Even or Odd without using a conditional statement (?)

Admiral_Snackbar

I think the quality of my function speaks for itself in regards to how seriously I take the subject of this thread, and the amount of time I took working on it was directly proportional to that seriousness and/or quality. Aside from that, however, I conceded your point about the loop technically constituting a conditional statement and amended my function accordingly in order to assuage any semantic misgivings.

Obligatory regular expression solution (with bonus javascript!):

function even(x) {
    var results = { true : ! (x % 2), false : NaN };
return results[ /^[+-]?[0-9]+$/.test(x) ];

}

var types = { true : "Even", false : "Odd", NaN : "Not an Integer" };

var x = prompt("Type a number");
alert(types[even(x)]);