After a quick web search, I came across a post by Paul Irish and it intrigued me. The first solution that he and some friends came up was this:

'#' + (function co(lor){   return (lor +=
  [0,1,2,3,4,5,6,7,8,9,'a','b','c','d','e','f'][Math.floor(Math.random()*16)])
  && (lor.length == 6) ?  lor : co(lor); })('');

It is actually quite a clever solution. My first suggestion was the last one that Paul Irish posted. It involved converting FFFFFF from hexadecimal to decimal and just generating a number from 0-16777215. The solution posted here however picks one of the possible 16 choices that exist in hexadecimal 6 times and concatenates them together to get the random color.

However, I was confused since right at the beginning, this function had a variable called “lor” that I had no idea where it was getting initialized. After some of my own testing and dissecting of this code, I learned that after a function header, after the closing curly bracket, you can pass a variable right off the bat. Like his example does the (‘‘) right at the end is an empty string that “lor” starts off as in this function.

Once it generates the random character the first time around, it checks if “lor” is six characters long. If it isn’t, then it recursively calls itself with the new “lor” variable. It is quite the ingenious solution.

In my next blog post, I will have to share the solution that Clay and I came up with to solve a recursive problem.