Ocaml syntax explanation please?

0
votes

I'm new to Ocaml and I'm trying to figure out how this function works:

let ccc c1 c2 c3 = fun (s1, s2, s3) n ->
     let (t1, r1) = (c1 s1 2) in
        if r1 = 0
        then let (t2, r2) = (c2 s2 n) in ((t1, t2, s3), r2)
        else let (t3, r3) = (c3 s3 n) in ((t1, s2, t3), r3) ;;

c1, c2, c3 are all "choosers". I'm aware that the purpose of the function is to take 3 choosers in and let the first chooser pick which of the other two choosers to use, but I'm confused by the syntax. Could anyone explain please? Thank you!

1
syntaxocaml

1 Answers

3
votes

It's hard to answer without knowing which part is confusing. Here are some simple examples that show some of the trickier parts.

# let f = fun x -> x + 1;;
val f : int -> int = <fun>
# f 3;;
- : int = 4

This defines f as a function that adds one to an integer. The expression fun args -> expr defines a function and the let binds the function to the name f.

# let f x = x + 1
val f : int -> int = <fun>
# f 3;;
- : int = 4

This defines the same function f. The meaning is exactly the same, it's just a slightly friendlier notation.

For whatever reason, your code is using both of these notations. I'm not sure I see a good reason to do this, but it does emphasize that if you pass three functions to ccc you'll get a function back.

The other pieces are pretty straightforward (though maybe they take some getting used to):

Function calls are formed just by writing things next to each other:

c1 s1 2
c2 s2 n
c3 s3 n

These are just calls to c1, c2, and c3.

Tuples are formed using commas (and conventionally parentheses also). So (t1, r1) is a pair of values that is returned by the call to c1.

I assume you understand if / then / else, and let in. If not, they aren't hard to explain.

I'd actually suggest reading a tutorial on OCaml. It should be more efficient than asking questions one at a time on SO. There are good tutorials at ocaml.org.