if (obj !== obj) does somethig?

In this page you can see the following example of how to implement an indexOf for arrays:

if (!Array.prototype.indexOf)
{
  Array.prototype.indexOf = function(searchElement /*, fromIndex */)
  {
    "use strict";

    if (this === void 0 || this === null)
      throw new TypeError();

    var t = Object(this);
    var len = t.length >>> 0;
    if (len === 0)
      return -1;

    var n = 0;
    if (arguments.length > 0)
    {
      n = Number(arguments[1]);
      if (n !== n)    // <-- code of interest
        n = 0;
      else if (n !== 0 && n !== (1 / 0) && n !== -(1 / 0))
        n = (n > 0 || -1) * Math.floor(M开发者_如何转开发ath.abs(n));
    }

    if (n >= len)
      return -1;

    var k = n >= 0
          ? n
          : Math.max(len - Math.abs(n), 0);

    for (; k < len; k++)
    {
      if (k in t && t[k] === searchElement)
        return k;
    }
    return -1;
  };
}

My question is about the line:

if (n !== n)

in which case would this boolean expression return true?

it's a shortcut for verifying if the number is NaN.
say if you have n = Number("string"); then n !== nwould evaluate to true.
in this case you could've use if(isNaN(n)) instead of if(n !== n).

That is how you check for NaN. They are probably doing this as a precaution because it is possible to over write the global function isNaN.

// most of the time you don't need to try to simulate the specification so exactly-

if(!Array.prototype.indexOf){ 
    Array.prototype.indexOf= function(what, i){
        if(i==undefined || isNaN(i)) i= 0;
        var L= this.length;
        while(i< L){
            if(this[i]=== what) return i;
            ++i;
        }
        return -1;
    }
}