JavaScript function.prototype.bind polyfill

Question

为什么我们需要在这里使用 instanceOf 而在这种情况下结果将为真？

if (!Function.prototype.bind) {
  Function.prototype.bind = function(oThis) {
    if (typeof this !== 'function') {
      // closest thing possible to the ECMAScript 5
      // internal IsCallable function
      throw new TypeError('Function.prototype.bind - what is trying to be bound is not callable');
    }

    var aArgs   = Array.prototype.slice.call(arguments, 1),
        fToBind = this,
        fNOP    = function() {},
        fBound  = function() {
          return fToBind.apply(this instanceof fNOP
                 ? this
                 : oThis,
                 aArgs.concat(Array.prototype.slice.call(arguments)));

//这里为什么要用instanceOf };

    if (this.prototype) {
      // Function.prototype doesn't have a prototype property
      fNOP.prototype = this.prototype; 
    }
    fBound.prototype = new fNOP();

    return fBound;
  };
}

//什么情况下结果为真？

Answer 1

我认为这是下面问题的重复，只是变量的命名在 2011 年提出时略有不同，所以很容易没有注意到它:

mozilla's bind function question

查看该问题以获得完整解释(如果您同意这是您问题的答案，请点赞答案)，但关键点在于:

Its allows you to call the bound function as a constructor without being bound to the original object. In other words the "bound" function will still work just like the original, unbound version if you call it with new.