototype .

Object.prototype.hasOwnProperty.call === Function.prototype.call // true

When you call call on a function, that function becomes the this value for call , so call can call the function. This is the case of your first code, which works.

However, in your second code, you don't call call as a method of an object. Therefore, its this value will be undefined in strict mode, or the global object in non-strict mode. Neither undefined nor the global object are callable, so call will throw.

In fact, your code is equivalent to

var hasProp = Function.prototype.call;
hasProp({a:1}, 'a');

As you can see, there is no reference to hasOwnProperty , so it can't work.

You can fix that using call to call call with hasOwnProperty as the this value:

var call = Function.prototype.call;
call.call(Object.prototype.hasOwnProperty, {a:1}, 'a');

But a better idea would be creating a new function that behaves like call but has its this value bound to hasOwnProperty . You can use bind to achieve this:

var hasProp = Function.prototype.call.bind(Object.prototype.hasOwnProperty);
hasProp({a:1}, 'a');

Answer 2

您可以通过这种方式分离方法以使其工作：

var hasProp = Function.call.bind(Object.prototype.hasOwnProperty)