Where did that leave me? I knew how to write nice object hierarchies, tight structural/procedural code. However, Boost wasn’t a big deal yet, and I had only barely dabbled in anything else, so I was missing a huge piece of the puzzle: functional programming. I used lisp for a couple of programming assignments in graduate school because I wanted to learn the language – and I did learn the basics, but what I didn’t learn was important functional concepts.

Now, as a working professional programmer, I’m still trying to learn functional concepts. Some are pretty easy: immutability and recursive computation. These are very basic – even recursive functions were taught in Java programming classes.

But the most interesting element of functional programming is higher order functions, which are the basis for some very powerful forms of abstraction. This is what I was really missing in my repertoire.

So now, the point: I’ve been trying to learn a web framework called Weblocks. It’s freaking cool – but it makes heavy use of continuations for control flow (one of its major features). I had never learned about continuations before, so this was (and still is, to some degree) making it hard to learn the framework. So I was reading about continuations online (here ,here, and here) and came across a reference to Paul Graham’s book, On Lisp. I have a copy that I printed for myself, so I looked up the chapter on Continuations to gain some insight, and found the best explanation I’ve seen so far:

A continuation is a function representing the future of a computation. Whenever an expression is evaluated, something is waiting for the value it will return. For example, in

(/ (- x 1) 2)

when (- x 1) is evaluated, the outer / expression is waiting for the value, and something else is waiting for its value, and so on and so on, all the way back to the toplevel…

We can think of the continuation at any given time as a function of one argument. If the previous expression were typed into the toplevel, then when the subexpression (- x 1) was evaluated, the continuation would be:

(lambda (val) (/ val 2))

That is, the remainder of the computation could be duplicated by calling this function on the return value.

I found the little example shown here to be the best way try to explain continuations. Trying to understand the applications of continuations is a beast, and without this basic understanding it’s very easy to get lost.

Now…back to Weblocks…