Looking at Clojure Koans

I worked on the Clojure koans.

The sixth file had this function:

(defn square [n] (* n n))

Then one of the problems was like this;

(= 25 (___ square))

And the answer was this:

(= 25 ((fn [f] (f 5)) square))

I admit that I copied and pasted it.

I am having a hard time figuring this one out. If you call square, you would do it like this:

(square 6)

But in the answer, “square” is at the end. It seems like to get inline functions with higher-order functions you have to alternate between reading left-to-right and right-to-left. So I will try to parse it myself.

Here is the function:

((fn [f] (f 5)) square)

So let’s get started.

So first we have the inline function:

(fn [f] (f 5))

So we are declaring a function that takes an argument of 5. And so “square” is the argument. I guess that even though the name of the function is a string, that if the string matches a function name then Clojure will realize it’s a function.

So let’s see what might happen with this code runs (keep in mind I have not actually looked at the Clojure source code or tried to run any sort of debugger; I am not sure if Clojure has or needs a debugger).

Here is the function again:

((fn [f] (f 5)) square)

So let’s get started.

Let’s put the argument into the definition:

(fn [square] (f 5)) 

Now let’s put the argument into the function body:

(square 5)

Somehow this exercise is not as enlightening as I thought it would be. I don’t know if this sequence could be used by someone to figure out the koan.

To paraphrase Barbi: Clojure is hard.